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a formula. 5.2 Permutations We can make a permutation by taking a number of objects and arranging them in a line. For example, the two possible permutations of the digits 5 and 9 are the numbers 59 and 95. Although there are several methods that we can use to find the number of possible permutations of objects, all methods involve use of the factorial function. 125 Permutations of n distinct objects The number of permutations of n distinct objects is denoted by Pn n, and there are n! permutations that can be made. For example, there are P 2! 2 digits 5 and 9, as we have just seen. 2 = 2 = permutations of the two KEY POINT 5.2 The number of permutations of n distinct objects is P n integer n > 0)( n − 2)… 3 2 1, for × × × any Consider all the three-digit numbers that can be made by arranging the digits 5, 6 and 7. In this simple case, we can make a list to show there are six possible three-digit numbers. These are 567, 576, 657, 675, 756 and 765. The following tree diagram gives another method of showing the six possible arrangements of the three digits. left middle right number............ 567............ 576............ 657............ 675............ 756............ 765 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Unfortunately, writing out lists and constructing tree diagrams to find numbers of possible arrangements of objects are suitable methods only for small numbers of objects. Imagine listing all the possible arrangements of seven different letters; there would be over 5000 on the list and a tree diagram would have over 5000 branches at its right-hand side! Clearly, a more practical method for finding numbers of arrangements is needed. This is the primary use of the factorial function. We can show that six three-digit numbers can be made from 5, 6 and 7 by considering how many choices we have for the digit that we place in each position in the arrangement. If we first place a digit at the left side, we have three choices. Next, we place a digit in the middle (two |
choices). Finally, we place the remaining digit at the right side, as shown in the following diagram. 3 choices × 2 choices × 1 choices The numbers above the lines in the diagram are not the digits that are being arranged – they are the numbers of choices that we have for placing the three digits. The three digits can be arranged in 3 × × = 2 1 3! = 3 P 3 = ways. 6 The seven letters mentioned previously can be arranged in 7! 5040 7 = × × × × × × = ways. 4 5 1 3 6 2 WORKED EXAMPLE 5.1 126 In how many different ways can five boys be arranged in a row? TIP Answer! = 5 P 5 = 120 ways. We multiply together the number of choices for each of the five positions, working from left to right. We could just as easily work from right to left, giving 1 5! 2 × × × × = ways. 4 5 3 WORKED EXAMPLE 5.2 In how many ways can nine elephants and four mice be arranged in a line? Answer 13! = 13 P 13 = 6227 020800 ways. The nine elephants and four mice are distinct, so we are arranging 13 different animals. TIP Large number answers can be given more accurately than to 3 significant figures. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations EXERCISE 5B 1 In how many ways can the six letters A, B, C, D, E and F be arranged in a row? 2 From a standard deck of 52 playing cards, find how many ways there are of arranging in a row: a all 52 cards b the four kings c the 13 diamonds. 3 In how many different ways can the following stand in a line? a two women b six men c eight adults. 4 In how many different ways can the following sit in a row on a bench? a four girls b three boys c four girls and three boys. 5 Seven cars and x vans can be parked in a line in 39 916 800 ways. Find the number of ways in which five cars and x 2+ vans can be parked in a line. 6 A woman has 10 children. She arranges 11 chairs in a row and sits on the chair |
in the middle. If her youngest child sits on the adjacent chair to her left, in how many ways can the remaining children be seated? PS 7 A group of n boys can be arranged in a line in a certain number of ways. By adding two more boys to the group, the number of possible arrangements increases by a factor of 420. Find the value of n. Permutations of n objects with repetitions When n objects include repetitions (i.e. when they are not all distinct), there will be fewer than Pn n permutations, so an adjustment to the use of the factorial function is needed. Consider making five-letter arrangements with A, A, B, C and D. To simplify the problem, we can distinguish the repeated A by writing the letters as A, A, B, C and D. AABCD is the same arrangement as AABCD. DACAB is the same arrangement as DACAB, and so on. Each time we swap A and A, we obtain the same arrangement. 127 If the five letters were distinct, there would be P 5! 120 arrangements, but the number 5 5 = = is reduced to half 1 2! of the total arrangements because the two repeated letters can be placed in 2! ways in any particular arrangement without changing that arrangement. Letters to be arranged 5= ; number of the same letter 2=. 5 There are 2 P 5 P 2 = 5! 2! = 60 five-letter arrangements that can be made. KEY POINT 5.3 The number of permutations of n objects, of which p are of one type, q are of another type, r are of another type, and so on, is =! ×... p! × q!! n × r! ×... where. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 5.3 The capital of Burkina Faso is OUAGADOUGOU. Find the number of distinct arrangements of all the letters in this word. Answer 11! 3! 3! 2! 2! × × × = 277200 11 letters are to be arranged, with repeats of three Os, |
three Us, two As and two Gs. In the formula of Key point 5.3, excluding 1! for the D in the denominator does not change our answer. EXERCISE 5C 1 Find the number of distinct arrangements of all the letters in these words: a TABLE b TABLET c COMMITTEE d MISSISSIPPI e HULLABALLOO. 2 Find how many six-digit numbers can be made from these sets of digits: a 1, 1, 1, 1, 1 and 3 b 2, 2, 2, 7, 7 and 7 c 5, 6, 6, 6, 7 and 7 d 8, 8, 9, 9, 9 and 9. 128 3 A girl has 20 plastic squares. There are five identical red squares, seven identical blue squares and eight identical green squares. By placing them in a row, joined edge-to-edge, find how many different arrangements she can make using: a one square of each colour b the five red squares only c all of the blue and green squares d all of the 20 squares. 4 Two students are asked to find how many ways there are to plant two trees and three bushes in a row. The first student gives 5! 120, and the second gives = 5! 2! 3! × = 10. Decide who you agree with and explain the error made by the other student. 5 Ten coins are placed in a row on a table, each showing a head or a tail. a How many different arrangements of heads and/or tails are possible? b Of the arrangements in part a, find how many have: i five heads and five tails showing ii more heads than tails showing. 6 There are 420 possible arrangements of all the letters in a particular seven-letter word. Give a description of the letters in this word. 7 Find the number of distinct five-letter arrangements that can be made from: a c two As and three Bs b two identical vowels and three Bs two identical vowels and any three identical consonants. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations EXPLORE 5.2 Consider the number of distinct arrangements of the 16 letters in the word COUNTERCLOCKWISE – there are close to 8. |
72 1011 numbers in our daily lives, so we are likely to see this as just a very large number whose true size we cannot really comprehend until it is put into some human context. For example, if everyone on Earth over the age of 14 (i.e. about 5.46 109 contributed one new arrangement of the word every day starting on 1st January, we would complete the list of arrangements around 9th June.. We rarely meet such people) × × The calculation for this is 11 8.72 10 × 5.46 10 × ≈ 160 days. 9 Devise a way of expressing the number of distinct arrangements of the letters in the word PNEUMONOULTRAMICROSCOPICSILICOVOLCANOCONIOSIS (which is the full name for the disease known as silicosis, and is the longest word in any major English language dictionary) in a way that is meaningful to human understanding. There are, for example, 3.15 107 × Earth, and the masses of the Earth and Sun are 5.97 1024 respectively. seconds in a year, about 7.48 109× people on 30 and 1.99 10 kg × ×, WEB LINK For other options, perform a web search for large numbers. Permutations of n distinct objects with restrictions The number of possible arrangements of objects is reduced when restrictions are put in place. As a general rule, the number of choices for the restricted positions should be investigated first, and then the unrestricted positions can be attended to. 129 WORKED EXAMPLE 5.4 Find the number of ways of arranging six men in a line so that: a the oldest man is at the far-left side b the two youngest men are at the far-right side c the shortest man is at neither end of the line. Answer Without restrictions, the six men can be arranged in P 6! than 720 arrangements. 6 = 6 = 720 ways. So, with restrictions, there will be fewer! 120 = arrangements 4P4 P = 2 4 P 4 × 2 4! 2! × = 2P2 arrangements 48 The oldest man must be at the far-left side (one choice), and the other five men can be arranged in the remaining five positions in P5 5 ways. The two spaces at the right are reserved for the two youngest men, who can be placed there in P2 2 ways. The other four men can be arranged in the remaining four positions in P4 4 ways, |
as shown. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 4P4! 4 × = 480 arrangements. There are only five men who can be placed at the far-left side, so there are only four men who can be placed at the far right. The remaining four positions can be filled by any of the other four men (one of whom is the shortest man) in P4 4 ways, as shown. Alternatively, the shortest man can be placed in one of four positions, and the other five positions can be filled in 5P5 ways, so 4 × 5! = 480. WORKED EXAMPLE 5.5 How many odd four-digit numbers greater than 3000 can be made from the digits 1, 2, 3 and 4, each used once? Answer Restrictions affect the digits in the thousands column and in the units column. The digit at the far left (i.e. thousands column) can be only 3 or 4, and the digit at the far right (i.e. units column) can be only 1 or 3. The 3 can be placed in either of the restricted positions, so we can investigate separately the four-digit numbers that start with 3, and the four-digit numbers that start with 4. 130 1 × 2 × 1 × 1 2P2 × = numbers 2P2 × = numbers 2 4 1 × 2 P 2 Start with 3: We must place 1 at the far right (one choice), and the remaining two positions can be filled by the other two digits in P2 2 ways, as shown. Start with 4: We can place 1 or 3 at the far right (two choices), and the remaining two positions can be filled by the other two digits in P2 2 ways, as shown. 4 6 + = odd numbers greater 2 than 3000 can be made. TIP Alternatively, we could solve this problem by investigating separately the numbers that end with 1, and the numbers that end with 3. TIP These six numbers are 3241, 3421, 4123, 4213, 4213, 4231 and 4321. WORKED EXAMPLE 5.6 Find how many ways two mangoes M( a line if the five fruits are |
distinguishable and the mangoes: ) and three watermelons W( ) can be placed in a must not be separated b must be separated Answer a 2P2 M1 M2 W1 W2 W3 1 object 4P4 2 P 2 × 4 P 4 = 48 ways The two mangoes can be placed next to each other in P2 2 ways. This pair is now considered as a single object to be arranged with the three watermelons, giving a total of four objects to arrange, as shown. TIP Objects that must not be separated are treated as a single object when arranged with others. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations b 120 − 48 = 72 ways EXERCISE 5D With no restrictions, the five items can be 5 arranged in P 5! 120 = know that the mangoes are not separated in 48 of these. ways, and we 5 = 1 Find how many five-digit numbers can be made using the digits 2, 3, 4, 5 and 6 once each if: a b there are no restrictions the five-digit number must be: i odd ii even iii odd and less than 40 000. 2 Find how many ways four men and two women can stand in a line if: a b c d the two women must be at the front there must be a woman at the front and a man at the back the two women must be separated the four men must not be separated e no two men may stand next to each other. 3 Find the ratio of odd-to-even six-digit numbers that can be made using the digits 1, 2, 3, 4, 5 and 7. 131 4 Find how many ways 10 books can be arranged in a row on a shelf if: a b the two oldest books must be in the middle two positions the three newest books must not be separated. 5 Five cows and one set of twin calves can be housed separately in a row of seven stalls in P =5040 ways. Find 7 7 in how many of these arrangements: a b the two calves are not in adjacent stalls the two calves and their mother, who is one of the 5 cows, are in adjacent stalls c each calf is in a stall adjacent to its mother. 6 Find |
how many of the six-digit numbers that can be made from 1, 2, 2, 3, 3 and 3: a begin with a 2 b are not divisible by 2. 7 Find the number of distinct arrangements that can be made from all the letters in the word THEATRE when the arrangement: a begins with two Ts and ends with two Es b has H as its middle letter c ends with the three vowels E, A and E. PS 8 The following diagram shows a row of post boxes with the owners’ names beneath. Five parcels, one for the owner of each box, have arrived at the post office. If one parcel is randomly placed in each box, find the number of ways in which: Mr A Ms B Mr C Ms D Mrs E Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 a the five parcels can all be placed in the correct boxes b exactly one parcel can be placed in the wrong box c the correct parcels can be placed in Mr A’s and one other person’s box only d exactly two parcels can be placed in the correct boxes. P 9 There are x boys and y girls to be arranged in a line. Find the relationship between x and y if it is not possible to separate all the boys. Permutations of r objects from n objects So far, we have dealt only with permutations in which all of the objects are selected and arranged. We can now take this a step further and look at permutations in which only some of the objects are selected and arranged. When we select and arrange r objects in a particular order from n distinct objects, we call this a permutation of r from n. Suppose, for example, we wish to select and arrange three letters from the five letters A, B, C, D and E. We have five choices for the first letter, four for the second, and three choices for the third. This gives us a total of 5 permutations, which is effectively 5! but 2! are missing. 60 × × = 3 4 There are 5! (5 3)! − = 5! 2! = 60 permutations altogether. 132 WORKED EXAMPLE 5.7 How many three-digit numbers can |
be made from the seven digits 3, 4, 5, 6, 7, 8 and 9, if each is used at most once? Answer KEY POINT 5.4 n ( = There are P r! n n r − permutations of r objects from n distinct objects. )! 7 P 3 = 7! (7 3)! − 7! 4! 7 6 5 = × × 210 = = three-digit numbers We select and arrange just three of the seven distinct digits (and ignore four of them). TIP The choices we have for the first, second and third digits are 7 × × = 210. 5 6 WORKED EXAMPLE 5.8 In how many ways can five playing cards from a standard deck of 52 cards be arranged in a row? Answer 52 P 5 = = 52! 47! 311875 200 ways We select and arrange five of the 52 playing cards (and ignore 47 of them). Copyright Material - Review Only - Not for Redistribution TIP The choices we have are 52 51 50 × × × 311 875200. 49 48 × = Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations WORKED EXAMPLE 5.9 In how many ways can 4 out of 18 girls sit on a four-seat sofa when the oldest girl must be given one of the seats? Answer 4 × 17 P 3 = 16320 ways WORKED EXAMPLE 5.10 Four ways for the oldest girl to occupy a seat, and P17P17 arrange three of the remaining 17 girls to sit with her. 3 ways to select and In how many ways can four boys and three girls stand in a row when no two girls are allowed to stand next to each other? Answer 4P4 ways to arrange 4 boys in a row B1 B2 B3 B4 Arrange the girls in 3 of these 5 spaces 5P3 4 5 P 4 × P 1440 = 3 ways 4 ways to arrange the four P4 boys in a row. 3 ways to select three of the P5 five spaces between or to the side of the boys and arrange the three girls in them, as shown. TIP Objects that must be separated are individually placed between or beyond the objects that can be separated. 133 EXERCISE 5E 1 Find how many permutations there are of |
: a five from seven distinct objects b four from nine distinct objects. 2 From 12 books, how many ways are there to select and arrange exactly half of them in a row on a shelf? 3 In how many ways can gold, silver and bronze medals be awarded for first, second and third places in a race between 20 athletes? You may assume that no two athletes tie in these positions. 4 a Find the number of ways in which Alvaro can paint his back door and his front door in a different colour if he has 14 colours of paint to choose from. b In how many ways could Alvaro do this if he also considered painting them the same colour? 5 Find how many of the arrangements of four letters from A, B, C, D, E and F: a begin with the letter A b contain the letter A. 6 From a group of 10 boys and seven girls, two are to be chosen to act as the hero and the villain in the school play. Find in how many ways this can be done if these two roles are to be played by: a any of the children b two girls or two boys c a boy and a girl. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 7 From a set of 10 rings, a jeweller wishes to display seven of them in their shop window. The formation of the display is shown in the diagram opposite. Find the number of possible displays if, from the set of 10: a b the ring with the largest diamond must go at the top of the display the most expensive ring must go at the top with the two least expensive rings adjacent to it. 8 Using each digit not more than once, how many even four-digit numbers can be made from the digits 1, 2, 3, 4, 5, 6 and 7? 9 Find how many three-digit numbers can be made from the digits 0, 1, 2, 3 and 4, used at most once each, if the three-digit number: a must be a multiple of 10 b cannot begin with zero. 10 Give an example of a practical situation where the calculation Pn r = 120 might arise. P 11 a Under what condition is Given that, find an expression for k |
in terms of n and r. 12 Five playing cards are randomly selected from a standard deck of 52 cards. These five cards are shuffled, and then the top three cards are placed in a row on a table. How many different arrangements of three of the 52 cards are possible? PS 13 Seven chairs, A to G, are arranged as shown. C D E 134 In how many ways can the chairs be occupied by 7 of a group of 12 people if three particular people are asked to sit on chairs B, D and F, in any order? PS 14 A minibus has 11 passenger seats. There are six seats in a row on the sunny side and five seats in a row on the shady side, as shown in the following diagram. B A F G Find how many ways eight passengers can be arranged in these seats if: a there are no restrictions b one particular passenger refuses to sit on the sunny side c two particular passengers refuse to sit in seats that are either next to each other or one directly in front of the other. shady sunny DID YOU KNOW? The rule to determine the number of permutations of n objects was known in Indian culture at least as early as 1150 and is explained in the Līlāvatī by Indian mathematician Bhaskara II. In his books Campanalogia and Tintinnalogia, Englishman Fabian Stedman in 1677 described factorials when explaining the number of permutations of the ringing of church bells. A complete peal of changes of n bells is made when they are rung in!n sequences without repetition. The speed at which church bells ring cannot be changed very much by the ringers and this may be why there are at most five bells in most churches. For example, 10 bells can be rung in 3628800 different sequences and it would take the ringers over 3 months to ring a complete peal of changes of 10 bells! Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations 5.3 Combinations A combination is simply a selection, where the order of selection is not important. Choosing strawberries and ice cream from a menu is the same combination as choosing ice cream and strawberries. When we select r objects in no particular order from n objects, |
we call this is a combination. KEY POINT 5.5 A combination of r objects which are then arranged in order is equivalent to a permutation. We write nCr to mean the number of combinations of r objects from n. Since there are rPr = r! ways of arranging the r objects, we have!( r n − ( )! r )! Suppose we wish to select three children from a group of five. We can view this task as ‘choosing three and ignoring two’ or as ‘choosing to ignore two and remaining with three’. Regardless of how we view it, choosing three from five and choosing two from five can be done in an equal number of ways, and so C 5 C. 5 = 3 2 The following three points should be noted. n n n C = r øC r n C n P r n r– C r =! n!( r n − = r )! No. we select from! No. selected! No. not selected! × WORKED EXAMPLE 5.11 There are! n!( r n − n C = r r )! combinations of r objects from n distinct objects. FAST FORWARD n r = ! n!( r n − r )! We will use this more modern notation in Chapter 7. However, most calculators use the Cn r notation, so we will use this in the current chapter. 135 In how many ways can three fish be selected from a bowl containing seven fish and two potatoes? Answer 7 C 3 = 7! × 4! 3! = 35 ways The two potatoes are irrelevant. We select three fish from seven fish. WORKED EXAMPLE 5.12 In how many ways can five books and three magazines be selected from eight books and six magazines? Answer 8 6 C 56 5 = ways C 3 = 20 ways 8 C 5 × 6 C 3 = = 56 20 × 1120 ways We select five from eight books and we select three from six magazines. TIP The books and the magazines are selected independently, so we multiply the numbers of combinations. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A |
Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 5.13 A team of five is to be chosen from six women and five men. Find the number of possible teams in which there will be more women than men. Answer From 6 women From 5 men No. teams 3 4 5 or or = 200 6 C 4 × 5 C 75 = 200 + 75 6 + = 281 teams with more women than men. The table shows the possible make-up of the team when it has more women than men in it; and also the number of ways in which those teams can be chosen. WORKED EXAMPLE 5.14 How many distinct three-digit numbers can be made from five cards, each with one of the digits 5, 5, 7, 8 and 9 written on it? Answer The 5 is a repeated digit, so we must investigate three situations separately. 136 No 5s selected: P = 6 numbers. 3 3 One 5 selected: C three-digit numbers. 2 × 3 Two 5s selected: C three-digit numbers. 1 × 3 three-digit The digits 7, 8 and 9 are selected and arranged. 3! 18 = 3! 2! 9 = Two digits from 7, 8 and 9 are selected and arranged with a 5. One digit from 7, 8 and 9 is selected and arranged with two 5s. 6 18 9 + = + be made. 33 three-digit numbers can EXERCISE 5F FAST FORWARD You will learn about probability distributions for the number of objects that can be selected in Chapter 6, such as the number of women selected for this team. TIP The selections in these three situations are mutually exclusive, so we add together the numbers of three-digit numbers. 1 Find the number of ways in which five apples can be selected from: a eight apples b nine apples and 12 oranges. 2 From seven men and eight women, find how many ways there are to select: a four men and five women b three men and six women c at least 13 people. 3 a How many different hands of five cards can be dealt from a standard deck of 52 playing cards? b How many of the hands in part a consist of three of the 26 red cards and two of the 26 black cards? 4 a From the 26 letters of the English alphabet, find how many ways there are to choose: i six different letters ii 20 different letters. b Use your results from part a to find the condition under which C x = y x C z, where |
x is a positive integer. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations 5 In a classroom there are four lights, each operated by a switch that has an on and an off position. How many possible lighting arrangements are there in the classroom? 6 From six boys and seven girls, find how many ways there are to select a group of three children that consists of more girls than boys. 7 A bag contains six red fuses, five blue fuses and four yellow fuses. Find how many ways there are to select: a three fuses of different colours b three fuses of the same colour c 10 fuses in exactly two colours d nine fuses in exactly two colours. 8 The diagram opposite shows the activities offered to children at a school camp. If children must choose three activities to fill their day, how many sets of three activities are there to choose from? Today’s Activities Morning: acting, painting or singing Afternoon: swimming, tennis, golf or cricket Evening: night-hike, star-gazing or drumming Afternoon swimming can be done at the pool or at the lake 9 Two taxis are hired to take a group of eight friends to the airport. One taxi can carry five passengers and the other can carry three passengers. What information is given in this situation by the fact that C 5 8 = 8 C 3 =? 56 10 Ten cars are to be parked in a car park that has 20 parking spaces set out in two rows of 10. Find how many different patterns of unoccupied parking spaces are possible if: a b c d the cars can be parked in any of the 20 spaces the cars are parked in the same row the same number of cars are parked in each row two more cars are parked in one row than in the other. 137 11 A boy has eight pairs of trousers, seven shirts and six jackets. In how many ways can he dress in trousers, shirt and jacket if he refuses to wear a particular pair of red trousers with a particular red shirt? 12 A girl has 11 objects to arrange on a shelf but there is room for only seven of them. In how many ways can she arrange seven of the objects in a row along the shelf, if her clock must be included? 13 A Mathematics teacher |
has 10 different posters to pin up in their classroom but there is enough space for only five of them. They have three posters on algebra, two on calculus and five on trigonometry. In how many ways can they choose the five posters to pin up if: a b c there are no restrictions they decide not to pin up either of the calculus posters they decide to pin up at least one poster on each of the three topics algebra, calculus and trigonometry? 14 As discussed at the beginning of this chapter in Explore 5.1 about encrypting letters, it states that there are over 27 million possibilities for the password encrypted as UJSNOL. How many possibilities are there? 15 How many distinct three-digit numbers can be made from 1, 2, 2, 3, 4 and 5, using each at most once? Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PS 16 From three sets of twins and four unrelated girls, find how many selections of five people can be made if exactly: a two sets of twins must be included b one set of twins must be included. EXPLORE 5.3 different ways. The Two women and three men can sit on a five-seater bicycle in 5! 120 photo shows an arrangement in which the two women are separated and the three men are also separated. = Consider, separately, the arrangements in which the women, and in which the men, are all separated from each other. 138 a Women separated from each other. Women next to each other 2!= Arrange three men with the women as a single object 4!= There are 2! the women are not separated. 72 (2! So there are 5! arrangements in which the women are separated from each other. 4!× arrangements in which 4!) × − = b Men separated from each other. Men next to each other 3!= Arrange two women with the men as a single object 3!= There are 3! 3!× arrangements in which the men are not separated. So there are 5! – (3! 3!) arrangements in which the men are separated from each other. 84 × = The calculations in a and b follow the same steps; however, the logic in one of |
them is flawed. Which of the two answers is correct? Can you explain why the other answer is not correct? 5.4 Problem solving with permutations and combinations Permutations and combinations can be used to find probabilities for certain events. If an event consists of a number of favourable permutations that are equiprobable, or a number of favourable combinations that are equiprobable, then P(event) = No. favourable permutations No. possible permutations P(event) = No. favourable combinations No. possible combinations Using either of the previous given methods can greatly reduce the amount of working required to solve problems in probability. Nevertheless, we must decide carefully which of them, if any, it is appropriate to use. Copyright Material - Review Only - Not for Redistribution REWIND From the introduction to this chapter, we know that the order of selection matters in a permutation but does not matter in a combination. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations WORKED EXAMPLE 5.15 There are 15 identical tins on a shelf. None of the tins are labelled but it is known that eight contain soup S( ) and three contain peas P( ), four contain beans B( ). If seven tins are randomly selected without replacement, find the probability that exactly five of them contain soup. Answer Favourable selections are when five tins of soup and two tins that are not soup are selected. (It is not important whether these two tins contain beans or peas.) We denote the 15 tins by S8 and S7 ′, where S ′ represents not soup. 8 C 5 × 7 C 2 favourable combinations Selecting S5 from S8 and S ′2 from S ′7. C15 7 possible combinations Selecting seven from 15 tins. P(select 5tinsof soup 15 C 7 21 56 × 6435 392 2145 or 0.183. 2 TIP In the numerator we have 8 7 15 and + = + =. 7 5 2 WORKED EXAMPLE 5.16 139 A girl has a bag containing 13 red cherries R( five cherries from the bag at random. Find the probability that she takes more red cherries than black cherries. ) and seven black cherries B( ). She takes |
Answer From 13 red From 7 black Number of ways 5 4 3 or or 0 1 2 13 C 5 × 7 C 1287 = 0 13 C 4 × 7 C 5005 = 1 13 C 3 × 7 C 2 = 6006 Total 12298 = 20 C 15504 5 = ways P(more red than black) = 12298 15504 or 0.793. The table shows the possible make-up of the selected cherries when there are more red than black; and also the number of ways in which those cherries can be chosen. Selecting five from 20 cherries. REWIND From Chapter 4, Section 4.2, recall that ) P( or or A B C ) P( ) P( P( ) B A = + + for mutually exclusive events. C Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 EXPLORE 5.4 We can, of course, find the solution to Worked example 5.16 using conditional probabilities. ● There is one way to select R5 and B0. ● There are five ways to select R4 and B1. ● There are 10 ways to select R3 and B2. Complete the calculations using conditional probabilities. Note how much working is involved and how long the calculations take. Compare the two approaches to solving this problem and decide for yourself which you prefer. REWIND We studied conditional probabilities in Chapter 4, Section 4.4. WORKED EXAMPLE 5.17 A minibus has seats for the driver (D) and seven passengers, as shown. When seven passengers are seated in random order, find the probability that two particular passengers, A and B, are sitting on: D a the same side of the minibus b opposite sides of the minibus. 140 Answer a P3 2 ways P4 2 ways P7 2 ways A and B both sitting on the driver’s side. A and B both not sitting on the driver’s side. A and B sitting in any two of the seven seats. P(same side) P(both ondriver’sside) P(both not on driver’s side + 12 42 7 3 P 2 P 2 6 42 3 7 b P(oppos |
itesides) 1 P(same side The events ‘sitting on the same side’ and ‘sitting on opposite sides’ are complementary. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations EXERCISE 5G 1 Two children are selected at random from a group of six boys and four girls. Use combinations to find the probability of selecting: a two boys b two girls c one boy and one girl. 2 Three chocolates are selected at random from a box containing 10 milk chocolates and 15 dark chocolates. Find the probability of selecting exactly: a two dark chocolates b two milk chocolates c two dark chocolates or two milk chocolates. 3 Four bananas are randomly selected from a crate of 17 yellow and 23 green bananas. Find the probability that: a no green bananas are selected b less than half of those selected are green. 4 A curator has 36 paintings and 44 sculptures from which they will randomly select eight items to display in their gallery. Find the probability that the display consists of at least three more paintings than sculptures. 5 Five people are randomly selected from a group of 67 women and 33 men. Find the probability that the selection consists of an odd number of women. 6 In a toolbox there are 25 screwdrivers, 16 drill bits, 38 spanners and 11 chisels. Find the probability that a random selection of four tools contains no chisels. 7 Five clowns each have a red wig and a blue wig, which they are all equally likely to wear at any particular time. Find the probability that, at any particular time: a exactly two clowns are wearing red wigs b more clowns are wearing blue wigs than red wigs. 141 8 A gardener has nine rose bushes to plant: three have red flowers and six have yellow flowers. If they plant them in a row in random order, find the probability that: a a yellow rose bush is in the middle of the row b the three red rose bushes are not separated c no two red rose bushes are next to each other. 9 A farmer has 50 animals. They have 24 sheep, of which three are male, and they have 26 cattle, of which 20 are female. A veterinary surgeon wishes |
to test six randomly selected animals. Find the probability that the selection consists of: a equal numbers of cattle and sheep b more females than males. 10 a How many distinct arrangements of the letters in the word STATISTICS are there? b Find the probability that a randomly selected arrangement begins with: i three Ts ii three identical letters. 11 Three skirts, four blouses and two jackets are hung in random order on a clothes rail. Find the probability that: a b the three skirts occupy the middle section of the arrangement the two jackets are not separated. 12 In a group of 180 people, there are 88 males, nine of whom are left-handed, and there are 85 females who are not left-handed. If six people are selected randomly from the group, find the probability that exactly four of them are left-handed or female. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 13 A small library holds 1240 books: 312 of the 478 novels N( ) have hard 1240 covers H( ), and there are 440 books that do not have hard covers. Some of this information is shown in the Venn diagram opposite. a Find the value of a, of b and of c. b A random selection of 25 of these books is to be donated to a charity group. The charity group hopes that at least 22 of the books will be novels or hard covers. Calculate the probability that the charity group gets what they hope for. 478 N H a 312 c b PS 14 A netball team of seven players is to be selected at random from five men and 10 women. Given that at least five women are selected for the team, find the probability that exactly two men are selected. PS 15 Two items are selected at random from a box that contains some tags and some labels. Selecting two tags is five times as likely as selecting two labels. Selecting one tag and one label is six times as likely as selecting two labels. Find the number of tags and the number of labels in the box. P 16 A photograph is to be taken of a pasta dish and n pizzas. The items are arranged in a line in random order. Event X is ‘the pasta dish is between two pizzas� |
�. 142 a Investigate the value of XP( b Hence, express the value of for any value of n ⩾ 2? ) for values of n from 2 to 5. P( P( X ′ X ) ) in terms of n. Can you justify your answer WEB LINK You will find a range of interesting and challenging probability problems (with hints and solutions) in Module 16 on the NRICH website. Checklist of learning and understanding ● n! = ( n n − 1)( n 0! 1= − … × × ×, for any integer n 0>. 3 2 1 2) ● A key word that points to a permutation is arranged. A permutation is a way of selecting and arranging objects in a particular order. ● Key words that point to a combination are chosen and selected. A combination is a way of selecting objects in no particular order. ● From n distinct objects, there are: n P n!= permutations of all n objects )! ( n permutations of r objects!( r n − permutations in which there are p q r,,, … of each type.... combinations of r objects. r )! Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations END-OF-CHAPTER REVIEW EXERCISE 5 1 The word MARMALADE contains four vowels and five consonants. Find the number of possible arrangements of its nine letters if: a there are no restrictions on the order b the arrangement must begin with the four vowels. 2 Five men, four children and two women are asked to stand in a queue at the post office. Find how many ways they can do this if: a the women must be separated b all of the children must be separated from each other. 3 Find the probability that a randomly selected arrangement of all the letters in the word PALLETTE begins and ends with the same letter. 4 Eight-digit mobile phone numbers issued by the Lemon Network all begin with 79. a How many different phone numbers can the network issue? b Find the probability that a randomly selected number issued by this network: i ends with the digits 97 ii reads the same left to right as right to left. 5 There are 12 books on a shelf. Five books are 15cm |
tall; four are 20 cm tall and three are 25cm tall. Find the number of ways that the books can be arranged on the shelf so that none of them is shorter than the book directly to its right. 6 The 11 letters of the word REMEMBRANCE are arranged in a line. i Find the number of different arrangements if there are no restrictions. ii Find the number of different arrangements which start and finish with the letter M. iii Find the number of different arrangements which do not have all 4 vowels (E, E, A, E) next to each other. 4 letters from the letters of the word REMEMBRANCE are chosen. iv Find the number of different selections which contain no Ms and no Rs and at least 2 Es. [1] [2] [2] [3] [3] [1] [2] [2] [2] [1] [2] [3] [3] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q6 November 2013 7 Find how many ways 15 children can be divided into three groups of five if: a there are no restrictions b two of the children are brothers who must be in the same group. 8 An entertainer has been asked to give a performance consisting of four items. They know three songs, five jokes, two juggling tricks and can play one tune on the mandolin. Find how many different ways there are for them to choose the four items if: a there are no restrictions on their performance b they decide not to sing any songs c they are not allowed to tell more than two jokes. 9 From a group of nine people, five are to be chosen at random to serve on a committee. In how many ways can this be done if two particular people refuse to serve on the committee together? [2] [3] [1] [2] [3] [3] Copyright Material - Review Only - Not for Redistribution 143 Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 10 Twenty teams have entered a tournament. In order to reduce the number of teams to eight, they are put into groups of five and the teams in each group play each other twice. The top two teams in each group progress to the next round. From this point on, teams are |
paired up, playing each other once with the losing team being eliminated. How many games are played during the whole tournament? [3 11 A bank provides each account holder with a nine-digit card number that is arranged in three blocks, as shown in the example opposite. Find, in index form, the number of card numbers available if: a there are no restrictions on the digits used b none of the three blocks can begin with 0 c the two digits in the second block must not be the same d the three-, two- and four-digit numbers on the card are even, odd and even, respectively. 12 A basket holds nine flowers: two are pink, three are yellow and four are red. Four of these flowers are chosen at random. Find the probability that at least two of them are red. 13 Find the number of ways in which 11 different pieces of fruit can be shared between three boys so that each boy receives an odd number of pieces of fruit. PS 14 A bakery wishes to display seven of its 14 types of cake in a row in its shop window. There are six types of 144 sponge cake, five types of cheesecake and three types of fruitcake. Find the number of possible displays that can be made if the bakery places: a a sponge cake at each end of the row and includes no fruitcakes in the display b a fruitcake at one end of the row with sponge cakes and cheesecakes placed alternately in the remainder of the row. PS 15 Five cards, each marked with a different single-digit number from 3 to 7, are randomly placed in a row. Find the probability that the first card in the row is odd and that the three cards in the middle of the row have a sum of 15. PS 16 Two ordinary fair dice are rolled and the two faces on which they come to rest are hidden by holding the dice together, as shown, and lifted off the table. The sum of the numbers on the 10 visible faces of the dice is denoted by T. a Find the number of possible values of T, and find the most likely value of T. b Calculate the probability that øT 38. PS 17 Three ordinary fair dice are rolled. Find the number of ways in which the number rolled with the first die can exceed the sum of the numbers rolled with the second and third dice. Hence, find the probability that this event does not occur in two successive rolls of the three dice. Copyright Material - Review Only - Not for Redistribution [1] [2 |
] [2] [3] [4] [5] [2] [4] [4] [4] [3] [6] Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 5: Permutations and combinations PS 18 How many even four-digit numbers can be made from the digits 0, 2, 3, 4, 5 and 7, each used at most once, when the first digit cannot be zero? 19 a i Find how many numbers there are between 100 and 999 in which all three digits are different. ii Find how many of the numbers in part i are odd numbers greater than 700. b A bunch of flowers consists of a mixture of roses, tulips and daffodils. Tom orders a bunch of 7 flowers from a shop to give to a friend. There must be at least 2 of each type of flower. The shop has 6 roses, 5 tulips and 4 daffodils, all different from each other. Find the number of different bunches of flowers that are possible. [4] [3] [4] [4] Cambridge International AS & A Level Mathematics 9709 Paper 61 Q6 June 2016 20 Three identical cans of cola, 2 identical cans of green tea and 2 identical cans of orange juice are arranged in a row. Calculate the number of arrangements if i ii the first and last cans in the row are the same type of drink, the 3 cans of cola are all next to each other and the 2 cans of green tea are not next to each other. [3] [5] Cambridge International AS & A Level Mathematics 9709 Paper 63 Q4 June 2010 145 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 CROSS-TOPIC REVIEW EXERCISE 2 1 Each of the eight players in a chess team plays 12 games against opponents from other teams. The total number of wins, draws and losses for the whole team are denoted by X Y, and Z, respectively. a State the value of X Y Z +. + b Find the |
least possible value of Z X−, given that Y 25=. c Given that none of the players drew any of their games and that X Z 50, find the exact mean number − = of games won by the players. 2 Six books are randomly given to two girls so that each receives at least one book. a In how many ways can this be done? [1] [1] [2] [3] 3 4 b Are both girls more likely to receive an odd number or an even number of books? Give a reason for your answer. [2] The 60 members of a ballroom dance society wish to participate in a competition but the coach that has been hired has seats for only 57 people. In how many ways can 57 members be selected if the society’s president and vice president must be included? [2] Four discs in two colours and in four sizes are placed in any order on either of two sticks. The following illustration shows one possible arrangement of the four discs. 146 a Find the number of ways in which the four discs can be arranged so that: i they are all on the same stick ii there are two discs on each stick. b In how many ways can the discs be placed if there are no restrictions? [2] [2] [2] 5 A fair triangular spinner with sides numbered 1, 2 and 3 is spun three times and the numbers that it comes to rest on are written down from left to right to form a three-digit number. a How many possible three-digit numbers are there? b Find the probability that the three-digit number is: i even ii odd and greater than 200. 6 A book of poetry contains seven poems, three of which are illustrated. In how many different orders can all the poems be read if no two illustrated poems are read one after the other? 7 Find the number of ways that seven goats and four sheep can sleep in a row if: a all the goats must sleep next to each other b no two sheep may sleep next to each other. 8 A teacher is looking for 6 pupils to appear in the school play and has decided to select them at random from a group of 11 girls and 13 boys. a Find the number of ways in which the teacher can select the 6 pupils. [1] [1] [2] [3] [2] [3] [1] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - |
Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cross-topic review exercise 2 b Two roles in the play must be played by girls; three roles must be played by boys, but the fool can be played by a girl or by a boy. If the first pupil selected is to play the role of the fool, find the probability that the fool is played by i a particular girl ii a boy. c If instead, the pupil who is to play the fool is the last of the six pupils selected, investigate what effect this change in the order of selection has on the probability that the fool is played by: i a particular girl ii a boy. [1] [1] [3] [3] 9 A radio presenter has enough time at the end of their show to play five songs. She has 13 songs by four groups to choose from: five songs by The Anvils, four by The Braziers, three by The Chisels and one by The Dustbins. Find the number of ways she can choose five songs to play if she decides: a that there should be no restrictions b to play all three songs by The Chisels c to play at least one song by each of the four groups. [1] [2] [4] 10 Students enrolling at an A Level college must select three different subjects to study from the six that are available. One subject must be chosen from each of the option groups A, B and C, as shown in the following table. Group A Physics Chemistry History Group B Biology Physics Group C Mathematics Biology Mathematics Computing a One student has chosen to study History and Mathematics. How many subjects do they have to choose from to complete their selection? b How many combinations of three subjects are available to a student who enrols at this college? 11 Four ordinary fair dice are arranged in a row. Find the number of ways in which this can be done if the four numbers showing on top of the dice: a are all odd b have a sum that is less than 7. 12 At company V, 12.5% of the employees have a university degree. At company W, 85% of the employees do not have a university degree. There are 112 employees at company V and 120 employees at company W. a One employee is randomly selected. Find the probability that they: i work for company V ii have a university degree. b Five employees from company W are selected at random. |
Find the probability that none of them has a university degree. 13 One hundred qualified drivers are selected at random. Out of these 100 drivers, of the 40 drivers who wear spectacles, 30 passed their driving test at the first attempt. Altogether, 25 of the drivers did not pass at their first attempt. a Show the data given about the drivers in a clearly labelled table or diagram. b Did these drivers pass the test at their first attempt independently of whether or not they wear spectacles? Explain your answer. Copyright Material - Review Only - Not for Redistribution 147 [1] [2] [1] [3] [1] [2] [2] [3] [3] Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 14 A conference hall has 24 overhead lights. Pairs of lights are operated by switches next to the main entrance, and each switch has three numbered settings: 0 (off), 1 (dim), 2 (bright). Find the number of possible lighting arrangements in the hall if: a there are no restrictions b two particular pairs of lights must be on setting 2 c three lights that are not operated by the same switch and five pairs of lights that are operated by the same switch are not working. 15 Twelve chairs in two colours are arranged, as shown. A B C D 1 2 3 Find in how many ways nine people can sit on these chairs if: a the two blue chairs in column C must remain unoccupied b all of the green chairs must be occupied c more blue chairs than green chairs must be occupied 148 d at least one of the chairs in row 2 must remain unoccupied. 16 In a certain country, vehicle registration plates consist of seven characters: a letter, followed by a three-digit number, followed by three letters. [1] [1] [2] [2] [3] [2] [5] For example: B 474 PQR The first letter cannot be a vowel; the three-digit number cannot begin with 0; and the first of the last three letters cannot be a vowel or any of the letters X, Y or Z. a Find the number of registration plates available. [2] b Find the probability that a randomly selected registration plate is unassigned, given that there are 48 |
.6 million vehicle owners in the country, and that each owns, on average, 1.183 registered vehicles. [3] 17 Seats for the guests at an awards ceremony are arranged in two rows of eight and ten, divided by an aisle, as shown. e l s i a Seats are randomly allocated to 18 guests. a Find the probability that two particular guests are allocated seats: i on the same side of the aisle ii in the same row iii on the same side of the aisle and in the same row. [3] [3] [4] b Give a reason why the answer to part a iii is not equal to the product of the answers to part a i and part a ii. [1] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6 Probability distributions In this chapter you will learn how to: 149 ■ identify and use a discrete random variable ■ construct a probability distribution table that relates to a given situation involving a discrete random variable, X, and calculate its expectation, XE( ), and its variance, Var( )X. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PREREQUISITE KNOWLEDGE Where it comes from What you should be able to do Check your skills IGCSE / O Level Mathematics Use the fact that A P( ) 1 P( = − A ′. ) Chapters 4 and 5, sections 4.3, 4.4, 4.5 and 5.4 Distinguish between independent and dependent events, and calculate probabilities accordingly. 1 A game can be won W( Given that W ′ = )D. find P( P( ), lost ( )L or drawn (, 0.65 ) ′ = )D. ) 0.46 and LP( 2 Two cubes are selected at random from a bag of three red cubes and three blue cubes. Show that the selected cubes are more likely to both be red when the selections are made with replacement than when the selections are made without replacement. |
Tools of the trade Suppose a trading company is planning a new marketing campaign. The campaign will probably go ahead only if the most likely outcome is that sales will increase. However, the company also needs to be aware of worst-case and best-case outcomes, as sales may decrease or decrease dramatically, stay the same or increase dramatically. The company will be able to make informed decisions based on its estimates of the probabilities of these possible outcomes. The likelihood of these outcomes will be based on an analysis of a probability distribution for the changes in sales. The probability distribution described above acts as a prediction for future sales and the risks involved. Suppose the company is considering entering a new line of business but needs to generate at least $50 000 in revenue before it starts to make a profit. If their probability distribution tells them that there is a 40 % chance that revenues will be less than $50 000, then the company knows roughly what level of risk it is facing by entering that new line of business. 150 6.1 Discrete random variables A variable is said to be discrete and random if it can take only certain values that occur by chance. For example, when we buy a carton of six eggs, some may be broken; the number of broken eggs in a carton is a discrete random variable that can take values 0, 1, 2, 3, 4, 5 or 6. Discrete random variables may arise from independent trials. For example, if we roll four. dice then the number of 6s obtained, S, is a discrete random variable with S {0, 1, 2, 3, 4} ∈ Situations where selections are made without replacement, can also generate discrete random variables. For example, if we randomly select three children from a group of four boys and two girls, the number of boys selected, B, and the number of girls selected, G, are discrete random variables with B {1, 2, 3}. and G {0, 1, 2} ∈ ∈ 6.2 Probability distributions The probability distribution of a discrete random variable is a display of all its possible values and their corresponding probabilities. The usual method of display is by tabulation in a probability distribution table. The probability distribution also can be represented in a vertical line graph or in a bar chart. Copyright Material - Review Only - Not for Redistribution TIP A variable is denoted by an upper-case letter and its possible values by the same lower-case letter. If X can take values of 1, |
2 and 3, we write X ∈ {1, 2, 3}, where the symbol ∈ means ‘is an element of’. REWIND We learnt how to find probabilities for selections with and without replacement in Chapters 4 and 5, Sections 4.3, 4.4, 4.5 and 5.4. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6: Probability distributions Consider tossing two fair coins, where we can obtain 0, 1 or 2 heads.. The number of heads obtained in each trial, X, is a discrete random variable and X {0, 1, 2} ∈ XP( = 0) P(tails and tails) = = 0.5 0.5 × = 0.25 XP( = 1) P(heads and tails) P(tails and heads) + = = (0.5 0.5) × + (0.5 0.5) × = 0.5 XP( = 2) P(heads and heads) = = 0.5 0.5 × = 0.25 The probability distribution for X is displayed in the following table. x P( )==X x 0 0.25 1 0.5 2 0.25 The probabilities for the possible values of X are equal to the relative frequencies of the values. We would expect 25% of the tosses to produce zero heads; 50% to produce one head and 25% to produce two heads. WORKED EXAMPLE 6.1 A fair square spinner with sides labelled 1, 2, 3 and 4 is spun twice. The two scores obtained are added together to give the total, X. Draw up the probability distribution table for X. Answer 1st spin The grid shows the 16 equally likely outcomes for the discrete random variable X, where. X {2, 3, 4, 5, 6, 7, 8} ∈ 2 1 16 3 2 16 4 3 16 5 4 16 6 3 16 7 2 16 8 1 16 Sum 1= The probability distribution for X is shown in the table. P( )== X x 151 TIP P( )=X x is equal to the relative frequency of each particular value of X. TIP Note that Σ P( X x = =. ) 1 WORKED EXAMPLE 6.2 The following table shows |
the probability distribution for the random variable V. v P( )==V v 2 0.05 3 2c 4 5 6 0.1+c 2 +c 0.05 0.16 Find the value of the constant c and find P( V >. 4) Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Answer 0.05 + c 2 c + + 0.1 2 + c ( c − c 0.2= or = − c 3.2 0.05 0.16 1 = + + 2 c + 0.2)( 3 – 0.64 c c + 3.2) The valid solution is c 0.2=. ∴ V P( > 4) P( = = V 5) P( + (2 0.2) 0.05 0.16 6) V = + + = = × 0.61 = = 0 0 Σ = to form and solve We use p 1 an equation in c. = − Note that if c P( = and VP( 3.2, then 3) 10.24, P( 4) V =. 6.35 5) = = = − V = − 3.1 KEY POINT 6.1 A probability distribution shows all the possible values of a variable and the sum of the probabilities is Σ = p 1 TIP Do check whether the solutions are valid. Remember that a probability cannot be less than 0 or greater than 1. WORKED EXAMPLE 6.3 There are spaces for three more passengers on a bus, but eight youths, one man and one woman wish to board. The bus driver decides to select three of these people at random and allow them to board. Draw up the probability distribution table for Y, the number of youths selected. 152 Answer )= Selections are made without replacement, so we can use combinations to find Y y P(. Possible values of Y are 1, 2 and 3 At least one youth will be selected because there are only two non-youths, who we denote by Y ′. C10 3 possible selections. Selecting three from 10 people. P( Y = 1) = 2 C 8 C × 1 10 C 3 2 = 1 |
15 Selecting one from Y8, and two from Y2 ′. P( Y = 2) = 8 C × 2 10 C 2 C 1 3 = 7 15 P( Y = 3) = 8 C × 3 10 C 2 C 0 3 = 7 15 y P( )==Y y 1 1 15 2 7 15 3 7 15 Selecting two from Y8, and one from Y2 ′. Selecting three from Y8, and none from Y2 ′. The table shows the probability distribution for Y. TIP Always check that 1pΣ =. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6: Probability distributions EXERCISE 6A 1 The discrete random variable V is such that V {1, 2, 3}. Given that V P( ∈ = 1) P( = V = 2) = × 2 P( V = 3), draw up the probability distribution table for V. 2 The probability distribution for the random variable X is given in the following table. x P( )==X x 2 p 3 p2 4 1 2 p 5 p3 Find the value of p and work out P(2 < X <. 5) 3 The probability distribution for the random variable W is given in the following table. w P( )==W w 3 k2 6 k2 9 k 2 12 4 5 3k− 15 13 50 a Form an equation using k, then solve it. b Explain why only one of your solutions is valid. c Find øW < P(6 10). 4 The probability that a boy succeeds with each basketball shot is random variable S represents the number of successful shots. 7 9. He takes two shots and the discrete 153 Show that SP( = 0) = 4 81 and draw up the probability distribution table for S. 5 At a garden centre, there is a display of roses: 25 are red, 20 are white, 15 are pink and 5 are orange. Three roses are chosen at random. a Show that the probability of selecting three red roses is approximately 0.0527. b Draw up the probability distribution table for the number of red roses selected. c Find the probability that at least one red rose is selected. 6 Three vehicles from a company� |
�s six trucks, five vans, three cars and one motorbike are randomly selected and tested for roadworthiness. a Show that the probability of selecting three vans is 2. 91 b Draw up the probability distribution table for the number of vans selected. c Find the probability that, at most, one van is selected. 7 Five grapes are randomly selected without replacement from a bag containing one red grape and six green grapes. Name and list the possible values of two discrete random variables in this situation. State the relationship between the values of your two variables. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 8 A pack of five DVDs contains three movies and two documentaries. Three DVDs are selected and the following table shows the probability distribution for M, the number of movies selected. m 1 2 3 P( ==M m ) 0.3 0.6 0.1 Draw up the probability distribution table for D, the number of documentaries selected. 9 In a particular country, 90% of the population is right-handed and 40% of the population has red hair. Two people are randomly selected from the population. Draw up the probability distribution for X, the number of right-handed, red-haired people selected, and state what assumption must be made in order to do this. 10 A fair 4-sided die, numbered 1, 2, 3 and 5, is rolled twice. The random variable X is the sum of the two numbers on which the die comes to rest. a Show that XP( = 8) 1 =. 8 b Draw up the probability distribution table for X, and find XP( 6)>. 11 There are eight letters in a post box, and five of them are addressed to Mr Nut. Mr Nut removes four letters at random from the box. a Find the probability that none of the selected letters are addressed to Mr Nut. b Draw up the probability distribution table for N, the number of selected letters that are addressed 154 to Mr Nut. c Describe one significant feature of a vertical line graph or bar chart that could be used to represent the probability distribution for N. 12 A discrete random variable Y is such that Y {8, 9, 10} ∈ )=. Given that Y y P( = k |
y, find the value of the constant k.. 13 Q is a discrete random variable and Q {3, 4, 5, 6} ∈ a Given that Q q P( = ) = cq 2, find the value of the constant c. b Hence, find QP( 4)>. 14 Four books are randomly selected from a box containing 10 novels, 10 reference books and 5 dictionaries. The random variable N represents the number of novels selected. a Find the value of NP( 2)=, correct to 3 significant figures. b Without further calculation, state which of N 0= or N 4= is more likely. Explain the reasons for your answer. 15 In a game, a fair 4-sided spinner with edges labelled 0, 1, 2 and 3 is spun. If a player spins 1, 2 or 3, then that is their score. If a player scores 0, then they spin a fair triangular spinner with edges labelled 0, 1 and 2, and the number they spin is their score. Let the variable X represent a player’s score. a Show that XP( = 0) = 1 12. b Draw up the probability distribution table for X, and find the probability that X is a prime number. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6: Probability distributions 16 A biased coin is tossed three times. The probability distribution for H, the number of heads obtained, is shown in the following table. h 0 1 2 P( )==H h 0.512 0.384 0.096 3 a a Find the probability of obtaining a head each time the coin is tossed. b Give another discrete random variable that is related to these trials, and calculate the probability that its value is greater than the value of H. 17 Two ordinary fair dice are rolled. A score of 3 points is awarded if exactly one die shows an odd number and there is also a difference of 1 between the two numbers obtained. A player who rolls two even numbers is awarded a score of 2 points, otherwise a player scores 1 point. a Draw up the probability distribution table for S, the number of points awarded. b Find the probability that a player scores 3 points, given that the sum of the numbers on their two dice is greater |
than 9. 18 The discrete random variable R is such that R {1, 3, 5, 7}. 1) ( k r + 2 r + a Given that R r, find the value of the constant k. P( ∈ ) = = 4)øR. b Hence, find P( EXPLORE 6.1 Consider the probability distribution for X, the number of heads obtained when two fair coins are tossed, which was given in the table presented in the introduction of Section 6.2. Sketch or simply describe the shape of a bar chart (or vertical line graph) that can be used to represent this distribution. In this activity, you will investigate how the shape of the distribution of X is altered when two unfair coins are tossed; that is, when the probability of obtaining heads is p. 0.5≠ Consider the case in which p probability distribution of X, the number of heads obtained. 0.4= for both coins. Draw a bar chart to represent the Next consider the case in which p represent the probability distribution of X. 0.6= for both coins, and draw a bar chart to What do you notice about the bar charts for p 0.4= and p? 0.6= Investigate other pairs of probability distributions for which the values of p add up to 1, such as p results.. Make general comments to summarise your and p 0.3= 0.7= M Investigate how the value of XP( represent this graphically. On the same diagram, show how the values of XP( and XP( changes as p increases from 0 to 1, and then 0)= change as p increases from 0 to 1. 1)= 2)= Copyright Material - Review Only - Not for Redistribution 155 WEB LINK This can be done manually or using the Coin Flip Simulation on the GeoGebra website. FAST FORWARD We will learn how to extend this Explore activity to more than two coins in Chapter 7. We will see how to represent the probability distribution for a continuous random variable in Chapter 8, Section 8.1. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 DID YOU KNOW? n The simple conjecture of Fermat’s Last Theorem, which is that x n 2>, defeated |
the greatest mathematicians for 350 years. has no positive integer solutions for any integer = + y z n n The theorem is simple in that it says ‘a square can be divided into two squares, but a cube cannot be divided into two cubes, nor a fourth power into two fourth powers, and so on’. Pierre de Fermat himself claimed to have a proof but only wrote in his notebook that ‘this margin is too narrow to contain it’! Fermat’s correspondence with the French mathematician, physicist, inventor and philosopher Blaise Pascal helped to develop a very important concept in basic probability that was revolutionary at the time; namely, the idea of equally likely outcomes and expected values. Since his death in 1665, substantial prizes have been offered for a proof, which was finally delivered by Briton Andrew Wiles in 1995. Wiles’ proof used highly advanced 20th century mathematics (i.e. functions of complex numbers in hyperbolic space and the doughnut-shaped solutions of elliptic curves!) that was not available to Fermat. 156 Pierre de Fermat, 1607–1665. Andrew Wiles 6.3 Expectation and variance of a discrete random variable Values of a discrete random variable with high probabilities are expected to occur more frequently than values with low probabilities. When a number of trials are carried out, a frequency distribution of values is produced, and this distribution has a mean or expected value. Expectation The mean of a discrete random variable X is referred to as its expectation, and is written XE( ). Suppose we have a biased spinner with which we can score 0, 1, 2 or 3. The probabilities for these scores, X, are as given in the following table and are also represented in the graph. x P( )==X x 0 0.1 1 0.3 2 0.4 3 0.2 However many times it is spun, we expect to score 0 with 10% of the spins; 1 with 30%; 2 with 40% and 3 with 20%. The expected frequencies of the scores in 1600 trials are shown in the following table. P(X = x) 0.4 0.2 0 0 1 2 x Expected frequency ( )f 0 1 2 0.1 1600 160 × = 0.3 1600 × = 480 0.4 1600 × = 640 0.2 1600 × = 320 Copyright Material - Review Only - Not for Redistribution x 3 3 TIP We can think of XE( ) |
as being the long-term average value of X over a large number of trials. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6: Probability distributions From this table of expected frequencies, we can calculate the mean (expected) score in 1600 trials. KEY POINT 6.2 The expectation of a discrete random ) variable is E( = Σ X xp TIP The denominator is Σ =p 1, so we can omit it from our calculation of XE( ). TIP An alternative way to write the formula for expectation is E( X x = = Σ P( X x × ) [ ) ]. 157 REWIND We can remember variance from Chapter 3, Section 3.3 as ‘mean of the squares minus square of the mean’. Mean E( = X ) = (0 160) × + xf Σ f Σ = (1 480) × (2 640) × + 1600 + (3 320) × = 1.7 We obtain the same value for XE( instead of frequencies. ) if relative frequencies (i.e. probabilities) are used Mean E( = X ) = xp Σ p Σ = (0 0.1) × + (1 0.3) × (2 0.4) × + (3 0.2) × = 1.7 + 1 EXPLORE 6.2 Adam and Priya each have a bag of five cards, numbered 1, 2, 3,4 and 5. They simultaneously select a card at random from their bag and place it face-up on a table. The numerical difference between the numbers on their cards, X, is recorded, where X {0, 1, 2, 3, 4} probability distribution table for X.. They repeat this 200 times and use their results to draw up a ∈ Adam suggests a new experiment in which the procedure will be the same, except that each of them can choose the card that they place on the table. He says the probability distribution for X will be very different because the cards are not selected at random. Priya disagrees, saying that it will be very similar, or may even be exactly the same. Do you agree with Adam or Priya? Explain your reasoning. Variance The variance and standard deviation of a discrete random variable give a measure |
of the spread of values around the mean, XE( using probabilities in place of frequencies. ). These measures, like XE( ), can be calculated If we replace f by p and replace x by XE( ) in the second of the two formulae for variance, we obtain Σ 2 x p p Σ − X{E( KEY POINT 6.3 )} 2, which simplifies to Σ 2 x p − X{E( )} 2 because Σ =p 1. The variance of a discrete random variable is VarE( X 2. )} WORKED EXAMPLE 6.4 The following table shows the probability distribution for X. Find its expectation, variance and standard deviation. x P( )==X x 0 1 12 5 3 12 15 5 12 20 3 12 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Answer XE( ) = = = = × × 0 1 12 1 12 12.5 × 1 12 + × 5 3 12 + 15 × 5 12 + 20 × 3 12 [(0 1) × + (5 3) × + (15 5) × + (20 3)] × 150 1 12 + 2 5 × 3 12 + 2 15 × 5 12 + 2 20 × 3 12 2 {12.5} − [(0 1) × + (25 3) × + (225 5) × + (400 3)] 156.25 − × Var 12 2400 12 43.75 − 156.25 XSD( ) = 43.75 = 6.61, correct to 3 significant figures. TIP The working is simpler when all fractions have the same denominator. TIP Remember to subtract the square of XE( when calculating variance. ) Substitute values of X and X x |
)= P( into the formula ) E( X = Σ xp. Substitute into the formula for Var(X). Take the square root of the variance. EXERCISE 6B 158 1 The probability distribution for the random variable X is given in the following table. x 0 1 2 3 P( )==X x 0.10 0.12 0.36 0.42 Calculate XE( ) and XVar( ). 2 The probability distribution for the random variable Y is given in the following table. y 0 P( )==Y y 0.03 1 2 p 2 0.32 3 p 4 0.05 a Find the value of p. b Calculate YE( ) and the standard deviation of Y. 3 The random variable T is such that T {1, 3, 6, 10}. Given that the four possible values of T are equiprobable, ∈ find TE( ) and TVar( ). 4 The following table shows the probability distribution for the random variable V. v 1 3 9 m P( )==V v 0.4 0.28 0.14 0.18 Given that VE( ) = 5.38, find the value of m and calculate VVar( ). 5 R is a random variable such that R {10, 20, 70, 100} ∈. Given that R r P( )= is proportional to r, show that RE( ) 77= and find RVar( ). Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6: Probability distributions 6 The probability distribution for the random variable W is given in the following table. w 2 7 a P( )==W w 0.3 0.3 0.1 24 0.3 Given that W a ) =, find a and evaluate WVar( E( ). M 7 The possible outcomes from a business venture are graded from 5 to 1, as shown in the following table. Grade 5 4 3 2 1 Outcome High profit Fair profit No loss Small loss Heavy loss Probability 0.24 0.33 0.24 0.11 0.08 a Calculate the expected grade and use it to describe the expected outcome of the venture. Find the standard deviation and explain what it gives a measure of in |
this case. b Investigate the expected outcome and the standard deviation when the grading is reversed (i.e. high profit is graded 1, and so on). Compare these outcomes with those from part a. 8 Two ordinary fair dice are rolled. The discrete random variable X is the lowest common multiple of the two numbers rolled. a Draw up the probability distribution table for X. b Find XE( ) and X P c Calculate XVar( [ ). > E( X ) ]. 159 9 In a game, a player attempts to hit a target by throwing three darts. With each throw, a player has a 30% chance of hitting the target. a Draw up the probability distribution table for H, the number of times the target is hit in a game. b How many times is the target expected to be hit in 1000 games? 10 Two students are randomly selected from a class of 12 girls and 18 boys. a Find the expected number of girls and the expected number of boys. b Write the ratio of the expected number of girls to the expected number of boys in simplified form. What do you notice about this ratio? c Calculate the variance of the number of girls selected. 11 A sewing basket contains eight reels of cotton: four are green, three are red and one is yellow. Three reels of cotton are randomly selected from the basket. a Show that the expected number of yellow cotton reels is 0.375. b Find the expected number of red cotton reels. c Hence, state the expected number of green cotton reels. 12 A company offers a $1000 cash loan to anyone earning a monthly salary of at least $2000. To secure the loan, the borrower signs a contract with a promise to repay the $1000 plus a fixed fee before 3 months have elapsed. Failure to do this gives the company a legal right to take $1540 from the borrower’s next salary before returning any amount that has been repaid. From past experience, the company predicts that 70% of borrowers succeed in repaying the loan plus the fixed fee before 3 months have elapsed. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 a Calculate the fixed fee that ensures the company an expected 40% profit from each |
$1000 loan. b Assuming that the company charges the fee found in part a, how would it be possible, without changing the loan conditions, for the company’s expected profit from each $1000 loan to be greater than 40%? 13 When a scout group of 8 juniors and 12 seniors meets on a Monday evening, one scout is randomly selected to hoist a flag. Let the variable X represent the number of juniors selected over n consecutive Monday evenings. a By drawing up the probability distribution table for X, or otherwise, show that XE( ) 1.2= when n 3=. b Find the number of Monday evenings over which 14 juniors are expected to be selected to hoist the flag. 14 An ordinary fair die is rolled. If the die shows an odd number then S, the score awarded, is equal to that number. If the die shows an even number, then the die is rolled again. If on the second roll it shows an odd number, then that is the score awarded. If the die shows an even number on the second roll, the score awarded is equal to half of that even number. a List the possible values of S and draw up a probability distribution table. [ P b Find S E S > ( ) ]. c Calculate the exact value of SVar( ). PS 15 A fair 4-sided spinner with sides labelled A B B B,,, is spun four times. a Show that there are six equally likely ways to obtain exactly two Bs with the four spins. 160 b By drawing up the probability distribution table for X, the number of times the spinner comes to rest on B, find the value of Var( X ) E( X ). c What, in the context of this question, does the value found in part b represent? EXPLORE 6.3 In this activity we will investigate a series of trials in which each can result in one of two possible outcomes. A ball is dropped into the top of the device shown in the diagram. When a ball hits a nail (which is shown as a red dot), there are two equally likely outcomes: it can fall to the left or it can fall to the right. Using L and R (to indicate left and right), list all the ways that a ball can fall into each of the cups A B, and C. Use your lists to tabulate the probabilities of a ball falling into each of the cups. Give all probabilities with denominator 4. A B C F |
AST FORWARD We will study the expectation of two special discrete random variables, and the variance of one of them, in Chapter 7, Sections 7.1 and 7.2. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6: Probability distributions The diagram shows a similar device with four cups labelled A to D. List all the ways that a ball can fall into each of the cups. Use your lists to tabulate the probabilities of a ball falling into each of the cups. Give all probabilities with denominator 8. A B C D Can you explain how and why the values probabilities in your tables? 2 11 2 and 3 11 2 are connected with the The next device in the sequence has 10 nails on four rows. Tabulate the probabilities of a ball falling into each of its five cups, A to E. FAST FORWARD We will study independent trials that have only two possible outcomes, such as left or right and success or failure, in Chapter 7, Sections 7.1 and 7.2. Checklist of learning and understanding ● A discrete random variable can take only certain values and those values occur in a certain random manner. ● A probability distribution for a discrete random variable is a display of all its possible values 161 and their corresponding probabilities. ● For the discrete random variable X : Σ = p 1 E( X ) = Σ xp VarE( X 2 )} Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 END-OF-CHAPTER REVIEW EXERCISE 6 1 Find the mean and the variance of the discrete random variable X, whose probability distribution is given in the [3] following table. x 1 2 3 4 P( )==X x 1 – k 2 – 3k 3 – 4k 4 – 6k 2 The following table shows the probability distribution for the random variable Y. y P( )==Y y 1 0.2 |
10 0.4 q 0.2 101 0.2 a Given that YVar( ) 1385.2 =, show that q 2 – 61 q + 624 = and solve this equation. 0 b Find the greatest possible value of YE( ). [4] [2] 3 An investment company has produced the following table, which shows the probabilities of various percentage profits on money invested over a period of 3 years. Profit ( )% 1 5 10 15 20 30 40 45 50 Probability 0.05 0.10 0.50 0.20 0.05 0.04 0.03 0.02 0.01 a Calculate the expected profit on an investment of $50 000. 162 b A woman considers investing $50 000 with the company, but decides that her money is likely to earn more when invested over the same period in a savings account that pays r% compound interest per annum. Calculate, correct to 2 decimal places, the least possible value of r. [3] [3] 4 A chef wishes to decorate each of four cupcakes with one randomly selected sweet. They choose the sweets at random from eight toffees, three chocolates and one jelly. Find the variance of the number of cupcakes that will [6] be decorated with a chocolate sweet. 5 The faces of a biased die are numbered 1, 2, 3, 4, 5 and 6. The random variable X is the score when the die is thrown. The probability distribution table for X is given. x P( )==.2 6 0.2 The die is thrown 3 times. Find the probability that the score is at least 4 on at least 1 of the 3 throws. [5] Cambridge international AS & A Level Mathematics 9709 Paper 61 Q2 June 2016 [Adapted] 6 A picnic basket contains five jars: one of marmalade, two of peanut butter and two of jam. A boy removes one jar at random from the basket and then his sister takes two jars, both selected at random. a Find the probability that the sister selects her jars from a basket that contains: i exactly one jar of jam ii exactly two jars of jam. [1] [1] b Draw up the probability distribution table for J, the number of jars of jam selected by the sister, and show that JE( ). 0.8= [4] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University |
Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 6: Probability distributions 7 Two ordinary fair dice are rolled. The product and the sum of the two numbers obtained are calculated. The score awarded, S, is equal to the absolute (i.e. non-negative) difference between the product and the sum. For example, if 5 and 3 are rolled, then S (5 3) × = − (5 3) + =. 7 a State the value of S when 1 and 4 are rolled. b Draw up a table showing the probability distribution for the 14 possible values of S, and use it to calculate SE( ). 8 A fair triangular spinner has sides labelled 0, 1 and 2, and another fair triangular spinner has sides labelled –1, 0 and 1. The score, X, is equal to the sum of the squares of the two numbers on which the spinners come to rest. a List the five possible values of X. b Draw up the probability distribution table for X. c Given that X 4<, find the probability that a score of 1 is obtained with at least one of the spinners. d Find the exact value of a, such that the standard deviation of X is 1 a × E( X ). 9 A discrete random variable X, where X {2, 3, 4, 5}, is such that X x P − 30 a Calculate the two possible values of b. b Hence, find P(2 < X <. 5) 10 Set A consists of the ten digits 0, 0, 0, 0, 0, 0, 2, 2, 2, 4. Set B consists of the seven digits 0, 0, 0, 0, 2, 2, 2. One digit is chosen at random from each set. The random variable X is defined as the sum of these two digits. i Show that XP( = 2) 3 =. 7 ii Tabulate the probability distribution of X. iii Find XE( ) and XVar( ). iv Given that X 2=, find the probability that the digit chosen from set A was 2. [1] [5] [1] [3] [2] [3] [3] [2] [2] [2] [3] [2] 163 Cambridge International AS & A Level Mathematics 9709 Paper 63 Q5 June 2010 PS PS 11 The discrete random variable |
Y is such that Y {4, 5, 8, 14, 17} ∈ )= and Y y P( is directly proportional to Find YP( 4)>. 12 X is a discrete random variable and X {0, 1, 2, 3}, find ∈. 0)> 0 or 2) øX 0.62 XP( P( 2 | X = =. Given that XP( > 1) = 0.24, P(0 < X < 3) = 0.5 and 1. 1+ y [4] [5] 13 Four students are to be selected at random from a group that consists of seven boys and x girls. The variables B and G are, respectively, the number of boys selected and the number of girls selected. a Given that B P( = 1) P( = B = 2), find the value of x. b Given that G 3≠, find the probability that G 4=. [3] [3] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 14 A box contains 2 green apples and 2 red apples. Apples are taken from the box, one at a time, without replacement. When both red apples have been taken, the process stops. The random variable X is the number of apples which have been taken when the process stops. 1 3 ii Draw up the probability distribution table for X. i Show that XP( [3] [3] 3) = =. Another box contains 2 yellow peppers and 5 orange peppers. Three peppers are taken from the box without replacement. iii Given that at least 2 of the peppers taken from the box are orange, find the probability that all 3 peppers are orange. [5] Cambridge International AS & A Level Mathematics 9709 Paper 63 Q7 November 2014 120 15 In a particular discrete probability distribution the random variable X takes the value r r 45, where r takes all integer values from 1 to 9 inclusive. i Show that XP( 1 15 ii Construct the probability distribution table for X. 40) = =. iii Which is the modal value of X? iv Find the probability that X lies between 18 and 100. with probability [2] [3] [1] |
[2] 164 Cambridge International AS & A Level Mathematics 9709 Paper 62 Q5 November 2009 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7 The binomial and geometric distributions 165 In this chapter you will learn how to: ■ use formulae for probabilities for the binomial and geometric distributions, and recognise ■ use formulae for the expectation and variance of the binomial distribution and for the expectation practical situations in which these distributions are suitable models of the geometric distribution. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PREREQUISITE KNOWLEDGE Where it comes from What you should be able to do Check your skills Chapter 4 Calculate expectation in a fixed number of repeated independent trials, given the probability that a particular event occurs. 1 Two ordinary fair dice are rolled 378 times. How many times can we expect the sum of the two numbers rolled to be greater than 8? IGCSE / O Level Mathematics Pure Mathematics 1 Expand products of algebraic expressions. Use the expansion of ( + ) where n is a positive integer. a + b n ab a b + 2 Given that ( 2 b a +, find the four fractions in the 3 and confirm 1 + 4 that their sum is equal to 1. expansion of 3 4 3 a 2 Two special discrete distributions Seen in very simple terms, all experiments have just two possible outcomes: success or failure. A business investment can make a profit or a loss; the defendant in a court case is found innocent or guilty; and a batter in a cricket match is either out or not! In most real-life situations, however, there are many possibilities between success and failure, but taking this yes/no view of the outcomes does allow us to describe certain situations using a mathematical model. 166 Two such situations concern discrete random variables that arise as a result of repeated independent trials, where the probability of success in each trial is constant. of independent trials. • A binomial distribution can be |
used to model the number of successes in a fixed number • A geometric distribution can be used to model the number of trials up to and including the first success in an infinite number of independent trials. 7.1 The binomial distribution Consider an experiment in which we roll four ordinary fair dice. In each independent trial, we can obtain zero, one, two, three or four 6s. Let the variable R be the number of 6s rolled, then ∈{0, 1, 2, 3, 4}. R To find the probability distribution for R, we must calculate P( possible values. )=R r for all of its Using 6 to represent a success and X to represent a failure in each trial, we have: P(success) P(6) = = 1 6 and P(failure) P(X) = = 5 6. Calculations to find P( R r are shown in the following table. ) = Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions r Ways to obtain r successes No. ways 0 (XXXX) 1 (6XXX), (X6XX), (XX6X), (XXX6) 2 3 (66XX), (6X6X), (6XX6), (X66X), (X6X6), (XX66) (666X), (66X6), (6X66), (X666) 4 0C = 1 4 1C = 4 4 2C = 6 4 3C = 4 4 4 =C 1 P( )== R r 4 C 0 1 6 0 4 5 6 3 1 1 6 5 6 4 C 1 4 C 2 1 6 4 C 3 1 6 4 C 4 � |
�� 1 6 2 3 4 2 1 0 5 6 5 6 5 6 4 Cr. These probabilities are the terms in the 4 (6666) In the table, we see that P( R r = ) = 1 + binomial expansion of 6 5 6 4. notation, the five probabilities shown in the previous table are given by Using the R r = P( = ) n r 4 discrete random variable that meets the following criteria is said to have a binomial distribution and it is defined by its two parameters, n and p. • There are n repeated independent trials. • n is finite. • There are just two possible outcomes for each trial (i.e. success or failure). • The probability of success in each trial, p, is constant. The random variable is the number of trials that result in a success. A discrete random variable, X, that has a binomial distribution is denoted by X ~ B(, n p. ) KEY POINT 7.1 If ~ B(, n p then the probability of r successes is X ) p r = n r r p (1 − p ) n r −. TIP Values of n r are the coefficients of the terms in a binomial expansion, and give the number of ways of obtaining r successes in n trials. r p (1 – ) – p n r is the probability for each way of obtaining r successes and ( – ) n r failures. For example, if the variable X ~ B(3, probabilities. p, then X {0, 1, 2, 3} ) ∈, and we have the following Copyright Material - Review Only - Not for Redistribution TIP Each way of obtaining a particular number of 6s has the same probability. T |
IP n is For work involving binomial expansions, the notation Cr rarely used nowadays. Your calculator may use this notation but it has mostly been 167 replaced by n r . REWIND We met a series of independent events with just two possible outcomes in the Explore 6.3 activity in Chapter 6, Section 6.3. TIP Coefficients for power 3 are 1, 3, 3 and 1. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 P( X = 0) = P( X = 2) = 3 0 × 3 2 × ( X = 1( X = 3) = 3 × 1 3 × 3 = The coefficients in all binomial expansions are symmetric strings of integers. When arranged in rows, they form what has come to be known as Pascal’s triangle (named after the French thinker Blaise Pascal). Part of this arrangement is shown in the following diagram, which includes the coefficient for power 0 for completeness. 1 2 6 1 3 1 4 11 3 1 4 10 10 15 20 15 power 0 1 2 3 4 5 6 WORKED EXAMPLE 7.1 168 A regular pentagonal spinner is shown. Find the probability that 10 spins produce exactly three As. Answer XP( = 3) = 10 3 × 10! × 7! 3! = = 3 0.4 7 0.6 × 3 0.4 × × 7 0.6 0.215 to 3 significant figures. Let the random variable X be the number of As obtained. We have 10 independent trials with a constant 0.4. P(A) probability of a success, So, and we require ~ B(10, 0.4) three successes and seven failures. X = WORKED EXAMPLE 7.2 |
~ B(8, 0.7), find P( X > 6), correct to 3 significant figures. Given that X Answer P( X > 6) P( = X = 7) P( + X = 8) X ~ B(8, 0.7) tells us that = = = 8 7 0.7 × 0.197650 8 1 0.3 + × 0.057648 … 8 7 × … + 0.7 8 0 0.3 × q p = = 8,, 0.3 0.7, n = and that. X {0, 1, 2, 3, 4, 5, 6, 7, 8} ∈ 0.255 Copyright Material - Review Only - Not for Redistribution REWIND We saw in Chapter 5, Section 5.3 that n r = n C r =! n!( r n r −. )! A B D C A TIP Remember that X represents the number of successes, so it can take integer values from 0 to n. TIP Premature rounding of probabilities in the working may lead to an incorrect final answer. Here, 0.198 + 0.0576 = 0.256. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions WORKED EXAMPLE 7.3 In a particular country, 85% of the population has rhesus-positive +(R ) blood. Find the probability that fewer than 39 people in a random sample of 40 have rhesus-positive blood. Answer P( X < 39) 1– [P( = X = 39) P( + X = 40)] 40 39 1 – × 1– 0.010604 [ 0.988 = = = 0.85 39 1 0.15 × + 40 40 × 0.85 40 |
0.15 0 × … + 0.001502 ] … Let the random variable X be the number in the sample with R+ blood, then X. ~ B(40, 0.85) REWIND Recall from Chapter 4, Section 4.1 that P( A ′ 1 P( − ) A ). = EXPLORE 7.1 Binomial distributions can be investigated using the Binomial Distribution resource on the GeoGebra website. We could, for example, check our answer to Worked example 7.3 as follows. 169 Click on the distribution tab and select binomial from the pop-up menu at the bottomleft. Select the parameters p = probability distribution will be generated., and a bar chart representing the and 0.85 n = 40 < 39), enter into the boxes To find XP( 38 ) and, by tapping the chart, P( 0 the value for this probability is displayed. (At the right-hand side you will see a list of the probabilities for the 41 possible values of X in this distribution.) ø øX WORKED EXAMPLE 7.4 Given that ~ B(, 0.4) X n and that XP( = 0) < 0.1, find the least possible value of n. TIP Answer P ( = X 0 ) = n 0 × 0 0.4 × n 0.6 n 0.6 = So we need 0.6 n < 0.1 n log 0.6 log 0.1 < log 0.6 log 0.1 < n n > log 0.1 log 0.6 n > 4.50... n = 5 is the least possible value of. n We first express P( in terms of n. X = 0) This leads to an inequality, which we can solve using base 10 logarithms. a>,. We must Recall from IGCSE / O Level that if x then x –< – a reverse the inequality sign when we multiply or divide by a negative number, such as 0.6 log. 10 TIP n takes integer values only. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review |
Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 3… = 0.6 0.216; 0.6 4 = 5 0.1296; 0.6 = 0.07776. The least possible value is =n 5. Alternatively, we can solve 0.6 n < 0.1 by trial and improvement. We know that n is an integer, so we 3 1 evaluate 0.6, 0.6, 0.6, 2 … up to the first one whose value is less than 0.1. EXERCISE 7A 1 The variable X has a binomial distribution with n = and 4 p = 0.2. Find: a P( X = 4) b P( X = 0) c P( X = 3) d P( X = 3 or 4). 2 Given that ~ B(7, 0.6) Y, find: a P( Y = 7) b P( Y = 5) c P( Y ≠ 4) d P(3 < <. 6) Y 3 Given that W ~ B(9, 0.32), find: a P( 5)W = b P( 5)W ≠ c WP( 2)< d WP(0 < 9) <. 170 4 Given that ~ B 8, V 2 7 , find: a P( V = 4) b ùVP( 7) c VP( 2)ø d P(3 ø, 6) V e P(V is an odd number). 5 Find the probability that each of the following events occur. a Exactly five heads are obtained when a fair coin is tossed nine times. b Exactly two 6s are obtained with 11 rolls of a fair die. 6 A man has five packets and each contains three brown sugar cubes and one white sugar cube. He randomly selects one cube from each packet. Find the probability that he selects exactly one brown sugar cube. 7 A driving test is passed by 70% of people at their first attempt. Find the probability that exactly five out of eight randomly selected people pass at their first attempt. 8 Research shows that the owners of 63% of all saloon cars are male. Find the probability that exactly 20 out of 30 randomly selected saloon cars are owned by: a males b females. 9 In a particular country, 58% of the adult population is married. Find the probability that exactly 12 out of 20 randomly selected adults |
are married. 10 A footballer has a 95% chance of scoring each penalty kick that she takes. Find the probability that she: a b scores from all of her next 10 penalty kicks fails to score from exactly one of her next seven penalty kicks. 11 On average, 13% of all tomato seeds of a particular variety fail to germinate within 10 days of planting. Find the probability that 34 or 35 out of 40 randomly selected seeds succeed in germinating within 10 days of planting. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions 12 There is a 15% chance of rain on any particular day during the next 14 days. Find the probability that, during the next 14 days, it rains on: a exactly 2 days b at most 2 days. 13 A factory makes electronic circuit boards and, on average, 0.3% of them have a minor fault. Find the probability that a random sample of 200 circuit boards contains: a exactly one with a minor fault b fewer than two with a minor fault. 14 There is a 50% chance that a six-year-old child drops an ice cream that they are eating. Ice creams are given to 5 six-year-old children. a Find the probability that exactly one ice cream is dropped. b 45 six-year-old children are divided into nine groups of five and each child is given an ice cream. Calculate the probability that exactly one of the children in at most one of the groups drops their ice cream. 15 A coin is biased such that heads is three times as likely as tails on each toss. The coin is tossed 12 times. The variables H and T are, respectively, the number of heads and the number of tails obtained. Find the value of P( P( H T = = 7) 7). 16 Given that ~ B(, 0.3) Q n and that P( Q = 0) 0.1 >, find the greatest possible value of n. 17 The variable ~ B(, 0.96) T n and it is given that P( T n= ) 0.5 >. Find the greatest possible value of n. 18 Given that ~ B(, 0.8) R n and that P( R n> − |
< 1) 0.006, find the least possible value of n. M 19 The number of damaged eggs, D, in cartons of six eggs have been recorded by an inspector at a packing depot. The following table shows the frequency distribution of some of the numbers of damaged eggs in 150 000 boxes. 171 No. damaged eggs ( )D 0 1 No. cartons ( )f 141 393 8396 The distribution of D is to be modelled by ~ B(6, D p. ) a Estimate a suitable value for p, correct to 4 decimal places. b Calculate estimates for the value of a and of b. c Calculate an estimate for the least number of additional cartons that would need to be inspected for there to be at least 8400 cartons containing one damaged egg. M 20 The number of months during the 4-month monsoon season (June to September) in which the total rainfall was greater than 5 metres, R, has been recorded at a location in Meghalaya for the past 32 years, and is shown in the following table. No. months No. years ( )f 0 2 1 8 2 12 3 8 4 2 The distribution of R is to be modelled by ~ B(4, R p. ) a Find the value of p, and state clearly what this value represents. b Give a reason why, in real life, it is unlikely that a binomial distribution could be used to model these data accurately. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 21 In a particular country, 90% of both females and males drink tea. Of those who drink tea, 40% of the females and 60% of the males drink it with sugar. Find the probability that in a random selection of two females and two males: a all four people drink tea b an equal number of females and males drink tea with sugar. PS 22 It is estimated that 0.5% of all left-handed people and 0.4% of all right-handed people suffer from some form of colour-blindness. A random sample of 200 left-handed and 300 right-handed people is taken. Find the probability that there is exactly one person in the sample that |
suffers from colour-blindness. DID YOU KNOW? Although Pascal’s triangle is named after the 17th century French thinker Blaise Pascal, it was known about in China and in Persia as early as the 11th century. The earliest surviving display is of Jia Xian’s triangle in a work compiled in 1261 by Yang Hui, as shown in the photo. 172 EXPLORE 7.2 A frog sits on the bottom-left square of a 5 by 5 grid. In each of the other 24 squares there is a lily pad and four of these have pink flowers growing from them, as shown in the image. 1 The frog can jump onto an adjacent lily pad but it can only jump northwards (N) or eastwards (E). The four numbers on the grid represent the number of different routes the frog can take to get to those particular lily pads. For example, there are three routes to the lily pad with the number 3, and these routes are EEN, ENE and NEE. 3 5 1 N Sketch a 5 by 5 grid and write onto it the number of routes to all 24 lily pads. Describe any patterns that you find in the numbers on your grid. The numbers on the lily pads with pink flowers form a sequence. Can you continue this sequence and find an expression for its nth term? Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions Expectation and variance of the binomial distribution Expectation and standard deviation give a measure of central tendency and a measure of variation for the binomial distribution. We can calculate these, along with the variance, from the parameters n and p. Consider the variable following table. X ~ B(2, 0.6), whose probability distribution is shown in the REWIND We saw in Chapter 6, Section 6.3 that expectation is a variable’s long-term average value.16 1 0.48 2 0.36 Applying the formulae for E( )X and Var X( ) gives the following results. E( X ) = Σ xp = (0 × 0.16) + (1 0.48) × + (2 × 0.36) 1.2 |
= 2 x p ) – {E( Var( X = Σ Our experiment consists of should not be surprised to find that E( = × n = trials with a probability of success 2 = × 0.6 1.2 = np.16) 0.48) (0 (2 X 0.36) – 1.2 p = 2 )} 2 2 (1 2 2 = 0.6 0.48 in each, so we REWIND We saw in Chapter 4, Section 4.1 that event A is expected to occur times. ) P( A× n What may be surprising (and a very convenient result), is that the variance of X also can be found from the values of the parameters n and p. Var( X ) = np (1 – ) p 2 = × 0.6 × 0.4 = 0.48 KEY POINT 7.2 TIP The mean and variance of X ~ B(, n p ) are given by npµ= and 2 σ = np (1 – ) p = npq. WORKED EXAMPLE 7.5 Given that X ~ B(12, 0.3), find the mean, the variance and the standard deviation of X. Answer Note that and 2 σ = µ= Var( )X E(. ) X 173 E( X ) np= 12 = × 3.6 0.3 = ) X Var( SD( X ) = = = = = = np 12 (1 – ) p 0.3 × × 2.52 0.7 TIP np (1 − p ) 2.52 1.59 to 3 significant figures Copyright Material - Review Only - Not for Redistribution We can also write our answers as 2 3.6, σ µ =. 1.59 σ = 2.52 = and Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 7.6 The random variable X ~ B(, n p. Given that E( ) a the value of n and of p b P( X =. 11) Answer a q = p = 7.5 12 1 – = 0.625 q = 0.375 n = 12 0.375 = 32 X = ) 12 and Var( |
) X = 7.5, find: We use q = npq np = ) Var( X ) E( X to find p. E( ) =X np, so =n ) E( X p 0.375 11 × 0.625 21 X ~ B(32, 0.375) 32 11 × 0.138 b P( X = 11) = = EXERCISE 7B 1 Calculate the expectation, variance and standard deviation of each of the following discrete random 174 variables. Give non-exact answers correct to 3 significant figures. a V ~ B(5, 0.2) b W ~ B(24, 0.55) c X ~ B(365, 0.18) d Y ~ B(20, 0.5 ) 2 Given that X ~ B(8,0.25), calculate: a XE( ) and XVar( ) b P[ =X E( X )] c P[ <X E( X )]. 3 Given that Y ~ B(11, 0.23), calculate: a ≠YP( 3) b P[ <Y E( Y )]. 4 Given that X ~ B(, n p ), E( X ) = 20 and =XVar( ) 12, find: a the value of n and of p b =XP( 21). 5 Given that G ~ B(, n p ), E( G ) = 24 1 2 and GVar( ) 10 5 = 24, find: a the parameters of the distribution of G b =GP( 20). 6 W has a binomial distribution, where =WE( ) 2.7 and =WVar( ) 0.27. Find the values of n and p and use them to draw up the probability distribution table for W. 7 Give a reason why a binomial distribution would not be a suitable model for the distribution of X in each of the following situations. a X is the height of the tallest person selected when three people are randomly chosen from a group of 10. b X is the number of girls selected when two children are chosen at random from a group containing one girl and three boys. c X is the number of motorbikes selected when four vehicles are randomly picked from a car park containing 134 cars, 17 buses and nine bicycles. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy |
- Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions, and its standard deviation is one-third of its mean. Calculate the non-zero value of 8 The variable Q n and find P(5 < ~ B, n <Q 1 3 8). 9 The random variable H ~ B(192, p p and find the value of k, given that ), and HE( 2) H P( = ) is 24 times the standard deviation of H. Calculate the value of k = × 2–379. 10 It is estimated that 1.3% of the matches produced at a factory are damaged in some way. A household box contains 462 matches. a Calculate the expected number of damaged matches in a household box. b Find the variance of the number of damaged matches and the variance of the number of undamaged matches in a household box. c Show that approximately 10.4% of the household boxes are expected to contain exactly eight damaged matches. d Calculate the probability that at least one from a sample of two household boxes contains exactly eight damaged matches. 11 On average, 8% of the candidates sitting an examination are awarded a merit. Groups of 50 candidates are selected at random. a How many candidates in each group are not expected to be awarded a merit? b Calculate the variance of the number of merits in the groups of 50. c Find the probability that: i ii three, four or five candidates in a group of 50 are awarded merits 175 three, four or five candidates in both of two groups of 50 are awarded merits. 7.2 The geometric distribution Consider a situation in which we are attempting to roll a 6 with an ordinary fair die. How likely are we to get our first 6 on the first roll; on the second roll; on the third roll, and so on? We can answer these questions using the constant probabilities of success and failure: p and 1 – p. P(first 6 on first roll) = → success. a p P(first 6 on second roll) = (1 – ) p p → failure followed by a success. a P(first 6 on third roll) = (1 – ) 2 →p p two failures followed by a success. The distribution of X, the number of trials up to and including the first success in |
a series of repeated independent trials, is a discrete random variable whose distribution is called a geometric distribution. The following table shows the probability that the first success occurs on the rth trial. r P( )== X r 1 p 2 3 4 (1 – ) p p (1 – )2 p p (1 – )3 p p........ n (1 – ) p 1 p n−........ Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 P( =X r The values of with first term p and common ratio to infinity of the GP. ) in the previous table are the terms of a geometric progression (GP) p1 –. The sum of the probabilities is equal to the sum ∑ [P( X r = )] = S ∞ = first term 1 common ratio − = p 1 (1 − − p ) = 1. The sum of the probabilities in a geometric probability distribution is equal to 1. A discrete random variable, X, is said to have a geometric distribution, and is defined by its parameter p, if it meets the following criteria. • The repeated trials are independent. • The repeated trials can be infinite in number. • There are just two possible outcomes for each trial (i.e. success or failure). • The probability of success in each trial, p, is constant. KEY POINT 7.3 A random variable X that has a geometric distribution is denoted by probability that the first success occurs on the rth trial is X ~ Geo( p, and the ) =p r (1 – ) –1 p p r for = r 1, 2, 3, … 176 The binomial and geometric distributions arise in very similar situations. The significant difference is that the number of trials in a binomial distribution is fixed from the start and the number of successes are counted, whereas, in a geometric distribution, trials are repeated as many times as necessary until the first success occurs. For X ~ B(, n p ), there are n r ways to obtain r successes. ~ Geo( For X is when there are r – 1 failures followed by a success. p ), |
there is only one way to obtain the first success on the rth trial, and that REWIND We saw in Chapter 6, Section 6.2 that 1 Σ =p for a probability distribution. You will also have seen geometric progressions and geometric series in Pure Mathematics 1, Chapter 6. TIP r q –1 =, p X r = An alternative form of this formula, P( ) × where p = 1 − q, reminds us that the – 1r failures occur before the first success. REWIND Recall from Section 7.1 that = n C!( )! WORKED EXAMPLE 7.7 Repeated independent trials are carried out in which the probability of success in each trial is 0.66. Correct to 3 significant figures, find the probability that the first success occurs: a on the third trial b on or before the second trial c after the third trial. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions X = 2) Let X represent the number of trials up to and including the first success, then X ~ Geo(0.66), 0.66 and where =p =p1 – 0.34. Answer a P( X = 3) = (.66 0.34 × 0.0763 b øP( X 2) P( = X 1) P( + = (1 – ) p p p + 0.884 = = c P( X > 3) 1 – P( = X ø 3) = = = 1 – [P( X = 1) P( + X = X = 3)] 2) P( + 2 (1 – ) ] p p (.0393 Probabilities that involve inequalities can be found by summation for small values of r, as in parts b and c of Worked example 7.7. However, for larger values of r, the following results will be useful. P( X rø = ) P(success on one of the first r trials) = 1 – P(failure on the first r trials) P( X r> ) P(first success after the rth trial) = P(failure on |
the first r trials) = These two results are written in terms of q in Key point 7.4. WORKED EXAMPLE 7.8 KEY POINT 7.4 ~ Geo( p ) p, then When and X 1 –=q =<P( X r • ) 1 – • P( )> X r = qr qr 177 In a particular country, 18% of adults wear contact lenses. Adults are randomly selected and interviewed one at a time. Find the probability that the first adult who wears contact lenses is: a one of the first 15 interviewed b not one of the first nine interviewed. Answer a øP( X 15) 1 – = q 15 = − 1 0.82 15 = 0.949 b P( X > 9) = 9 q 0.82 9 0.168 = = Let X represent the number of adults interviewed up to and including the first one who wears contact lenses, then X ~ Geo(0.18) and = 1 – 0.18 0.82. = q Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 7.9 A coin is biased such that the probability of obtaining heads with each toss is equal to. The coin is tossed until the first head is obtained. Find the probability that the 5 11 coin is tossed: a at least six times b fewer than eight times. Answer a ùP( X 6) P( = X > 5) 5 q = 5 6 = 11 0.0483 = b P( X < 8) P( = X ø 7 11 = − 1 0.986 = 178 EXERCISE 7C Let X represent the number of times the coin is tossed up to and including the first heads, then 5 11 and X ~ Geo 6. 11 =q TIP ‘At least six times’ has the same meaning as ‘more than five times’. TIP ‘Fewer than eight times’ has the same meaning as ‘seven or fewer times’. 1 Given the discrete |
random variable X ~ Geo(0.2), find: a =XP( 7) b P( X ≠ 5) c >XP( 4). 2 Given that T ~ Geo(0.32), find: a =TP( 3) b øTP( 6) c >TP( 7). 3 The probability that Mike is shown a yellow card in any football match that he plays is probability that Mike is next shown a yellow card: 1 2. Find the a in the third match that he plays b before the fourth match that he plays. 4 On average, Diya concedes one penalty in every six hockey matches that she plays. Find the probability that Diya next concedes a penalty: a in the eighth match that she plays b after the fourth match that she plays. 5 The sides of a fair 5-sided spinner are marked 1, 1, 2, 3 and 4. It is spun until the first score of 1 is obtained. Find the probability that it is spun: a exactly twice b at most five times c at least eight times. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions 6 It is known that 80% of the customers at a DIY store own a discount card. Customers queuing at a checkout are asked if they own a discount card. a Find the probability that the first customer who owns a discount card is: i the third customer asked ii not one of the first four customers asked. b Given that 10% of the customers with discount cards forget to bring them to the store, find the probability that the first customer who owns a discount card and remembered to bring it to the store is the second customer asked. 7 In a manufacturing process, the probability that an item is faulty is 0.07. Items from those produced are selected at random and tested. a Find the probability that the first faulty item is: i the 12th item tested ii not one of the first 10 items tested iii one of the first eight items tested. b What assumptions have you made about the occurrence of faults in the items so that you can calculate the probabilities in part a? 8 Two independent random variables are X ~ Geo(0.3) and Y ~ Geo(0.7). Find: a =XP( |
2) b =YP( 2) c P( X = 1 and Y = 1). 9 On average, 14% of the vehicles being driven along a stretch of road are heavy goods vehicles (HGVs). A girl stands on a footbridge above the road and counts the number of vehicles, up to and including the first HGV that passes. Find the probability that she counts: a at most three vehicles b at least five vehicles. 179 10 The probability that a woman can connect to her home Wi-Fi at each attempt is 0.44. Find the probability that she fails to connect until her fifth attempt. 11 Decide whether or not it would be appropriate to model the distribution of X by a geometric distribution in the following situations. In those cases for which it is not appropriate, give a reason. a A bag contains two red sweets and many more green sweets. A child selects a sweet at random and eats it, selects another and eats it, and so on. X is the number of sweets selected and eaten, up to and including the first red sweet. b A monkey sits in front of a laptop with a blank word processing document on its screen. X is the number of keys pressed by the monkey, up to and including the first key pressed that completes a row of three letters that form a meaningful three-letter word. c X is the number of times that a grain of rice is dropped from a height of 2 metres onto a chessboard, up to and including the first time that it comes to rest on a white square. d X is the number of races in which an athlete competes during a year, up to and including the first race that he wins. 12 The random variable T has a geometric distribution and it is given that P( P( T T = = 2) 5) = 15.625. Find =TP( 3). PS PS 13 X ~ Geo( p ) and XP( = 2) = 0.2464. Given that <p 0.5, find >XP( 3). 14 Given that X ~ Geo( p ) and that øXP( 4) = 2385 2401, find P(1 ø X 4)<. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level |
Mathematics: Probability & Statistics 1 PS 15 Two ordinary fair dice are rolled simultaneously. Find the probability of obtaining: a b the first double on the fourth roll the first pair of numbers with a sum of more than 10 before the 10th roll. PS 16 X ~ Geo(0.24) and Y ~ Geo(0.25) are two independent random variables. Find the probability that Mode of the geometric distribution All geometric distributions have two features in common. These are clear to see when bar charts or vertical line graphs are used to represent values of ) for different values of the parameter p. You can do this manually or using a graphing tool such as GeoGebra. =X r P( X Y 4. + = =XP( The first common feature is that 1) has the greatest probability in all geometric distributions. This means that the most likely value of X is 1, so the first success is most likely to occur on the first trial. Secondly, the value of =X r ( This is because the common ratio between the probabilities q (1 – ) p p p ) decreases as r increases. = is less than 1: (1 – ) p p (1 – ) p p (( 3 2 4. > … > > > > The following table shows some probabilities for the distributions X ~ Geo(0.2) and X ~ Geo(0.7). In both distributions, we can see that probabilities decrease as the value of X increases. P( X == 1) P( X == 2) P( X == 3) P( X == 4) P( X == 5) Geo(0.2) 180 Geo(0.7) 0.2 0.7 0.16 0.21 0.128 0.063 0.1024 0.0189 0.08192 0.00567 Expectation of the geometric distribution Recall that the expectation or mean of a discrete random variable is its long-term average, which is given by E( X ). µ= = Σ xpx Applying this to the geometric distribution Geo(p), it turns out we find that the mean is equal to 1 p, the reciprocal of p. EXPLORE 7.3 KEY POINT 7.5 The mode of all geometric distributions is 1. KEY POINT 7.6 p then ) If X ~ Geo( 1 p µ =. Using algebra, we can prove that the mean of the geometric distribution is equal to For X ~ Geo(p), |
we have X {1, 2, 3, 4,...} and ∈ =p x {,, p pq pq 2, 3 pq,...}. Step 1 of the proof is to form an equation that expresses µ in terms of p and q. To do this we use µ = Σ xpx. 1 p. REWIND We studied the expectation of a discrete random variable in Chapter 6, Section 6.3. There are three more steps required to complete the proof, which you might like to try without any further assistance. However, some guidance is given below if needed. Step 2: Multiply the equation obtained in step 1 throughout by q to obtain a second equation. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions Step 3: Subtract one equation from the other. Step 4: If you have successfully managed steps 1, 2 and 3, you should need no help completing the proof! WORKED EXAMPLE 7.10 One in four boxes of Zingo breakfast cereal contains a free toy. Let the random variable X be the number of boxes that a child opens, up to and including the one in which they find their first toy. a Find the mode and the expectation of X. b Interpret the two values found in part a in the context of this question. Answer a The mode of X is 1. E child is most likely to find their first toy in the first box they open but, on average, a child will find their first toy in the fourth box that they open. WORKED EXAMPLE 7.11 The variable is X ~ Geo 1 4 . TIP An answer written ‘in context’ must refer to a specific situation; in this case, the situation described in the question. 181 The variable X follows a geometric distribution. Given that X = E( ) 3 1 2, find >XP( 6). Answer E( X ) = 1 p = 7 2, so p = 2 7 q = 1 – P( X > 2 7 6.133 = We find the parameter p and then we find q. We use P( X r > ) = qr from Key point 7.4. |
Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 7.12 Given that X ~ Geo( p ) and that øXP( 3) = 819 1331, find: a >XP( 3) b P(1 < ø X 3). Answer a P( X > 3) 1 P( = − ø 3) X 819 1331 1 = − = 512 1331 b 1 – q 3 = P( X ø 3 11 819 1331 819 1331 − and p = 3 11 r 1 – We use q find q and p. P( < X r ) to = TIP 182 P(1 ø< X 3) P( = X = 2) P( + X = 3) = = pq + 456 1331 2 pq or 0.343 = = = Alternatively, we can use < < P(1 3) X < − 3 11 P( X 819 1331 456 1331 3) P( − X = 1) EXERCISE 7D 1 Given that X ~ Geo(0.36), find the exact value of XE( ). 2 The random variable Y follows a geometric distribution. Given that YP( = 1) = 0.2, find YE( ). 3 Given that S ~ Geo( p ) and that SE( ) 4 1 2=, find =SP( 2). 4 Let T be the number of times that a fair coin is tossed, up to and including the toss on which the first tail is obtained. Find the mode and the mean of T. 5 Let X be the number of times an ordinary fair die is rolled, up to and including the roll on which the first 6 is obtained. Find XE( ) and evaluate P[ >X E( X )]. 6 A biased 4-sided die is numbered 1, 3, 5 and 7. The probability of obtaining each score is proportional to that score. a Find the expected number of times that the die will be rolled, up to and including the roll on which the first non-prime number is obtained. b Find the probability that the first prime number is obtained on the third roll |
of the die. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions 7 Sylvie and Thierry are members of a choir. The probabilities that they can sing a perfect high C note on each attempt are and, respectively. 4 7 5 8 a Who is expected to fail fewer times before singing a high C note for the first time? b Find the probability that both Sylvie and Thierry succeed in singing a high C note on their second attempts. PS 8 A standard deck of 52 playing cards has an equal number of hearts, spades, clubs and diamonds. A deck is shuffled and a card is randomly selected. Let X be the number of cards selected, up to and including the first diamond. a Given that X follows a geometric distribution, describe the way in which the cards are selected, and give the reason for your answer. b Find the probability that: i X is equal to XE( ) ii neither of the first two cards selected is a heart and the first diamond is the third card selected. PS 9 A study reports that a particular gene in 0.2% of all people is defective. X is the number of randomly selected people, up to and including the first person that has this defective gene. Given that X b >øP( XE( ) and find the smallest possible value of b. ) 0.865, find PS 10 Anouar and Zane play a game in which they take turns at tossing a fair coin. The first person to toss heads is the winner. Anouar tosses the coin first, and the probability that he wins the game is 1 0.5 5 0.5 0.5 0. Describe the sequence of results represented by the value 0.55 in this series. b Find, in a similar form, the probability that Zane wins the game. c Find the probability that Anouar wins the game. EXPLORE 7.4 In a game for two people that cannot be drawn, you are the stronger player with a 60% chance of winning each game. The probability distributions for the number of games won by you and those won by your opponent when a single game is played, X and Y, are shown. x P( X x== |
) 0 0.40 1 0.60 y 0 P( = ) Y y 0.60 1 0.40 Investigate the probability distributions for X and Y in a best-of-three contest, where the first player to win two games wins the contest. Who gains the advantage as the number of games played in a contest increases? What evidence do you have to support your answer? PS How likely are you to win a best-of-five contest? Copyright Material - Review Only - Not for Redistribution 183 Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Checklist of learning and understanding ● A binomial distribution can be used to model the number of successes in a series of n repeated independent trials where the probability of success on each trial, p, is constant. If X ~ B(, n p ) then p r = E( X ) µ= = np n r r p (1 – ) –. n r p Var( X ) 2 σ= = np (1 – ) p = npq, where geometric distribution can be used to model the number of trials up to and including the first success in a series of repeated independent trials where the probability of success on each trial, p, is constant. ● ● If X ~ Geo( XE( ) µ= = ) then p 1 p =p r (1 – ) –1 for r = p p r 1, 2, 3, … ● P X r =ø( ) 1 – qr and P( X r > ) = qr, where =q p1 –. ● The mode of all geometric distributions is 1. 184 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 7: The binomial and geometric distributions END-OF-CHAPTER REVIEW EXERCISE 7 1 Given that, ~ X B n 1 n find an expression for =XP( 1) in terms of n. |
2 A family has booked a long holiday in Skragness, where the probability of rain on any particular day is 0.3. Find the probability that: a b the first day of rain is on the third day of their holiday it does not rain for the first 2 weeks of their holiday. [2] [1] [2] 3 One plastic robot is given away free inside each packet of a certain brand of biscuits. There are four colours of plastic robot (red, yellow, blue and green) and each colour is equally likely to occur. Nick buys some packets of these biscuits. Find the probability that i he gets a green robot on opening his first packet, ii he gets his first green robot on opening his fifth packet. Nick’s friend Amos is also collecting robots. iii Find the probability that the first four packets Amos opens all contain different coloured robots. [1] [2] [3] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q3 November 2015 4 Weiqi has two fair triangular spinners. The sides of one spinner are labelled 1, 2, 3, and the sides of the other are labelled 2, 3, 4. Weiqi spins them simultaneously and notes the two numbers on which they come to rest. a Find the probability that these two numbers differ by 1. b Weiqi spins both spinners simultaneously on 15 occasions. Find the probability that the numbers on which they come to rest do not differ by 1 on exactly eight or nine of the 15 occasions. 5 A computer generates random numbers using any of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The numbers appear on the screen in blocks of five digits, such as 50119 26317 40068....... Find the probability that: a there are no 7s in the first block b the first zero appears in the first block c the first 9 appears in the second block. 6 Four ordinary fair dice are rolled. a In how many ways can the four numbers obtained have a sum of 22? b Find the probability that the four numbers obtained have a sum of 22. c The four dice are rolled on eight occasions. Find the probability that the four numbers obtained have a sum of 22 on at least two of these occasions. 185 [2] [3] [1] [1] [2] [2] [2] [3] 7 When a certain driver parks their car in the evenings, they are equally likely to |
remember or to forget to switch off the headlights. Giving your answers in their simplest index form, find the probability that on the next 16 occasions that they park their car in the evening, they forget to switch off the headlights: a 14 more times than they remember to switch them off b at least 12 more times than they remember to switch them off. [2] [3] M 8 Gina has been observing students at a university. Her data indicate that 60% of the males and 70% of the females are wearing earphones at any given time. She decides to interview randomly selected students and to interview males and females alternately. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 a Use Gina’s observation data to find the probability that the first person not wearing earphones is the third male interviewed, given that she first interviews: i a male ii a female iii a male who is wearing earphones. b State any assumptions made about the wearing of earphones in your calculations for part a. 9 In Restaurant Bijoux 13% of customers rated the food as ‘poor’, 22% of customers rated the food as ‘satisfactory’ and 65% rated it as ‘good’. A random sample of 12 customers who went for a meal at Restaurant Bijoux was taken. i Find the probability that more than 2 and fewer than 12 of them rated the food as ‘good’. On a separate occasion, a random sample of n customers who went for a meal at the restaurant was taken. ii Find the smallest value of n for which the probability that at least 1 person will rate the food as ‘poor’ is greater than 0.95. [2] [2] [2] [1] [3] [3] 10 A biased coin is four times as likely to land heads up compared with tails up. The coin is tossed k times so that Cambridge International AS & A Level Mathematics 9709 Paper 62 Q3 June 2012 the probability that it lands tails up on at least one occasion is greater than 99%. Find the least possible value of k. 186 PS 11 Given that X 1) k = ×. find the smallest value of n for which k |
25> ~ B(, 0.4) and that X P( = n P( X n = – 1), express the constant k in terms of n, and PS 12 A book publisher has noted that, on average, one page in eight contains at least one spelling error, one page in five contains at least one punctuation error, and that these errors occur independently and at random. The publisher checks 480 randomly selected pages from various books for errors. a How many pages are expected to contain at least one of both types of error? b Find the probability that: i ii the first spelling error occurs after the 10th page the first punctuation error occurs before the 10th page iii the 10th page is the first to contain both types of error. 13 Robert uses his calculator to generate 5 random integers between 1 and 9 inclusive. i Find the probability that at least 2 of the 5 integers are less than or equal to 4. Robert now generates n random integers between 1 and 9 inclusive. The random variable X is the number of these n integers which are less than or equal to a certain integer k between 1 and 9 inclusive. It is given that the mean of X is 96 and the variance of X is 32. ii Find the values of n and k. [4] [5] [2] [2] [2] [2] [3] [4] PS 14 Anna, Bel and Chai take turns, in that order, at rolling an ordinary fair die. The first person to roll a 6 wins the game. Cambridge International AS & A Level Mathematics 9709 Paper 62 Q4 June 2013 Find the ratio P(Anna wins) : P(Bel wins) : P(Chai wins), giving your answer in its simplest form. [7] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8 The normal distribution In this chapter you will learn how to: 187 distribution tables ■ sketch normal curves to illustrate distributions or probabilities ■ use a normal distribution to model a continuous random variable and use normal ■ solve problems concerning a normally distributed variable ■ recognise conditions under which the normal distribution can be used as an approximation to the binomial distribution, and use this approximation, with a continuity correction, in solving problems. Copyright Material - Review Only - Not for Redistribution |
Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PREREQUISITE KNOWLEDGE Where it comes from What you should be able to do Check your skills Chapter 7 Find and calculate with the expectation and variance of a binomial distribution. 1 Given XVar( X ~ B 45, 0.52, find XE( ). ( ) ) and 2 Given that X follows a binomial distribution with XE( Var( the distribution of X. 7.28 X = ) ) 11.2, find the parameters of and = Why are errors quite normal? If you study any of the sciences, you will be required at some time to measure a quantity as part of an experiment. That quantity could be a measurement of time, mass, distance, volume and so on. Whatever it is, any measurement you make of a continuous quantity such as these will be subject to error. The very nature of continuous quantities means that they cannot be measured precisely and, no matter how hard we try, inaccuracy is also likely because our tools lack perfect calibration and we, as human beings, add in a certain amount of unreliability. 188 However, small errors are more likely than large errors and our measurements are usually just as likely to be underestimates as overestimates. When repeated measurements are taken, errors are likely to cancel each other out, so the average error is close to zero and the average of the measurements is virtually error-free. This chapter serves as an introduction to the idea of a continuous random variable and the method used to display its probability distribution. We will later focus our attention on one particular type of continuous random variable, namely a normal random variable. The normal distribution was discovered in the late 18th century by the German mathematician Carl Friedrich Gauss through research into the measurement errors made in astronomical observations. Some key properties of the normal distribution are that values close to the average are most likely; the further values are from the average, the less likely they are to occur, and the distribution is symmetrical about the average. 8.1 Continuous random variables A continuous random variable is a quantity that is liable to change and whose infinite number of possible values are the numerical outcomes of a random phenomenon. Examples include the amount of sugar in an orange, the time required to run a marathon, measurements of height |
and temperature and so on. A continuous random variable is not defined for specific values. Instead, it is defined over an interval of values. Consider the mass of an apple, denoted by X grams. Within the range of possible masses, X can take any value, such as 111.2233…, or 137.8642…, or 145.2897…, or …. The probability that X takes a particular value is necessarily equal to 0, since the number of values that it can take is infinite. However, there will be a countable number of values in any chosen interval, such as < <X be found. 140, so a probability for each and every interval can 130 The probability distribution of a discrete random variable shows its specific values and their probabilities, as we saw in Chapters 6 and 7. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy The probability distribution of a continuous random variable shows its range of values and the probabilities for intervals within that range. Chapter 8: The normal distribution ● When X is a discrete random variable, we can represent X r ● When X is a continuous random variable, we can represent P( )=. ø ø a X b. ) P( Before looking at probability distributions for continuous random variables in detail, we will consider how we can represent the probability distribution of a set of collected or observed continuous data. Representation of a probability distribution A set of continuous data can be illustrated in a histogram, where column areas are proportional to frequencies. To illustrate the probability distribution of a set of data, we draw a graph that is based on the shape of a histogram, as we now describe. (relative frequency density If we change the frequency density values on the vertical axis to relative frequency density relative frequency values will represent relative frequencies, which are estimates of probabilities. The vertical axis of the diagram can now be labeled ‘probability density’. class width) then column areas = ÷ For equal-width class intervals, the process described above has no effect on the ‘shape’ of fΣ ’ to 1, which the diagram. The result is that the total area of the columns changes from ‘ is the sum of the probabilities of all the possible values. So we can draw a curved graph over |
the columns of an equal-width interval histogram (preferably one displaying large amounts of data with many classes) to model the probability distribution of a set of continuous data. In the case of a random variable, such a curved graph represents a function, = f( and is called a probability density function, abbreviated to PDF or pdf. The area under the graph of the PDF is also equal to 1. x, ) y A curved graph is sketched over each of the histograms in the diagram below area under curve = 1 y = f(x area under curve = 1 y = f(x) x If you were asked to describe these two curves, you may be tempted to say that the curve on the right is ‘a bit odd’ and that the curve on the left is ‘a bit more normal’… and you would be quite right in doing so, as you will see shortly. Copyright Material - Review Only - Not for Redistribution REWIND We saw how to display continuous data in a histogram in Chapter 1, Section 1.3. 189 TIP The word function should only be used when referring to a random variable. For data, we should rather use curve and/or graph. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy TIP The mode is located at the graph’s peak. The median is at the value where the area under the graph is divided into two equal parts; this value can be found by calculation from the histogram or estimated from a cumulative frequency graph. Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Three commonly occurring types of curved graph are shown in the diagrams below. negatively skewed symmetric positively skewed longer tail to the left even tails longer tail to the right EXPLORE 8.1 Three frequency distributions are shown in the tables below. Use a histogram to sketch a graph representing the probability distribution for each of w, x and y. w f x f y f w3 ø < 6 6 9<ø w 9 12<ø w 12 15<ø w 15 18<ø w 18 21<ø w 21 24<ø w 13 13 13 13 13 13 13 3 6<ø x 6 9<ø x 9 12<ø x 12 15<ø x 15 18<ø x 18 21<ø x 21 |
24<ø x 3 9 18 24 18 9 3 3 6<ø y 6 9<ø y 9 12<ø y 12 15<ø y 15 18<ø y 18 21<ø y 21 24<ø y 8 19 10 4 10 19 8 Discuss and describe the shapes of the three graphs. What feature do they have in common? Compare the measures of central tendency (averages) for w, x and y. 190 The normal curve The frequency distribution of x in the Explore 8.1 activity produces a special type of curved graph. It is a symmetric, bell-shaped curve, known as a normal curve. If a probability distribution is represented by a normal curve, then: ● Mean median mode = = ● The peak of the curve is at the mean ( )µ, and this is where we find the curve’s line of symmetry ● Probability density decreases as we move away from the mean on both sides, so the further the values are from the mean, the less likely they are to occur ● An increase in the standard deviation σ( ) means that values become more spread out from the mean. This results in the curve’s width increasing and its height decreasing, so that the area under the graph is kept at a constant value of 1. Graphs that represent probability distributions of related sets of data, such as the heights of the boys and the heights of the girls at your school, can be represented on the same diagram, so that comparisons can be made. The following diagram shows two pairs of normal curves with their means and standard deviations compared. Note that the areas under the graphs in each pair are equal. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution As we can see, A and B have the same mean, but the shapes of the normal curves are different because they do not have the same standard deviation. Curve B is obtained from curve A by stretching it both vertically (from the horizontal axis) and horizontally (from the line of symmetry). X and Y have identically-shaped normal curves because they have the same standard deviation, but their positions or locations are different because they have different means. Each curve can be obtained from the other by a horizontal translation. EXPLORE 8.2 191 You can |
investigate the effect of altering the mean and/or standard deviation on the location and shape of a normal curve by visiting the Density Curve of Normal Distribution resource on the GeoGebra website. Note that the area under the curve is always equal to 1, whatever the values of µ and σ. EXERCISE 8A 1 The probability distributions for A and B are represented in the diagram. Indicate whether each of the following statements is true or false. a µ µ>Α Β b σ σ<Α B c A and B have the same range of values. d σ 2 2 σ=Α Β e At least half of the values in B are greater than µΑ. f At most half of the values in A are less than µΒ Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 2 The diagram shows normal curves for the probability distributions of P and Q, that each contain n values. a Write down a statement comparing: i σP and σQ ii iii the median value for P and the median value for Q the interquartile range for P and the interquartile range for Q The datasets P and Q are merged to form a new dataset denoted by W. i Describe the range of W. ii Is the probability distribution for W a normal curve? Explain your answer. iii Copy the diagram above and sketch onto it a curved graph representing the probability distribution for W. Mark the relative positions of µP, µQ and µW along the horizontal axis. 192 3 The distributions of the heights of 1000 women and of 1000 men both produce normal curves, as shown. The mean height of the women is 160 cm and the mean height of the men is 180 cm. The heights of these women and men are now combined to form a new set of data. Assuming that the combined heights also produce a normal curve, copy the graph opposite and sketch onto it the curve for the combined heights of the 2000 women and men. 4 Probability distributions for the quantity of apple juice in 500 apple juice tins and for the quantity of peach juice in 500 peach juice tins are both represented by normal curves. The mean quantity of apple juice is 340 ml with |
variance 4 ml2, and the mean quantity of peach juice is 340 ml with standard deviation 4 ml. a Copy the diagram and sketch onto it the normal curve for the quantity of peach juice in the peach juice tins. b Describe the curves’ differences and similarities women men 160 180 Height (cm) apple juice 340 Volume (ml) 5 The masses of 444 newborn babies in the USA and 888 newborn babies in the UK both produce normal curves. For the USA babies, µ = 3.4 kg and σ = 200 g; for the UK babies, µ = 3.3 kg and σ = 36100 g 2. 2 a On a single diagram, sketch and label these two normal curves. b Describe the curves’ differences and similarities. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution 6 The values in two datasets, whose probability distributions are both normal curves, are summarised by the following totals: 2 =x Σ 2 =y Σ 35 000, Σ =x 12 000 and n = 5000. 72 000, Σ =y 26 000 and =n 10 000. a Show that the centre of the curve for y is located to the right of the curve for x. b On the same diagram, sketch a normal curve for each dataset. 8.2 The normal distribution In Section 8.1, we saw how a curved graph can be used to represent the probability distribution of a set of continuous data. A curved graph that represents the probability distribution of a continuous random variable, as stated previously, is called a probability density function or PDF. If we collect data on, say, the masses of a randomly selected sample of 1000 pineapples, we can produce a curved graph to illustrate the probabilities for the full and limited range of these masses. If there are no pineapples with masses under 0.2 kg or over 6 kg, then our 0 and graph will indicate that P(mass P(mass 0.2) 6) 0. > = = < However, the continuous random variable ‘the possible mass of a pineapple’ is a theoretical model for the probability distribution. In the model, masses of less than 0.2 kg and masses of more than 6 kg would be shown to be |
extremely unlikely, but not impossible. 0 and The continuous random variable would, therefore, indicate that P(mass P(mass 0.2) 6) 0. < > > > [Incidentally, the greatest ever recorded mass of a pineapple is 8.28 kg!] The probability distribution of a continuous random variable is a mathematical function that provides a method of determining probabilities for the occurrence of different outcomes or observations. If the random variable X is normally distributed with mean µ and variance σ2, then its equation is TIP 193 f( x ) = 1 2 π σ exp { − ( 2 x − ) µ 2 2 σ, for all real values of x. } The parameters that define a normally distributed random variable are its mean µ and its variance σ2. To describe the normally distributed random variable X, we write X ~ N, µ σ ) ( 2. KEY POINT 8.1 X ~ N, ( 2µ σ ) describes a normally distributed random variable. We read this as ‘X has a normal distribution with mean µ and variance σ2’ The probability that X takes a value between a and b is equal to the area under the curve between the x-axis and the boundary lines =x a and =x b. b The area under the graph of = f( y x can be found by integration: ) P( ø ø a X b ) f( x ) d x ∫= a Copyright Material - Review Only - Not for Redistribution exp { } means the number e 2.71828 = raised to the power in the bracket, and pe for any power p. …, 0> TIP The area under any part of the curve is the same, whether or not the boundary values are included. ø ø, 7) P(3 X ø < P(3 7) X, 7)ø< P(3 X and P(3 are 7) X < < indistinguishable. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Unfortunately, it is not possible to perform this integration accurately but, as we will see later, mathematicians have found ways to handle this challenge. Normal distributions have many interesting properties, some of which are detailed in the following |
table. Properties Half of the values are less than the mean. Half of the values are greater than the mean. Approximately 68.26% of the values lie within 1 standard deviation of the mean. Approximately 95.44% of the values lie within 2 standard deviations of the mean. Approximately 99.72% of the values lie within 3 standard deviations of the mean. Probabilities P( P( P.5 = 0.5 P( – µ σ ø ø X µ σ+ ) = 0.6826 P+ = 0.9544 P+ = 0.9972 In the following diagrams, the values 0, ±1 and ±2 represent numbers of standard deviations from the mean. 0.6826 0.8413 0.9544 0.9772 194 –1 –σμ 0 1 +σμμ 10 +σμμ –2 – 2σμ 0 2 + 2σμμ –2 – 2σμ 0 μ We can use the curve’s symmetry, along with the table and diagrams above, to find other probabilities, such as: We know that P(–1 ø ø X 1) = 0.6826, so TIP The probability that the values in a normal distribution lie within a certain number of standard deviations of the mean is fixed. 0.6826 + 0.5 ) ø ø = ) 0.5 2) X + 0.9544 P( X ø 1) (= 1 2 × We know that P(–2 P( X ø 2) (= 1 2 × EXPLORE 8.3 = 0.8413 = P( X ù − 1). 0.9544, so = 0.9772 = P( X ù − 2). Calculated estimates of the mean and variance of the continuous random variables A, B and C are given in the following table. Mean Variance A 40 64 B 72 144 C 123 121 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution Random observations from each distribution were made with the following results: For A: 8060 out of 13 120 observations lie in the interval from 32 to 48. For B: 8475 out of 12 420 observations lie in the |
interval from 60 to 84. For C: 8013 out of 10 974 observations lie in the interval from 112 to 134. Investigate this information (using the previous table showing properties and probabilities for normal distributions) and comment on the statement ‘The distributions of A, B and C are all normal’. The standard normal variable Z There are clearly an infinite number of values for the parameters of a normally distributed random variable. Nevertheless, most problems can be solved by transforming the random variable into a standard normal variable, which is denoted by Z, and which has mean 0 and variance 1. 0µ = and σ = 1 By substituting can find the equation of the PDF for Z ~ N(0, 1). This is denoted by φ( )z and its equation is into the equation for the normal distribution PDF, we z is shown below.. The graph of exp ( ) z y = φ( ) − = φ 2 1 2 π { 2z 2 } 0.4 0.2 y = ϕ(z) –3 –2 –1 0 1 2 3 z The mean of Z is 0. The axis of symmetry is a vertical line through the mean, as with every normal distribution. Z has a variance of 1 and, therefore, a standard deviation of 1., ±2 and ±3 represent values that are 1, 2 and 3 standard deviations above or below = ±1 z the mean. Any < 0 z represents a value that is less the mean. Any > 0 z represents a value that is greater the mean. For > 3 z and for < −3 z, φ ( ) z ≈ 0. The area under the graph of = φ( ) y z is equal to 1. A vertical line drawn at any value of Z divides the area under the curve into two parts: one representing Z z )ø and the other representing Z z. ) > P( P( FAST FORWARD Later in this section, we will see how any normal variable can be transformed to the standard normal variable by coding. KEY POINT 8.2 The standard normal (0, 1) variable is Z N~ 195 TIP φ and Φ are the lower and upper-case Greek letter phi. The value of Z z values by integration. Tables showing the value of )ø is denoted by P( z( )Φ and, as mentioned earlier, we do not find such z( )Φ for different |
values of z have been Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 compiled and appear in the Standard normal distribution function table at the end of the book. In addition, some modern calculators are able to give the value of inverse function z( )Φ and the directly. z( ) 1Φ− 0ù ) appear in the tables, the graph’s Although only zero and positive values of Z (i.e. z symmetry allows us to use the tables for positive and for negative values of z, as you will see after Worked example 8.2. = 2.999 Values of the standard normal variable appear as 4-figure numbers from = 0.000 in the tables. The first and second figures of z appear in the left-hand column; z the third and fourth figures appear in the top row. The numbers in the ‘ADD’ column for the fourth figures indicate what we should add to the value of z( )Φ in the body of the table. to z zΦ( ) zΦ( ) can be found for any given value of z, and z can be found for any given value of by using the tables in reverse (as shown in Worked example 8.4). In the critical values table, values for zΦ( ) are denoted by p. A section of the tables, from which we will find the value of Φ(0.274), is shown below. First and second figures Third figure Fourth figure.0 0.1 0.2 0.3 0.5000 0.5398 0.5793 0.6179 0.5040 0.5438 0.5832 0.6217 0.5080 0.5478 0.5871 0.6255 0.5120 0.5517 0.5910 0.6293 0.5160 0.5557 0.5949 0.6331 0.5199 0.5596 0.5987 0.6368 0.5239 0.5636 0.6026 0.6406 0.5279 0.5675 0.6064 0.6443 0.5319 0. |
5714 0.6103 0.6480 0.5359 0.5753 0.6141 0.6517 1 4 4 4 4 4 432 8765 9 ADD 8 8 8 8 7 12 12 12 12 11 16 16 16 15 15 20 20 20 19 19 24 24 24 23 22 28 28 28 27 26 32 32 32 31 30 36 36 36 35 34 TIP Critical values refer to probabilities of 75%, 90%, 95%, … and their complements 25%, 10%, 5%, … and so on. 196 We locate the first and second figures of z (namely 0.2) in the left-hand column. TIP We then locate the third figure of z (namely 7) along the top row… this tells us that Φ 0.6064. (0.27) = Next we locate the fourth figure of z (namely 4) at the top-right. In line with 0.6064, we see ‘ADD 15’, which means that we must add 15 to the last two figures of 0.6064 to obtain the value of Φ(0.274). Φ (0.274) = 0.6064 + 0.0015 = 0.6079 WORKED EXAMPLE 8.1 Given that Z ~ N(0, 1), find P( Z < 1.23) and ZP( ù 1.23). 0.6079 (0.274) Φ = can be expressed using inverse notation as –1Φ 0.274 (0.6079) =. Answer 0.8907 1 – 0.8907 = 0.1093 (1.23) Φ left of z = = is the area to the 0.8907. 1.23 Φ (1.23) 1 – = the right of z the graphs. 0.1093 1.23 = is the area to, as shown in 0 1.23 z 0 1.23 z ∴ < ZP( 1.23) = 0.8907 and ZP( ù 1.23) = 0.1093 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution WORKED EXAMPLE 8.2 |
Given that Z ~ N(0, 1), find P(0.4 ø Z < 1.7) correct to 3 decimal places. Answer 0.4 1.7 z Φ (1.7) = 0.9554 and Φ (0.4) = 0.6554 P(0.4 ø < Z 1.7) = P( 1.7) – P( Z < 0.4) Z < (1.7) – = Φ Φ (0.4) The required probability is equal to the difference between the area to the left of z left of z and the area to the, as illustrated. 1.7= 0.4 = Φ (1.70) (0.40) in the main body of the We find the values of Φ table, which means that we do not need to use the ADD section. and = = 0.9554 – 0.6554 0.300 As noted previously, the normal distribution function tables do not show values for < 0 However, we can use the symmetry properties of the normal curve, and the fact that the area under the curve is equal to 1, to find values of zΦ( ) when z is negative. z. 197 Situations in which > 0 z, and in which < 0 z, are illustrated in the two diagrams below. For a positive value, z b= : For a negative value, z a–= : The shaded area in this graph represents the value of bΦ( ). The shaded area in this graph represents the value of aΦ( ). Φ( b) = P(Z b)> Φ(a) = P(Z –a)> 0 b z –a 0 z Φ ( ) b = P( Z b )ø and 1 – Φ ( ) b = P( Z b )ù. Φ ( ) a = P( Z – )ù a and 1 – Φ ( ) a = P( Z – )ø a. From the tables, the one piece of information, Φ probabilities: (0.11) = 0.5438, actually tells us four P( Z ø P( Z ù 0.11) = 0.5438 and ZP( ù –0.11) =. 0.5438 0.11) = 1 – 0.5438 = 0.4562 and Z |
P( ø –0.11) = 1 – 0.5438 = 0.4562. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Information given about probabilities in a normal distribution should always be transferred to a sketched graph. Useful information, such as whether a particular value of z is positive or negative, will then be easy to see. This could, of course, also be determined by considering inequalities. If, for example, Z z P( ù > ) 0.5 P(, then Z z ø < ) 0.5 and, therefore, < 0 z. WORKED EXAMPLE 8.3 Given that Z ~ N(0, 1), find P(–1 ø < Z 2.115) correct to 3 significant figures. Answer The required probability is given by the difference between the area to the left of. –1= z and the area to the left of z 2.115 = The first of these areas is greater than 0.5 and the second is less than 0.5. –1 0 2.115 z = Φ (2.115) and ZP( < –1) = 1 – Φ. (1) ZP( P(–1 198 2.115) < ø < Z 2.115) = Φ (2.115) – [1 – (1)] Φ (1) – 1 + Φ 0.8413 – 1 = Φ (2.115) = = 0.9828 + 0.824 WORKED EXAMPLE 8.4 Given that Z ~ N(0, 1), find the value of a such that P( Z a < ) = 0.9072. Answer 0.9066 = Φ (1.32) 0.004 a = = 1.32 + 1.324 0.9072) (0.9066 + 0.0006) (0.9066) + 0.004 0.004 = = 1.32 + 1.324 To find a value of z, we use the tables in reverse and search for the value closest to 0.9072, which is 0.9066 |
. z( )Φ For our value of 0.9072, we need to add 0.0006 to 0.9066, so ‘ADD 6’ is required – this will be done if 1.32 is given a 4th figure of 4. We can check the value obtained for a by reading the tables in the usual way. (1.324) Φ ‘ADD 6’ (1.32) = Φ + = = 0.9066 + 0.0006 0.9072 Alternatively, we can show our working using inverse notation. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy WORKED EXAMPLE 8.5 Given that Z ~ N(0, 1), find the value of b such that Z b P( ≥ Chapter 8: The normal distribution ) =. 0.7713 Answer 0.7713 0.7713 –a 0 z 0 a z P( Z a ø = ) 0.7713, so 0.7713) (0.7704 + 0.0009) (0.7704) 0.003 + 0.003 = = = 0.740 + 0.743 –0.743 ∴ = b a– 0.5 ù > ) Z b tells us that b is P( negative, so on our diagram we can replace b by a–, resulting in the two situations shown. The value closest to 0.7713 in the tables is 0.7704. This requires the addition of 0.0009 to bring it up to 0.7713, and 9 is in the column headed ‘ADD 3’. EXERCISE 8B 199 1 Given that Z ~ N(0, 1), find the following probabilities correct to 3 significant figures. a e i P( Z < 0.567) P( Z > 0.817) P( Z < 1.96) b ZP( ø 2.468) f j ZP( ù 2.009) P( Z > 2.576) c g P( Z P( Z > < –1.53) –1.75) d h ZP( ZP( ù ø –0.07 |
7) –0.013) 2 The random variable Z is normally distributed with mean 0 and variance 1. Find the following probabilities, correct to 3 significant figures. a d g j P(1.5 < Z < 2.5) P(–2.807 < Z < –1.282) P(–1.2 ( P 2 < Z < ø <Z 1.2) ) 5 b e h P(0.046 < Z < 1.272) c P(1.645 < Z < 2.326) P(–1.777 < Z < –0.746) P(–1.667 < Z < 2.667) f i P( –1.008 0.337) < 8 5 ) 3 Given that Z ~ N(0, 1), find the value of k, given that: a e i P( Z k < ) = 0.9087 b P( Z k < ) = 0.5442 P( Z k < ) = 0.25 P(– k Z k < < ) = 0.9128 f j P( Z k < ) = 0.3552 P(– k Z k < < ) = 0.6994 c g P( Z k > ) = 0.2743 P( Z k > ) = 0.9296 d h P( Z k > ) = 0.0298 P( Z k > ) = 0.648 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 4 Find the value of c in each of the following where Z has a normal distribution with. 0µ = and ( c Z < < 1.638) = 0.2673 P(1 < Z c < ) = 0.1408 P( c Z < < 2) = 0.6687 P(–1.221 < Z c < ) = 0.888 P(–2.63 < Z c < ) = 0.6861 b d f h j P( c Z < < 2.878) = 0.4968 P(0.109 < Z c < ) = 0.35 P( c Z < < 1.85 |
) = 0.9516 P(–0.674 < Z c < ) = 0.725 P(–2.7 < Z c < ) = 0.0252 Standardising a normal distribution The probability distribution of a normally distributed random variable is represented by a normal curve. This curve is centred on the mean µ; the area under the curve is equal to 1, and its height is determined by the standard deviation σ. We already have a method for finding probabilities involving the standard normal variable Z ~ N(0, 1) using the normal distribution function tables. Fortunately, this same set of tables can be used to find probabilities involving any normal random variable, no matter what the values of µ and σ2. Although we have only learnt about coding data, it turns out that coding works in exactly the same way for normally distributed random variables: they behave in the way that we expect and remain normal after coding. If we code X by subtracting µ, then the PDF is translated horizontally by µ– units and is has mean 0 and standard deviation σ. now centred on 0. The new random variable µ–X µ–X by multiplying by If we now code deviation (and variance) will be equal to 1, while the mean remains 0. The coded random variable X µ (i.e. dividing by σ) then the standard − is normally distributed with mean 0 and variance 1. σ 1 σ Coding the random variable X in this way is called standardising, because it transforms the distribution )2 to Z ~ N(0, 1). X ~ N(, µ σ KEY POINT 8.3 200 When X ~ N(, µ σ )2 then Z = A standardised value z = x µ − σ has a standard normal distribution. X µ − σ tells us how many standard deviations x is from the mean. Probabilities involving values of X are equal to probabilities involving the corresponding values of Z, which can be found from the normal distribution function tables for Z ~ N(0, 1). For example, if X ~ N(20, 9), then P( X < 23) = P Z < 23 20 − 9 . Copyright Material - Review Only - Not for Redistribution REWIND In Chapter 2, Section 2.2 and in Chapter 3, Section 3.3, we saw how the coding of data by addition and/ |
or multiplication affects the mean and the standard deviation. FAST FORWARD We will learn more about coding random variables in the Probability & Statistics 2 Coursebook, Chapter 3. REWIND In the table showing properties and probabilities of normal distributions prior to Explore 8.3, we saw that probabilities are determined by the number of standard deviations from the mean. The properties given in that table apply to all normal random variables. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution WORKED EXAMPLE 8.6 Given that X ~ N(11, 25), find P( X < 18) correct to 3 significant figures. Answer z = P( X < 18) = 18 P( 11 − 25 Z < (1.4) = Φ 1.4) = 1.4 We standardise x 18= and find that it is 1.4σ above the mean of 11. = 0.919 WORKED EXAMPLE 8.7 Given that X ~ N(20, 7), find P( X < 16.6) correct to 3 significant figures. Answer 16.6 z = P( X ø 16.6) = = = − 7 ø 20 = –1.285 – 1.285) P( Z (1.285) 1 – Φ 0.0994 We standardise x below the mean of 20. = 16.6 and find that it is 1.285σ 201 WORKED EXAMPLE 8.8 Given that X ~ N(5, 5), find P(2 9)<Xø correct to 3 significant figures. Answer For x = z2, = For x = 91.342 = 1.789 The required area is shown in two parts in the diagram. 2 –1.342 5 0 9 1.789 x z Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Area to the right of z Φ (1.789) – Φ (0) = Φ 0= is (1.789) − = |
0.4633 Area to the left of z = Φ (–1.342) (0) – Φ Total area X< ∴ P(2 = = 0.4633 < 9) = 0= is [ 0.5 – 1 – 0.4102 0.4102 + 0.874 0.5 Φ (1.342) ] = 0.8735 Here, we find the two areas separately then add them to obtain our final answer, which is where we round to the accuracy specified in the question. TIP Where possible, always use a 4-figure value for z. Try solving this problem by the method shown in Worked example 8.3 (using the areas to the left of both z –1.342 z approach you prefer. and ) and decide which 1.789 = = Some useful results from previous worked examples are detailed in the following graphs. 0 For < a < b For – a < 0 < b For – a < 0 < a 202 P( 0 a b z –a 0 b z –(– (– WORKED EXAMPLE 8.9 Given that ~ N(, Y 2µ σ ), P( Y < 10) = 0.75 and YP( ù 12) = 0.1, find the values of µ and σ. Answer P( Y ù 12) < 0.5, so Y < P( 12) > 0.5, which means that µ>12. P( Y < 10) > 0.5, which means also that. µ>10 0.75 0.90 μ 0 10 za y z μ 0 12 zb y z 0.674 gives 10 – µ = 0.674 σ …… [1] 1.282 gives 12 – µ = 1.282 σ …… [2] µ = µ = 10 12 − σ − σ These simple sketch graphs allow us to locate the values 10 and 12 relative to µ. za zb = Φ = Φ –1 –1 (0.75) (0.90) = = 0.674 1.282 and Note that both 0.75 and 0.90 are critical values. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University |
Press - Review Copy Chapter 8: The normal distribution We subtract equation 1[ ] from 2[ ] to solve this pair of simultaneous equations. [2] [1]. 7.78 12 – µ = 1.282 σ 10 – µ = 0.674 σ 2 ∴σ = = 3.29 0.608 σ and µ = EXERCISE 8C 1 Standardise the appropriate value(s) of the normal variable X represented in each diagram, and find the required probabilities correct to 3 significant figures. a Find XP( 11)ø, given that X. ~ N(8, 25) b Find P( X < 69.1), given that X. ~ N(72, 11) 8 0 11...... x z 203 69.1....... 72 0 c Find P(3 < <X 7), given that X. ~ N(5, 5) 3.... 5 0 7.... x z x z From this point in the exercise, you are strongly advised to sketch a diagram to help answer each question. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 2 Calculate the required probabilities correct to 3 significant figures. a Find XP( 9.7)ø and P( X > 9.7), given that X ~ N(6.2, 6.25). b Find XP( 5)ø and P( X > 5), given that X ~ N(3, 49). c Find P( X d Find P( X > < 33.4) and XP( 13.5) and XP( ø ù 33.4), given that X ~ N(37, 4). 13.5), given that X ~ N(20, 15). 91(1 P(2 and XP( 91)ø, given that X ~ N(80, 375). 21), given that X ~ N(11, 25). 5), given that X ~ N(3, 7). 8.8), given that X ~ N(7, 1.44). [Read carefully.] 28), given that X ~ N(25, 6). e Find P( f Find g Find h Find i Find P |
(6.2 P(26 Find P(8 10), given that X ~ N(12, 2.56). 3 a Find a, given that X ~ N(30, 16) and that P( X a< ) = 0.8944. b Find b, given that X ~ N(12, 4) and that P( X b< ) = 0.9599. c Find c, given that X ~ N(23, 9) and that P( X c > ) = 0.9332. d Find d, given that X ~ N(17, 25) and that P( X d > ) = 0.0951. e Find e, given that X ~ N(100, 64) and that P( X e > ) = 0.95. 204 4 a Find f, given that X ~ N(10, 7) and that P( f Xø < 13.3) = 0.1922. b Find g, given that X ~ N(45, 50) and that P( g Xø < 55) = 0.5486. c Find h, given that X ~ N(7, 2) and that P(8 d Find j, given that X ~ N(20, 11) and that P( ø X h < j Xø < ) = 0.216. 22) = 0.5. 5 X is normally distributed with mean 4 and variance 6. Find the probability that X takes a negative value. 6 Given that X 2) ( µ µ ~ N, 4 9 where 0µ >, find XP(. 2 )µ< 7 If ~ N(10, T )2σ and P( T > 14.7) = 0.04, find the value of σ. 8 It is given that ~ N(, 13)µ V and P( V < 15) = 0.75. Find the value of µ. 9 The variable W ~ N(, )2µ σ. Given that 4µ σ= and P( W < 83) = 0.95, find the value of µ and of σ. 10 X has a normal distribution in which σ µ= – 30 and XP( ù 12) = 0.9. Find the value of µ and of σ. 11 The variable )2µ σ Q and of σ, and calculate the ~ N(,. Given that � |
� < Q P(4 P( Q. 5) < 1.288) = 0.281 and P( Q < 6.472) = 0.591, find the value of µ 12 For the variable ~ N(, V )2µ σ, it is given that P( V = 0.7509 and P( V > 9.2) = 0.1385. Find the value of µ and of σ, and calculate VP( 8.4) < 10)ø. 13 Find the value of µ and of σ and calculate 0.6858 = and WP( 4.75) 2.25) WP( ø ù = 6.48) for the variable W ~ N(, )2µ σ, given that P( W >. 0.0489 14 X has a normal distribution, such that P( X > 147.0) = 0.0136 and XP( ø 59.0) = 0.0038. Use this information to calculate the probability that 80.0 ø < X. 130.0 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution 8.3 Modelling with the normal distribution The German mathematician Carl Friedrich Gauss showed that measurement errors made in astronomical observations were well modelled by a normal distribution, and the Belgian statistician and sociologist Adolphe Quételet later applied this to human characteristics when he saw that distributions of such things as height, weight, girth and strength were approximately normal. We are now in a position to apply our knowledge to real-life situations, and to solve more advanced problems involving the normal distribution. WORKED EXAMPLE 8.10 The mass of a newborn baby in a certain region is normally distributed with mean 3.35 kg and variance 0.0858 kg2. Estimate how many of the 1356 babies born last year had masses of less than 3.5 kg. TIP We cannot know the exact number of newborn babies from the model because it only gives estimates. However, we do know that the number of babies must be an integer. 205 Answer Φ Z( ) = Φ = Φ − 3.35 0.0858 3.5 � |
�� (0.512) = 0.6957 We standardise the mass of 3.5 kg. 0.6957 is a relative frequency equal to 69.57% < P(mass 69.57% of 1356 ∴ There were about 943 newborn babies. 0.6957 = 943.3692 3.5 kg) = WORKED EXAMPLE 8.11 A factory produces half-litre tins of oil. The volume of oil in a tin is normally distributed with mean 506.18 ml and standard deviation 2.96 ml. a What percentage of the tins contain less than half a litre of oil? b Find the probability that exactly 1 out of 3 randomly selected tins contains less than half a litre of oil. Answer a z = 500 506.18 − 2.96 = –2.088 Let X represent the amount of oil in a tin, then X ~ N(506.18, 2.96 )2. The graph shows the probability distribution for the amount of oil in a tin. 500 –2.088 506.18 0 x z Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 P( X < 500) = = = = P( Z –2.088) < (2.088) 1 – Φ 1 – 0.9816 0.0184 ∴ 1.84% of the tins contain less than half a litre of oil. b YP( = 1) = = 3 1 × 0.0532 1 0.0184 × 0.9816 2 EXERCISE 8D Let the discrete random variable Y be the number of tins containing less than half a litre of oil, then ~ B(3, 0.0184) Y. TIP A probability obtained from a normal distribution can be used as the parameter p in a binomial distribution. 1 The length of a bolt produced by a machine is normally distributed with mean 18.5 cm and variance 0.7 cm2. Find the probability that a randomly selected |
bolt is less than 18.85 cm long. 2 The waiting times, in minutes, for patients at a clinic are normally distributed with mean 13 and variance 16. a Calculate the probability that a randomly selected patient has to wait for more than 16.5 minutes. b Last month 468 patients attended the clinic. Calculate an estimate of the number who waited for less than 206 9 minutes. 3 Tomatoes from a certain producer have masses which are normally distributed with mean 90 grams and standard deviation 17.7 grams. The tomatoes are sorted into three categories by mass, as follows: Small: under 80 g; Medium: 80 g to 104 g; Large: over 104 g. a Find, correct to 2 decimal places, the percentage of tomatoes in each of the three categories. b Find the value of k such that P( k X ø < 104) = 0.75, where X is the mass of a tomato in grams. 4 The heights, in metres, of the trees in a forest are normally distributed with mean µ and standard deviation 3.6. Given that 75% of the trees are less than 10 m high, find the value of µ. 5 The mass of a certain species of fish caught at sea is normally distributed with mean 5.73 kg and variance 2.56 kg2. Find the probability that a randomly selected fish caught at sea has a mass that is: a less than 6.0 kg b more than 3.9 kg c between 7.0 and 8.0 kg 6 The distance that children at a large school can hop in 15 minutes is normally distributed with mean 199 m and variance 3700 m2. a Calculate an estimate of b, given that only 25% of the children hopped further than b metres. b Find an estimate of the interquartile range of the distances hopped. 7 The daily percentage change in the value of a company’s shares is expected to be normally distributed with mean 0 and standard deviation 0.51. On how many of the next 365 working days should the company expect the value of its shares to fall by more than 1%? 8 The masses, w grams, of a large sample of apples are normally distributed with mean 200 and variance 169., calculate an estimate of the number of 213 <w< Given that the masses of 3413 apples are in the range 187 apples in the sample. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy |
Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution 9 The ages of the children in a gymnastics club are normally distributed with mean 15.2 years and standard deviation σ. Find the value of σ given that 30.5% of the children are less than 13.5 years of age. 10 The speeds, in kmh–1, of vehicles passing a particular point on a rural road are normally distributed with mean µ and standard deviation 20. Find the value of µ and find what percentage of the vehicles are being driven at under 80 kmh–1, given that 33% of the vehicles are being driven at over 100 kmh–1. 11 Coffee beans are packed into bags by the workers on a farm, and each bag claims to contain 200 g. The actual mass of coffee beans in a bag is normally distributed with mean 210 g and standard deviation σ. The farm owner informs the workers that they must repack any bag containing less than 200 g of coffee beans. Find the value of σ, given that 0.5% of the bags must be repacked. 12 Colleen exercises at home every day. The length of time she does this is normally distributed with mean 12.8 minutes and standard deviation σ. She exercises for more than 15 minutes on 42 days in a year of 365 days. a Calculate the value of σ. b On how many days in a year would you expect Colleen to exercise for less than 10 minutes? 13 The times taken by 15-year-olds to solve a certain puzzle are normally distributed with mean µ and standard deviation 7.42 minutes. a Find the value of µ, given that three-quarters of all 15-year-olds take over 20 minutes to solve the puzzle. b Calculate an estimate of the value of n, given that 250 children in a random sample of n 15-year-olds fail to solve the puzzle in less than 30 minutes. 14 The lengths, cmX, of the leaves of a particular species of tree are normally distributed with mean µ and variance σ2. 207 a Find P Find the probability that a randomly selected leaf from this species has a length that is more than 2 standard deviations from the mean. c Find the value of µ and of σ, given that XP( < 7.5) = 0.75 and XP( < 8.5) =. 0. |
90 15 The time taken in seconds for Ginger’s computer to open a specific large document is normally distributed with mean 9 and variance 5.91. a Find the probability that it takes exactly 5 seconds or more to open the document. b Ginger opens the document on her computer on n occasions. The probability that it fails to open in less than exactly 5 seconds on at least one occasion is greater than 0.5. Find the least possible value of n. 16 The masses of all the different pies sold at a market are normally distributed with mean 400 g and standard deviation 61g. Find the probability that: a the mass of a randomly selected pie is less than 425 g b 4 randomly selected pies all have masses of less than 425 g c exactly 7 out of 10 randomly selected pies have masses of less than 425 g. 17 The height of a female university student is normally distributed with mean 1.74 m and standard deviation 12.3 cm. Find the probability that: a a randomly selected female student is between 1.71 and 1.80 metres tall b 3 randomly selected female students are all between 1.71 and 1.80 m tall c exactly 15 out of 50 randomly selected female students are between 1.71 and 1.80 metres tall. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 8.4 The normal approximation to the binomial distribution In Chapter 7, Section 7.1, we saw that the binomial distribution can be used to solve problems such as ‘Find the probability of obtaining exactly 60 heads with 100 tosses of a fair coin’, and that this is equal to 100 60 × 60 0.5 × 0.5 40. Therefore, to find the probability of obtaining 60 or more heads, we must find the probability for 60 heads, for 61 heads, for 62 heads and so on, and add them all together. Imagine how long it took to calculate binomial probabilities before calculators and computers! However, in certain situations, we can approximate a probability such as this by a method that involves far fewer calculations using the normal distribution. EXPLORE 8.4 Binomial probability distributions for 2, 4 |
, and 12 tosses of a fair coin are shown in the following diagrams. Notice that, as the number of coin tosses increases, the shape of the probability distribution becomes increasingly normal. 0.5 p = 0.5, n = 2 0.4 p = 0.5, n = 4 0.25 p = 0.5, n = 12 208 10 12 Does the binomial probability distribution maintain its normal shape for large values of n when p varies? Find out using the Binomial Distribution resource on the GeoGebra website. Select any ùn note of when the distribution loses its normal shape. Repeat this for other values of n. 20 then use the pause/play button or the slider to vary the value of p. Take Can you generalise as to when the binomial distribution begins to lose its normal shape? DID YOU KNOW? FAST FORWARD Abraham de Moivre, the 18th century statistician and consultant, was often asked to make long calculations concerning games of chance. He noted that when the number of events increased, the shape of the binomial distribution approached a very smooth curve, and saw that he would be able to solve these long calculation problems if he could find a mathematical expression for this curve: this is exactly what he did. The curve he discovered is now called the normal curve. Before the late 1870s, when the term normal was coined independently by Peirce, Galton and Lexis, this distribution was known – and still is by some – as the Gaussian distribution after the German mathematician Carl Friedrich Gauss. The word normal is not meant to suggest that all other distributions are abnormal! Abraham de Moivre 1667–1754 Copyright Material - Review Only - Not for Redistribution n = de Moivre’s theorem, sin ) (cos i θ θ + cos sin n n i θ θ + links trigonometry with complex numbers – a topic that we cover in the Pure Mathematics 2 & 3 Coursebook, Chapter 11. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution The following diagrams show the shapes of four binomial distributions for n 25=. p = 0.15, q = 0.85 p = 0.35, q = 0.65 p = 0.75, q = 0 |
.25 p = 0.95, q = 0.05 0 25 0 25 0 25 0 25 np = 3.75, nq = 21.25 np = 8.75, nq = 16.25 np = 18.75, nq = 6.25 np = 23.75, nq = 1.25 As you can see, the binomial distribution loses its normal shape when p is small and also when q is small. KEY POINT 8.4 ) ~ B(, n p can be X approximated by )2µ σ, where N(, 2σ = npq, µ = np and provided that n is large enough to ensure that np and nq > 5 > 5. 209 KEY POINT 8.5 Continuity corrections must be made when a discrete distribution is approximated by a continuous distribution. A more detailed investigation shows that the binomial distribution has an approximately normal shape if np and nq are both greater than 5. These are the values that we use to decide whether a binomial distribution can be well-approximated by a normal distribution. The larger the values of np and nq, the more accurate the approximation will be. As we can see from the above diagrams, the approximation is adequate (but not very good) when np and nq = 18.75 = 6.25. The distribution. because nq < 5 X ~ B(40, 0.9) cannot be well-approximated by a normal distribution The distribution = 50 because np X and ~ B(250, 0.2) = 200 nq can be well-approximated by a normal distribution, both of which are substantially greater than 5. = 13 13.5 must be treated as being represented by the class of continuous values. For this reason, When we approximate a discrete distribution by a continuous distribution, a discrete value such as X ø X 12.5 < in our probability calculations. Making this replacement is known as ‘making a = 13.5 X continuity correction’. Deciding whether to use depends on whether or not is included in the probability that we wish to find. must be replaced by either = 12.5 = 13.5 = 12.5 or by = 13 or X X X X X = 13 For example, if we wish to find. = 12.5 using X P( X < 13), where X = 13 is not included, we calculate If we wish to find øXP |
( 13), where X = 13 is included, we calculate using X. = 13.5 Further details of continuity corrections are given in Worked example 8.12. WORKED EXAMPLE 8.12 Given that ~ B(100, 0.4) X to find: a P( X < 43) b P( X > 43), use a suitable approximation and continuity correction Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 Answer 40 =np and µ= X approximated by N(40, 24). ~ B(100, 0.4) 2σ = can be npq = 24. The conditions for approximating a binomial distribution by a normal distribution are met because np and nq than 5. 40=, which are both greater 60 42 43 44 x 42.5 43.5 In the continuous distribution N(40, 24), 43 is represented by the class of continuous values 42.5 X ≤ diagram., as shown in the 43.5 < Possible continuity corrections for a discrete value of 43 are given below: For X < P( <XP( For 43), we would use the lower boundary value 42.5…… [part a] 43), we would use the upper boundary value 43.5 For XP( For > >XP(, we would use the upper boundary value 43.5 …… [part b] 43) 43), we would use the lower boundary value 42.5 210 = Φ a P( X < 43) ≈ P( 0.510) Z < (0.510) = 43) 0.6950 ≈ 0.695 X∴ P( < b P( X > 43) ≈ = = 43) X∴ P( > P(Z 0.714) > (0.714) 1 – Φ 0.2377 ≈ 0.238 For x = 42.5, z = 40 42.5 − 24 = 0.5103 For x = 43.5, z = 40 43.5 − 24 = 0.7144 FAST FORWARD We will also make continuity corrections when using the normal distribution as an approximation to the Poisson distribution in the Probability & Statistics 2 Coursebook |
, Chapter 2. TIP X a< means ‘X is fewer/less than a’. X a> means ‘X is more/greater than a’. X a< means ‘X is at most a’ and ‘X is not more than a’ and ‘X is a or less’. X a> means ‘X is at least a’ and ‘X is not less than a’ and ‘X is a or more’. WORKED EXAMPLE 8.13 Boxes are packed with 8000 randomly selected items. It is known that 0.2% of the items are yellow. Find, using a suitable approximation, the probability that: a a box contains fewer than 20 yellow items b exactly 2 out of 3 randomly selected boxes contain fewer than 20 yellow items. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 8: The normal distribution Answer a X ~ B(8000, 0.002) Let X be the number of yellow items in a box. X ~ B(8000, 0.002) can be approximated by N(16, 15.968). P( X < 20) P ≈ Z P( Z < 19.5 16 − 15.968 ) … 0.87587 = = Φ < (0.876) np = 16 and nq = 7984 are both greater than 5, so we can approximate the binomial distribution by a normal distribution using npµ= = 16 and = 15.968. 2 σ = npq P( To find X < calculate with the value x = 19.5. 20), we must = 0.8094 ∴ The probability that a box contains fewer than 20 yellow items is approximately 0.809. b P( Y = 2) = = 3 2 × 0.375 0.8094 2 1 0.1906 × Let Y be the number of boxes containing fewer than 20 yellow items, then Y ~ B(3, 0.8094). TIP Do not forget to |
make the continuity correction! TIP Although our answer to part a is only an approximation, we should not use a rounded probability, such as 0.8, in further calculations. 211 WORKED EXAMPLE 8.14 A fair coin is tossed 888 times. Find, by use of a suitable approximation, the probability that the coin lands heads-up at most 450 times. Answer X ~ B(888, 0.5) Let X represent the number of times the coin lands heads-up. X ~ B(888, 0.5) can be approximated by N(444, 222). np = 444, nq = 444 are both greater than 5, and npq = 222. P( X < 450) P ≈ Z P( Z = < 0.436) < 450.5 444 − 222 To find X øP( 450) we calculate with x = 450.5. = Φ (0.436) = 0.6686 ∴ P(at most 450 heads) ≈ 0.669 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 EXPLORE 8.5 By visiting the Binomial and Normal resource on the GeoGebra website you will get a clear picture of how the normal approximation to the binomial distribution works. Select values of n and p so that np and (1 – ) n p are both greater than 5. The binomial probability distribution is displayed with an overlaid normal curve ) p are displayed in red at the top-right). If (the value of µ = np and of you then check the probability box, adjustable values of =x a and =x b appear on the diagram, with the area between them shaded. Remember that a discrete variable is being approximated by a continuous variable, so appropriate continuity corrections are needed to find the best probability estimates. σ = np (1 − EXERCISE 8E 1 Decide whether or not each of the following binomial distributions can be well-approximated by a normal distribution. For those that can, state the values of the parameters µ and 2σ. For those that cannot, |
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