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9 19 h. P(A\|B) = P(AANDB) P(B) + 10 19 = 1) = 3 6, P(B\|A) = P(AANDB) P(A), No = 3 9 3.1 The sample space S is the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)). a. S = _____________________________ Let event A = the sum is even and event B = the first number is prime. b. A = _____________________, B = _____________________ c. P(A) = _____________, P(B) = ________________ d. A AND B = ____________________, A OR B = ________________ e. P(A AND B) = _________, P(A OR B) = _____________ f. B′ = _____________, P(B′) = _____________ g. P(A) + P(A′) = ____________ h. P(A\|B) = ___________, P(B\|A) = _____________; are the probabilities equal? Example 3.2 A fair, six-sided die is rolled. Describe the sample space S, identify each of the following events with a subset of S and compute its probability (an outcome is the number of dots that show up). a. Event T = the outcome is two. b. Event A = the outcome is an even number. c. Event B = the outcome is less than four. d. The complement of A. e. A GIVEN B f. B GIVEN A g. A AND B h. A OR B This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 167 i. A OR B′ j. Event N = the outcome is a prime number. k. Event I = the outcome is seven. Solution 3.2 a. T = {2}, P(T) = 1 6 b. A = {2, 4, 6}, P(A) = 1 2 c. B = {1, 2, 3}, P(B) = 1 2 d. A′ = {1, 3, 5}, P(A′) =
1 2 e. A\|B = {2}, P(A\|B) = 1 3 f. B\|A = {2}, P(B\|A) = 1 3 g. A AND B = {2}, P(A AND B) = 1 6 h. A OR B = {1, 2, 3, 4, 6}, P(A OR B) = 5 6 i. A OR B′ = {2, 4, 5, 6}, P(A OR B′) = 2 3 j. N = {2, 3, 5}, P(N) = 1 2 k. A six-sided die does not have seven dots. P(7) = 0. Example 3.3 Table 3.1 describes the distribution of a random sample S of 100 individuals, organized by gender and whether they are right- or left-handed. Right-handed Left-handed Males 43 Females 44 Table 3.1 9 4 Let’s denote the events M = the subject is male, F = the subject is female, R = the subject is right-handed, L = the subject is left-handed. Compute the following probabilities: a. P(M) b. P(F) c. P(R) d. P(L) e. P(M AND R) f. P(F AND L) g. P(M OR F) h. P(M OR R) 168 CHAPTER 3 \| PROBABILITY TOPICS i. P(F OR L) j. P(M') k. P(R\|M) l. P(F\|L) m. P(L\|F) Solution 3.3 a. P(M) = 0.52 b. P(F) = 0.48 c. P(R) = 0.87 d. P(L) = 0.13 e. P(M AND R) = 0.43 f. P(F AND L) = 0.04 g. P(M OR F) = 1 h. P(M OR R) = 0.96 i. P(F OR L) = 0.57 j. P(M') = 0.48 k. P(R\|M) = 0.8269 (rounded to four decimal places) l. P(F\|L) = 0.3077 (rounded to four decimal places) m. P(L\|F) = 0.0833 3
.2 \| Independent and Mutually Exclusive Events Independent and mutually exclusive do not mean the same thing. Independent Events Two events are independent if the following are true: • P(A\|B) = P(A) • P(B\|A) = P(B) • P(A AND B) = P(A)P(B) Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are NOT independent, then we say that they are dependent. Sampling may be done with replacement or without replacement. • With replacement: If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick. • Without replacement: When sampling is done without replacement, each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent. If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 169 Example 3.4 You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit. a. Sampling with replacement: Suppose you pick three cards with replacement. The first card you pick out of the 52 cards is the Q of spades. You put this card back, reshuffle the cards and pick a second card from the 52-card deck. It is the ten of clubs. You put this card back,
reshuffle the cards and pick a third card from the 52-card deck. This time, the card is the Q of spades again. Your picks are {Q of spades, ten of clubs, Q of spades}. You have picked the Q of spades twice. You pick each card from the 52-card deck. b. Sampling without replacement: Suppose you pick three cards without replacement. The first card you pick out of the 52 cards is the K of hearts. You put this card aside and pick the second card from the 51 cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining 50 cards in the deck. The third card is the J of spades. Your picks are {K of hearts, three of diamonds, J of spades}. Because you have picked the cards without replacement, you cannot pick the same card twice. 3.4 You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit. Three cards are picked at random. a. Suppose you know that the picked cards are Q of spades, K of hearts and Q of spades. Can you decide if the sampling was with or without replacement? b. Suppose you know that the picked cards are Q of spades, K of hearts, and J of spades. Can you decide if the sampling was with or without replacement? Example 3.5 You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. a. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are QS, 1D, 1C, QD. b. Suppose you pick four cards and put each card back before you pick the next card. Your cards are KH,
7D, 6D, KH. Which of a. or b. did you sample with replacement and which did you sample without replacement? Solution 3.5 a. Without replacement; b. With replacement 3.5 You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. Suppose that you sample four cards without replacement. Which of the following outcomes are possible? Answer the same question for sampling with replacement. a. QS, 1D, 1C, QD b. KH, 7D, 6D, KH 170 CHAPTER 3 \| PROBABILITY TOPICS c. QS, 7D, 6D, KS Mutually Exclusive Events A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P(A AND B) = 0. For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}. A AND B = {4, 5}. P(A AND B) = 2 10 and is not equal to zero. Therefore, A and B are not mutually exclusive. A and C do not have any numbers in common so P(A AND C) = 0. Therefore, A and C are mutually exclusive. If it is not known whether A and B are mutually exclusive, assume they are not until you can show otherwise. The following examples illustrate these definitions and terms. Example 3.6 Flip two fair coins. (This is an experiment.) The sample space is {HH, HT, TH, TT} where T = tails and H = heads. The outcomes are HH, HT, TH, and TT. The outcomes HT and TH are different. The HT means that the first coin showed heads and the second coin showed tails. The TH means that the first coin showed tails and the second coin showed heads. • Let A = the
event of getting at most one tail. (At most one tail means zero or one tail.) Then A can be written as {HH, HT, TH}. The outcome HH shows zero tails. HT and TH each show one tail. • Let B = the event of getting all tails. B can be written as {TT}. B is the complement of A, so B = A′. Also, P(A) + P(B) = P(A) + P(A′) = 1. • The probabilities for A and for B are P(A) = 3 4 and P(B) = 1 4. • Let C = the event of getting all heads. C = {HH}. Since B = {TT}, P(B AND C) = 0. B and C are mutually exclusive. (B and C have no members in common because you cannot have all tails and all heads at the same time.) • Let D = event of getting more than one tail. D = {TT}. P(D) = 1 4 • Let E = event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) E = {HT, HH}. P(E) = 2 4 • Find the probability of getting at least one (one or two) tail in two flips. Let F = event of getting at least one tail in two flips. F = {HT, TH, TT}. P(F) = 3 4 3.6 Draw two cards from a standard 52-card deck with replacement. Find the probability of getting at least one black card. Example 3.7 Flip two fair coins. Find the probabilities of the events. a. Let F = the event of getting at most one tail (zero or one tail). b. Let G = the event of getting two faces that are the same. c. Let H = the event of getting a head on the first flip followed by a head or tail on the second flip. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 171 d. Are F and G mutually exclusive? e. Let J = the event of getting all tails. Are J and H mutually exclusive? Solution 3.7 Look at the sample space in Example 3.6. a. Zero (0) or
one (1) tails occur when the outcomes HH, TH, HT show up. P(F) = 3 4 b. Two faces are the same if HH or TT show up. P(G) = 2 4 c. A head on the first flip followed by a head or tail on the second flip occurs when HH or HT show up. P(H) = 2 4 d. F and G share HH so P(F AND G) is not equal to zero (0). F and G are not mutually exclusive. e. Getting all tails occurs when tails shows up on both coins (TT). H’s outcomes are HH and HT. J and H have nothing in common so P(J AND H) = 0. J and H are mutually exclusive. 3.7 A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events: a. Let F = the event of getting the white ball twice. b. Let G = the event of getting two balls of different colors. c. Let H = the event of getting white on the first pick. d. Are F and G mutually exclusive? e. Are G and H mutually exclusive? Example 3.8 Roll one fair, six-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event A = a face is odd. Then A = {1, 3, 5}. Let event B = a face is even. Then B = {2, 4, 6}. • Find the complement of A, A′. The complement of A, A′, is B because A and B together make up the sample space. P(A) + P(B) = P(A) + P(A′) = 1. Also, P(A) = 3 6 and P(B) = 3 6. • Let event C = odd faces larger than two. Then C = {3, 5}. Let event D = all even faces smaller than five. Then D = {2, 4}. P(C AND D) = 0 because you cannot have an odd and even face at the same time. Therefore, C and D are mutually exclusive events. • Let event E = all faces less than five. E = {1, 2, 3, 4}. Are C and E mutually exclusive events? (Answer yes or
no.) Why or why not? Solution 3.8 No. C = {3, 5} and E = {1, 2, 3, 4}. P(C AND E) = 1 6. To be mutually exclusive, P(C AND E) must be zero. • Find P(C\|A). This is a conditional probability. Recall that the event C is {3, 5} and event A is {1, 3, 5}. To find P(C\|A), find the probability of C using the sample space A. You have reduced the sample space from the original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5}. So, P(C\|A) = 2 3. 172 CHAPTER 3 \| PROBABILITY TOPICS 3.8 Let event A = learning Spanish. Let event B = learning German. Then A AND B = learning Spanish and German. Suppose P(A) = 0.4 and P(B) = 0.2. P(A AND B) = 0.08. Are events A and B independent? Hint: You must show ONE of the following: • P(A\|B) = P(A) • P(B\|A) • P(A AND B) = P(A)P(B) Example 3.9 Let event G = taking a math class. Let event H = taking a science class. Then, G AND H = taking a math class and a science class. Suppose P(G) = 0.6, P(H) = 0.5, and P(G AND H) = 0.3. Are G and H independent? If G and H are independent, then you must show ONE of the following: • P(G\|H) = P(G) • P(H\|G) = P(H) • P(G AND H) = P(G)P(H) NOTE The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information. a. Show that P(G\|H) = P(G). Solution 3.9 P(G\|H) = P(G AND H) P(H) = 0.3 0.5 = 0.6 = P(G) b. Show P(G AND H) = P(G)P(H). Solution 3.9 P(G)
P(H) = (0.6)(0.5) = 0.3 = P(G AND H) Since G and H are independent, knowing that a person is taking a science class does not change the chance that he or she is taking a math class. If the two events had not been independent (that is, they are dependent) then knowing that a person is taking a science class would change the chance he or she is taking math. For practice, show that P(H\|G) = P(H) to show that G and H are independent events. 3.9 In a bag, there are six red marbles and four green marbles. The red marbles are marked with the numbers 1, 2, 3, 4, 5, and 6. The green marbles are marked with the numbers 1, 2, 3, and 4. • R = a red marble • G = a green marble • O = an odd-numbered marble • The sample space is S = {R1, R2, R3, R4, R5, R6, G1, G2, G3, G4}. S has ten outcomes. What is P(G AND O)? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 173 Example 3.10 Let event C = taking an English class. Let event D = taking a speech class. Suppose P(C) = 0.75, P(D) = 0.3, P(C\|D) = 0.75 and P(C AND D) = 0.225. Justify your answers to the following questions numerically. a. Are C and D independent? b. Are C and D mutually exclusive? c. What is P(D\|C)? Solution 3.10 a. Yes, because P(C\|D) = P(C). b. No, because P(C AND D) is not equal to zero. c. P(D\|C) = P(C AND D) = 0.3 P(C) = 0.225 0.75 3.10 A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) = 0.40, P(D
) = 0.30 and P(B AND D) = 0.20. a. Find P(B\|D). b. Find P(D\|B). c. Are B and D independent? d. Are B and D mutually exclusive? Example 3.11 In a box there are three red cards and five blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card. Let R = red card is drawn, B = blue card is drawn, E = even-numbered card is drawn. The sample space S = R1, R2, R3, B1, B2, B3, B4, B5. S has eight outcomes. • P(R) = 3 8. P(B) = 5 8. P(R AND B) = 0. (You cannot draw one card that is both red and blue.) • P(E) = 3 8. (There are three even-numbered cards, R2, B2, and B4.) • P(E\|B) = 2 5. (There are five blue cards: B1, B2, B3, B4, and B5. Out of the blue cards, there are two even cards; B2 and B4.) • P(B\|E) = 2 3. (There are three even-numbered cards: R2, B2, and B4. Out of the even-numbered cards, to are blue; B2 and B4.) • The events R and B are mutually exclusive because P(R AND B) = 0. • Let G = card with a number greater than 3. G = {B4, B5}. P(G) = 2 8. Let H = blue card numbered between one and four, inclusive. H = {B1, B2, B3, B4}. P(G\|H) = 1 4. (The only card in H that has a number greater than three is B4.) Since 2 8 = 1 4, P(G) = P(G\|H), which means that G and H are independent. 174 CHAPTER 3 \| PROBABILITY TOPICS 3.11 In a basketball arena, • 70% of the fans are rooting
for the home team. • 25% of the fans are wearing blue. • 20% of the fans are wearing blue and are rooting for the away team. • Of the fans rooting for the away team, 67% are wearing blue. Let A be the event that a fan is rooting for the away team. Let B be the event that a fan is wearing blue. Are the events of rooting for the away team and wearing blue independent? Are they mutually exclusive? Example 3.12 In a particular college class, 60% of the students are female. Fifty percent of all students in the class have long hair. Forty-five percent of the students are female and have long hair. Of the female students, 75% have long hair. Let F be the event that a student is female. Let L be the event that a student has long hair. One student is picked randomly. Are the events of being female and having long hair independent? • The following probabilities are given in this example: • P(F) = 0.60; P(L) = 0.50 • P(F AND L) = 0.45 • P(L\|F) = 0.75 NOTE The choice you make depends on the information you have. You could use the first or last condition on the list for this example. You do not know P(F\|L) yet, so you cannot use the second condition. Solution 1 Check whether P(F AND L) = P(F)P(L). We are given that P(F AND L) = 0.45, but P(F)P(L) = (0.60)(0.50) = 0.30. The events of being female and having long hair are not independent because P(F AND L) does not equal P(F)P(L). Solution 2 Check whether P(L\|F) equals P(L). We are given that P(L\|F) = 0.75, but P(L) = 0.50; they are not equal. The events of being female and having long hair are not independent. Interpretation of Results The events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair. 3.12 Mark is deciding which route to take to work. His choices are I = the Interstate and F = Fifth Street. • P(I) = 0.44 and P(F) = 0.55 •
P(I AND F) = 0 because Mark will take only one route to work. What is the probability of P(I OR F)? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 175 Example 3.13 a. Toss one fair coin (the coin has two sides, H and T). The outcomes are ________. Count the outcomes. There are ____ outcomes. b. Toss one fair, six-sided die (the die has 1, 2, 3, 4, 5 or 6 dots on a side). The outcomes are ________________. Count the outcomes. There are ___ outcomes. c. Multiply the two numbers of outcomes. The answer is _______. d. If you flip one fair coin and follow it with the toss of one fair, six-sided die, the answer in three is the number of outcomes (size of the sample space). What are the outcomes? (Hint: Two of the outcomes are H1 and T6.) e. Event A = heads (H) on the coin followed by an even number (2, 4, 6) on the die. A = {_________________}. Find P(A). f. Event B = heads on the coin followed by a three on the die. B = {________}. Find P(B). g. Are A and B mutually exclusive? (Hint: What is P(A AND B)? If P(A AND B) = 0, then A and B are mutually exclusive.) h. Are A and B independent? (Hint: Is P(A AND B) = P(A)P(B)? If P(A AND B) = P(A)P(B), then A and B are independent. If not, then they are dependent). Solution 3.13 a. H and T; 2 b. 1, 2, 3, 4, 5, 6; 6 c. 2(6) = 12 d. T1, T2, T3, T4, T5, T6, H1, H2, H3, H4, H5, H6 e. A = {H2, H4, H6}; P(A) = 3 12 f. B = {H3}; P(B) = 1 12 g
. Yes, because P(A AND B) = 0 h. P(A AND B) = 0.P(A)P(B) = ⎛ ⎝ ⎞ ⎠ 3 12 ⎛ ⎝ 1 12 ⎞ ⎠. P(A AND B) does not equal P(A)P(B), so A and B are dependent. 3.13 A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let T be the event of getting the white ball twice, F the event of picking the white ball first, S the event of picking the white ball in the second drawing. a. Compute P(T). b. Compute P(T\|F). c. Are T and F independent?. d. Are F and S mutually exclusive? e. Are F and S independent? 3.3 \| Two Basic Rules of Probability When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not. The Multiplication Rule If A and B are two events defined on a sample space, then: P(A AND B) = P(B)P(A\|B). 176 CHAPTER 3 \| PROBABILITY TOPICS This rule may also be written as: P(A\|B) = P(A AND B) P(B) (The probability of A given B equals the probability of A and B divided by the probability of B.) If A and B are independent, then P(A\|B) = P(A). Then P(A AND B) = P(A\|B)P(B) becomes P(A AND B) = P(A)P(B). The Addition Rule If A and B are defined on a sample space, then: P(A OR B) = P(A) + P(B) - P(A AND B). If A and B are mutually exclusive, then P(A AND B) = 0. Then P(A OR B) = P(A) + P(B) - P(A AND B) becomes P(A OR B) = P(A) + P(B). Example 3.14 Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B
= Alaska • Klaus can only afford one vacation. The probability that he chooses A is P(A) = 0.6 and the probability that he chooses B is P(B) = 0.35. • P(A AND B) = 0 because Klaus can only afford to take one vacation • Therefore, the probability that he chooses either New Zealand or Alaska is P(A OR B) = P(A) + P(B) = 0.6 + 0.35 = 0.95. Note that the probability that he does not choose to go anywhere on vacation must be 0.05. Example 3.15 Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P(A) = 0.65. B = the event Carlos is successful on his second attempt. P(B) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90. a. What is the probability that he makes both goals? Solution 3.15 a. The problem is asking you to find P(A AND B) = P(B AND A). Since P(B\|A) = 0.90: P(B AND A) = P(B\|A) P(A) = (0.90)(0.65) = 0.585 Carlos makes the first and second goals with probability 0.585. b. What is the probability that Carlos makes either the first goal or the second goal? Solution 3.15 b. The problem is asking you to find P(A OR B). P(A OR B) = P(A) + P(B) - P(A AND B) = 0.65 + 0.65 - 0.585 = 0.715 Carlos makes either the first goal or the second goal with probability 0.715. c. Are A and B independent? Solution 3.15 c. No, they are not, because P(B AND A) = 0.585. P(B)P(A) = (0.65)(0.65) = 0.423 0.423 ≠ 0.585 = P(B AND A) So, P(B AND A) is not equal to P(B)P(A). d. Are A
and B mutually exclusive? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 177 Solution 3.15 d. No, they are not because P(A and B) = 0.585. To be mutually exclusive, P(A AND B) must equal zero. 3.15 Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P(C) = 0.75. D = the event Helen makes the second shot. P(D) = 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws? Example 3.16 A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Fortyseven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly. a. What is the probability that the member is a novice swimmer? Solution 3.16 a. 28 150 b. What is the probability that the member practices four times a week? Solution 3.16 b. 80 150 c. What is the probability that the member is an advanced swimmer and practices four times a week? Solution 3.16 c. 40 150 d. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not? Solution 3.16 d. P(advanced AND intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time. e. Are being a novice swimmer and practicing four times a week independent events? Why or why not? Solution 3.16 e. No, these are not independent events. P(novice AND practices four times per week) = 0.0667 P(novice)P(practices four times per week) = 0.0996 0.0667
≠ 0.0996 178 CHAPTER 3 \| PROBABILITY TOPICS 3.16 A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is taking a gap year? Example 3.17 Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is 0.25. Let: M = math class, S = speech class, M\|S = math given speech a. What is the probability that Felicity enrolls in math and speech? Find P(M AND S) = P(M\|S)P(S). b. What is the probability that Felicity enrolls in math or speech classes? Find P(M OR S) = P(M) + P(S) - P(M AND S). c. Are M and S independent? Is P(M\|S) = P(M)? d. Are M and S mutually exclusive? Is P(M AND S) = 0? Solution 3.17 a. 0.1625, b. 0.6875, c. No, d. No 3.17 A student goes to the library. Let events B = the student checks out a book and D = the student check out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D\|B) = 0.5. a. Find P(B AND D). b. Find P(B OR D). Example 3.18 Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random. a. What is the probability that the woman develops breast cancer
? What is the probability that woman tests negative? Solution 3.18 a. P(B) = 0.143; P(N) = 0.85 b. Given that the woman has breast cancer, what is the probability that she tests negative? Solution 3.18 b. P(N\|B) = 0.02 c. What is the probability that the woman has breast cancer AND tests negative? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 179 Solution 3.18 c. P(B AND N) = P(B)P(N\|B) = (0.143)(0.02) = 0.0029 d. What is the probability that the woman has breast cancer or tests negative? Solution 3.18 d. P(B OR N) = P(B) + P(N) - P(B AND N) = 0.143 + 0.85 - 0.0029 = 0.9901 e. Are having breast cancer and testing negative independent events? Solution 3.18 e. No. P(N) = 0.85; P(N\|B) = 0.02. So, P(N\|B) does not equal P(N). f. Are having breast cancer and testing negative mutually exclusive? Solution 3.18 f. No. P(B AND N) = 0.0029. For B and N to be mutually exclusive, P(B AND N) must be zero. 3.18 A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is going to college and plays sports? Example 3.19 Refer to the information in Example 3.18. P = tests positive. a. Given that a woman develops breast cancer, what is the probability that she tests positive. Find P(P\|B) = 1 - P(N\|B). b. What is the probability that a woman develops breast cancer and tests positive. Find P(B AND P) = P(P\|B)P(B).
c. What is the probability that a woman does not develop breast cancer. Find P(B′) = 1 - P(B). d. What is the probability that a woman tests positive for breast cancer. Find P(P) = 1 - P(N). Solution 3.19 a. 0.98; b. 0.1401; c. 0.857; d. 0.15 3.19 A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D\|B) = 0.5. a. Find P(B′). b. Find P(D AND B). c. Find P(B\|D). d. Find P(D AND B′). e. Find P(D\|B′). 180 CHAPTER 3 \| PROBABILITY TOPICS 3.4 \| Contingency Tables A contingency table provides a way of portraying data that can facilitate calculating probabilities. The table helps in determining conditional probabilities quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. Later on, we will use contingency tables again, but in another manner. Example 3.20 Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data: Speeding violation in the last year No speeding violation in the last year Cell phone user Not a cell phone user Total Table 3.2 25 45 70 280 405 685 Total 305 450 755 The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755. Calculate the following probabilities using the table. a. Find P(Person is a car phone user). Solution 3.20 a. number of car phone users total number in study = 305 755 b. Find P(person had no violation in the last year). Solution 3.20 b. number that had no violation total number in study = 685 755 c. Find P(Person had no violation in the last year AND was a car phone user). Solution 3.20 c. 280 755 d. Find P(Person is a car phone user OR person had no violation in the last year). Solution 3.20 d. ⎛ +
685 ⎝ 755 305 755 ⎞ ⎠ − 280 755 = 710 755 e. Find P(Person is a car phone user GIVEN person had a violation in the last year). Solution 3.20 e. 25 70 (The sample space is reduced to the number of persons who had a violation.) This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 181 f. Find P(Person had no violation last year GIVEN person was not a car phone user) Solution 3.20 f. 405 450 (The sample space is reduced to the number of persons who were not car phone users.) 3.20 Table 3.3 shows the number of athletes who stretch before exercising and how many had injuries within the past year. Injury in last year No injury in last year Total Stretches 55 Does not stretch 231 Total 286 Table 3.3 295 219 514 350 450 800 a. What is P(athlete stretches before exercising)? b. What is P(athlete stretches before exercising\|no injury in the last year)? Example 3.21 Table 3.4 shows a random sample of 100 hikers and the areas of hiking they prefer. Sex The Coastline Near Lakes and Streams On Mountain Peaks Total Female 18 Male Total ___ ___ 16 ___ 41 Table 3.4 Hiking Area Preference ___ 14 ___ 45 55 ___ a. Complete the table. Solution 3.21 a. Sex The Coastline Near Lakes and Streams On Mountain Peaks Total Female 18 Male Total 16 34 16 25 41 Table 3.5 Hiking Area Preference 11 14 25 45 55 100 b. Are the events "being female" and "preferring the coastline" independent events? Let F = being female and let C = preferring the coastline. 182 CHAPTER 3 \| PROBABILITY TOPICS 1. Find P(F AND C). 2. Find P(F)P(C) Are these two numbers the same? If they are, then F and C are independent. If they are not, then F and C are not independent. Solution 3.21 b. 1. P(F AND C) = 18 100 = 0.18 2. P(F)P(C) = ⎛ ⎝ 45 100 ⎛ ⎞ ⎝ �
�� 34 100 ⎞ ⎠ = (0.45)(0.34) = 0.153 P(F AND C) ≠ P(F)P(C), so the events F and C are not independent. c. Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let M = being male, and let L = prefers hiking near lakes and streams. 1. What word tells you this is a conditional? 2. Fill in the blanks and calculate the probability: P(___\|___) = ___. 3. Is the sample space for this problem all 100 hikers? If not, what is it? Solution 3.21 c. 1. The word 'given' tells you that this is a conditional. 2. P(M\|L) = 25 41 3. No, the sample space for this problem is the 41 hikers who prefer lakes and streams. d. Find the probability that a person is female or prefers hiking on mountain peaks. Let F = being female, and let P = prefers mountain peaks. 1. Find P(F). 2. Find P(P). 3. Find P(F AND P). 4. Find P(F OR P). Solution 3.21 d. 1. P(F) = 45 100 2. P(P) = 25 100 3. P(F AND P) = 11 100 4. P(F OR P) = 45 100 + 25 100 - 11 100 = 59 100 3.21 Table 3.6 shows a random sample of 200 cyclists and the routes they prefer. Let M = males and H = hilly path. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 183 Gender Lake Path Hilly Path Wooded Path Total Female Male Total 45 26 71 Table 3.6 38 52 90 27 12 39 110 90 200 a. Out of the males, what is the probability that the cyclist prefers a hilly path? b. Are the events “being male” and “preferring the hilly path” independent events? Example 3.22 Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is 1 5 and the probability he is not caught
is 4 5. If he goes out the second door, the probability he gets caught by Alissa is 1 4 and the probability he is not caught is 3 4. The probability that Alissa catches Muddy coming out of the third door is 1 2 and the probability she does not catch Muddy is 1 2 Muddy will choose any of the three doors so the probability of choosing each door is 1 3. It is equally likely that. Caught or Not Door One Door Two Door Three Total Caught Not Caught 1 15 4 15 1 12 3 12 1 6 1 6 Total ____ ____ ____ ____ ____ 1 Table 3.7 Door Choice • The first entry 1 15 = ⎛ ⎝ is P(Door One AND Caught) • The entry 4 15 = ⎛ ⎝ Verify the remaining entries. is P(Door One AND Not Caught) a. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1. Solution 3.22 a. Caught or Not Door One Door Two Door Three Total Caught 1 15 1 12 1 6 19 60 Table 3.8 Door Choice 184 CHAPTER 3 \| PROBABILITY TOPICS Caught or Not Door One Door Two Door Three Total Not Caught Total 4 15 5 15 Table 3.8 Door Choice 3 12 4 12 1 6 2 6 41 60 1 b. What is the probability that Alissa does not catch Muddy? Solution 3.22 b. 41 60 c. What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa? Solution 3.22 c. 9 19 Example 3.23 Table 3.9 contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the U.S. Year Robbery Burglary Rape Vehicle Total 2008 145.7 2009 133.1 2010 119.3 2011 113.7 Total 732.1 717.7 701 702.2 29.7 29.1 27.7 26.8 314.7 259.2 239.1 229.6 Table 3.9 United States Crime Index Rates Per 100,000 Inhabitants 2008–2011 TOTAL each column and each row. Total data = 4,520.7 a. Find P(2009 AND Robbery). b. Find P(2010 AND Burglary). c. Find P(2010 OR Burglary). d. Find P(2011
\|Rape). e. Find P(Vehicle\|2008). Solution 3.23 a. 0.0294, b. 0.1551, c. 0.7165, d. 0.2365, e. 0.2575 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 185 3.23 Table 3.10 relates the weights and heights of a group of individuals participating in an observational study. Weight/Height Tall Medium Short Totals 18 20 12 28 51 25 14 28 9 Obese Normal Underweight Totals Table 3.10 a. Find the total for each row and column b. Find the probability that a randomly chosen individual from this group is Tall. c. Find the probability that a randomly chosen individual from this group is Obese and Tall. d. Find the probability that a randomly chosen individual from this group is Tall given that the idividual is Obese. e. Find the probability that a randomly chosen individual from this group is Obese given that the individual is Tall. f. Find the probability a randomly chosen individual from this group is Tall and Underweight. g. Are the events Obese and Tall independent? 3.5 \| Tree and Venn Diagrams Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities. Tree Diagrams A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram. Example 3.24 In an urn, there are 11 balls. Three balls are red (R) and eight balls are blue (B). Draw two balls, one at a time, with replacement. "With replacement" means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows. 186 CHAPTER 3 \| PROBABILITY TOPICS Figure 3.2 Total = 64 + 24 + 24 + 9 = 121 The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes
is distinct. In fact, we can list each red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the nine RR outcomes can be written as: R1R1; R1R2; R1R3; R2R1; R2R2; R2R3; R3R1; R3R2; R3R3 The other outcomes are similar. There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space. a. List the 24 BR outcomes: B1R1, B1R2, B1R3,... Solution 3.24 a. B1R1; B1R2; B1R3; B2R1; B2R2; B2R3; B3R1; B3R2; B3R3; B4R1; B4R2; B4R3; B5R1; B5R2; B5R3; B6R1; B6R2; B6R3; B7R1; B7R2; B7R3; B8R1; B8R2; B8R3 b. Using the tree diagram, calculate P(RR). Solution 3.24 b. P(RR) = ⎛ 3 ⎝ 11 ⎞ ⎛ ⎝ ⎠ ⎞ ⎠ 3 11 = 9 121 c. Using the tree diagram, calculate P(RB OR BR). Solution 3.24 c. P(RB OR BR) = ⎛ ⎝ 3 11 ⎞ ⎛ ⎝ ⎠ ⎞ ⎠ 8 11 + ⎛ ⎝ 8 11 ⎞ ⎛ ⎝ ⎠ ⎞ ⎠ 3 11 = 48 121 d. Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw). Solution 3.24 d. P(R on 1st draw AND B on 2nd draw) = P(RB) = ⎛ ⎝ 3 11 ⎛ ⎞ ⎝ ⎠
⎞ ⎠ 8 11 = 24 121 e. Using the tree diagram, calculate P(R on 2nd draw GIVEN B on 1st draw). Solution 3.24 e. P(R on 2nd draw GIVEN B on 1st draw) = P(R on 2nd\|B on 1st) = 24 88 = 3 11 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 187 This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB). Twenty-four of the 88 possible outcomes are BR. 24 88 = 3 11. f. Using the tree diagram, calculate P(BB). Solution 3.24 f. P(BB) = 64 121 g. Using the tree diagram, calculate P(B on the 2nd draw given R on the first draw). Solution 3.24 g. P(B on 2nd draw\|R on 1st draw) = 8 11 There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B on the second draw. The probability is then 24 33. 3.24 In a standard deck, there are 52 cards. 12 cards are face cards (event F) and 40 cards are not face cards (event N). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate P(FF). Figure 3.3 Example 3.25 An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. "Without replacement" means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the ⎞ two corresponding branches, for example, ⎛ ⎠ = 6 ⎝ 110 3 11 2 10 ⎛ ⎞ ⎠ ⎝
. 188 CHAPTER 3 \| PROBABILITY TOPICS Figure 3.4 Total = 56 + 24 + 24 + 6 110 = 110 110 = 1 NOTE If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement, so that on the second draw there are ten marbles left in the urn. Calculate the following probabilities using the tree diagram. a. P(RR) = ________ Solution 3.25 a. P(RR) = ⎛ 3 ⎝ 11 ⎛ ⎞ ⎠ ⎝ 2 10 ⎞ ⎠ = 6 110 b. Fill in the blanks: P(RB OR BR) = ⎛ ⎝ 3 11 ⎛ ⎞ ⎠ ⎝ 8 10 ⎞ ⎠ + (___)(___) = 48 110 Solution 3.25 b. P(RB OR BR) = ⎛ ⎝ 3 11 ⎛ ⎞ ⎠ ⎝ ⎞ ⎠ 8 10 + ⎛ ⎝ 8 11 ⎛ ⎞ ⎠ ⎝ ⎞ ⎠ 3 10 = 48 110 c. P(R on 2nd\|B on 1st) = Solution 3.25 c. P(R on 2nd\|B on 1st) = 3 10 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 189 d. Fill in the blanks. P(R on 1st AND B on 2nd) = P(RB) = (___)(___) = 24 100 Solution 3.25 d. P(R on 1st AND B on 2nd) = P(RB) = ⎛ ⎝ 3 11 ⎛ ⎞ ⎠ ⎝ ⎞ ⎠ 8 10 = 24 100 e. Find P(BB). Solution 3.25 e. P(BB) = ⎛ 8 ⎝ 11 ⎛ ⎞ ⎠ ⎝ ⎞ ⎠ 7 10 f. Find P
(B on 2nd\|R on 1st). Solution 3.25 f. Using the tree diagram, P(B on 2nd\|R on 1st) = P(R\|B) = 8 10. If we are using probabilities, we can label the tree in the following general way. • P(R\|R) here means P(R on 2nd\|R on 1st) • P(B\|R) here means P(B on 2nd\|R on 1st) • P(R\|B) here means P(R on 2nd\|B on 1st) • P(B\|B) here means P(B on 2nd\|B on 1st) 3.25 In a standard deck, there are 52 cards. Twelve cards are face cards (F) and 40 cards are not face cards (N). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities. 190 CHAPTER 3 \| PROBABILITY TOPICS Figure 3.5 a. Find P(FN OR NF). b. Find P(N\|F). c. Find P(at most one face card). Hint: "At most one face card" means zero or one face card. d. Find P(at least on face card). Hint: "At least one face card" means one or two face cards. Example 3.26 A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 191 a. What is the probability that both kittens are tabby? a. What is the probability that one kitten of each coloring is selected? a. What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first? d. What is the probability of choosing two kittens of the same color? Solution 3.26 a. c, b. d, c. 4 8, d. 32 72 3.26 Suppose there are four red balls and three yellow balls in a box. Three balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected? Venn Di
agram A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. Example 3.27 Suppose an experiment has the outcomes 1, 2, 3,..., 12 where each outcome has an equal chance of occurring. Let event A = {1, 2, 3, 4, 5, 6} and event B = {6, 7, 8, 9}. Then A AND B = {6} and A OR B = {1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is as follows: Figure 3.6 3.27 Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event C = {green, blue, purple} and event P = {red, yellow, blue}. Then C AND P = {blue} and C OR P = {green, blue, purple, red, yellow}. Draw a Venn diagram representing this situation. 192 CHAPTER 3 \| PROBABILITY TOPICS Example 3.28 Flip two fair coins. Let A = tails on the first coin. Let B = tails on the second coin. Then A = {TT, TH} and B = {TT, HT}. Therefore, A AND B = {TT}. A OR B = {TH, TT, HT}. The sample space when you flip two fair coins is X = {HH, HT, TH, TT}. The outcome HH is in NEITHER A NOR B. The Venn diagram is as follows: Figure 3.7 3.28 Roll a fair, six-sided die. Let A = a prime number of dots is rolled. Let B = an odd number of dots is rolled. Then A = {2, 3, 5} and B = {1, 3, 5}. Therefore, A AND B = {3, 5}. A OR B = {1, 2, 3, 5}. The sample space for rolling a fair die is S = {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation. Example 3.29 Forty percent of the students at a local college belong to a club and 50% work part time. Five percent of the students work part time and belong to a club. Draw a Venn diagram showing the relationships
. Let C = student belongs to a club and PT = student works part time. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 193 Figure 3.8 If a student is selected at random, find • • • • • the probability that the student belongs to a club. P(C) = 0.40 the probability that the student works part time. P(PT) = 0.50 the probability that the student belongs to a club AND works part time. P(C AND PT) = 0.05 the probability that P(C\|PT) = the student belongs = 0.05 0.50 P(C AND PT) P(PT) to a club given that the student works part time. = 0.1 the probability that the student belongs to a club OR works part time. P(C OR PT) = P(C) + P(PT) - P(C AND PT) = 0.40 + 0.50 - 0.05 = 0.85 3.29 Fifty percent of the workers at a factory work a second job, 25% have a spouse who also works, 5% work a second job and have a spouse who also works. Draw a Venn diagram showing the relationships. Let W = works a second job and S = spouse also works. Example 3.30 A person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any blood type. Four percent of African Americans have type O blood and a negative RH factor, 5−10% of African Americans have the Rh- factor, and 51% have type O blood. 194 CHAPTER 3 \| PROBABILITY TOPICS Figure 3.9 The “O” circle represents the African Americans with type O blood. The “Rh-“ oval represents the African Americans with the Rh- factor. We will take the average of 5% and 10% and use 7.5% as the percent of African Americans who have the Rhfactor. Let O = African American with Type O blood and R = African American with Rh- factor. a. P(O) = ___________ b. P(R) = ___________ c. P(O AND R) = ___________ d. P(O OR R) = ____________ e.
f. In the Venn Diagram, describe the overlapping area using a complete sentence. In the Venn Diagram, describe the area in the rectangle but outside both the circle and the oval using a complete sentence. Solution 3.30 a. 0.51; b. 0.075; c. 0.04; d. 0.545; e. The area represents the African Americans that have type O blood and the Rh- factor. f. The area represents the African Americans that have neither type O blood nor the Rh- factor. 3.30 In a bookstore, the probability that the customer buys a novel is 0.6, and the probability that the customer buys a non-fiction book is 0.4. Suppose that the probability that the customer buys both is 0.2. a. Draw a Venn diagram representing the situation. b. Find the probability that the customer buys either a novel or anon-fiction book. c. In the Venn diagram, describe the overlapping area using a complete sentence. d. Suppose that some customers buy only compact disks. Draw an oval in your Venn diagram representing this event. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 195 3.1 Probability Topics Class time: Names: Student Learning Outcomes • The student will use theoretical and empirical methods to estimate probabilities. • The student will appraise the differences between the two estimates. • The student will demonstrate an understanding of long-term relative frequencies. Do the Experiment Count out 40 mixed-color M&Ms® which is approximately one small bag’s worth. Record the number of each color in Table 3.11. Use the information from this table to complete Table 3.12. Next, put the M&Ms in a cup. The experiment is to pick two M&Ms, one at a time. Do not look at them as you pick them. The first time through, replace the first M&M before picking the second one. Record the results in the “With Replacement” column of Table 3.13. Do this 24 times. The second time through, after picking the first M&M, do not replace it before picking the second one. Then, pick the second one. Record the results in the “Without Replacement” column section of Table 3.14.
After you record the pick, put both M&Ms back. Do this a total of 24 times, also. Use the data from Table 3.14 to calculate the empirical probability questions. Leave your answers in unreduced fractional form. Do not multiply out any fractions. Color Quantity Yellow (Y) Green (G) Blue (BL) Brown (B) Orange (O) Red (R) Table 3.11 Population With Replacement Without Replacement P(2 reds) P(R1B2 OR B1R2) P(R1 AND G2) P(G2\|R1) P(no yellows) P(doubles) P(no doubles) Table 3.12 Theoretical Probabilities 196 CHAPTER 3 \| PROBABILITY TOPICS NOTE G2 = green on second pick; R1 = red on first pick; B1 = brown on first pick; B2 = brown on second pick; doubles = both picks are the same colour. With Replacement Without Replacement ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) ( __, __ ) Table 3.13 Empirical Results With Replacement Without Replacement P(2 reds) P(R1B2 OR B1R2) P(R1 AND G2) P(G2\|R1) P(no yellows) P(doubles) P(no doubles) Table
3.14 Empirical Probabilities Discussion Questions 1. Why are the “With Replacement” and “Without Replacement” probabilities different? 2. Convert P(no yellows) to decimal format for both Theoretical “With Replacement” and for Empirical “With Replacement”. Round to four decimal places. a. Theoretical “With Replacement”: P(no yellows) = _______ b. Empirical “With Replacement”: P(no yellows) = _______ c. Are the decimal values “close”? Did you expect them to be closer together or farther apart? Why? 3. If you increased the number of times you picked two M&Ms to 240 times, why would empirical probability values change? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 4. Would this change (see part 3) cause the empirical probabilities and theoretical probabilities to be closer together or farther apart? How do you know? 5. Explain the differences in what P(G1 AND R2) and P(R1\|G2) represent. Hint: Think about the sample space for each probability. CHAPTER 3 \| PROBABILITY TOPICS 197 198 CHAPTER 3 \| PROBABILITY TOPICS KEY TERMS Conditional Probability the likelihood that an event will occur given that another event has already occurred contingency table the method of displaying a frequency distribution as a table with rows and columns to show how two variables may be dependent (contingent) upon each other; the table provides an easy way to calculate conditional probabilities. Dependent Events If two events are NOT independent, then we say that they are dependent. Equally Likely Each outcome of an experiment has the same probability. Event a subset of the set of all outcomes of an experiment; the set of all outcomes of an experiment is called a sample space and is usually denoted by S. An event is an arbitrary subset in S. It can contain one outcome, two outcomes, no outcomes (empty subset), the entire sample space, and the like. Standard notations for events are capital letters such as A, B, C, and so on. Experiment a planned activity carried out under controlled conditions Independent Events The occurrence of one event has no effect on the probability of the occurrence of another event. Events A
and B are independent if one of the following is true: 1. P(A\|B) = P(A) 2. P(B\|A) = P(B) 3. P(A AND B) = P(A)P(B) Mutually Exclusive Two events are mutually exclusive if the probability that they both happen at the same time is zero. If events A and B are mutually exclusive, then P(A AND B) = 0. Outcome a particular result of an experiment Probability a number between zero and one, inclusive, that gives the likelihood that a specific event will occur; the foundation of statistics is given by the following 3 axioms (by A.N. Kolmogorov, 1930’s): Let S denote the sample space and A and B are two events in S. Then: • 0 ≤ P(A) ≤ 1 • If A and B are any two mutually exclusive events, then P(A OR B) = P(A) + P(B). • P(S) = 1 Sample Space the set of all possible outcomes of an experiment Sampling with Replacement If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. Sampling without Replacement When sampling is done without replacement, each member of a population may be chosen only once. The AND Event An outcome is in the event A AND B if the outcome is in both A AND B at the same time. The Complement Event The complement of event A consists of all outcomes that are NOT in A. The Conditional Probability of One Event Given Another Event P(A\|B) is the probability that event A will occur given that the event B has already occurred. The Conditional Probability of A GIVEN B P(A\|B) is the probability that event A will occur given that the event B has already occurred. The OR of Two Events An outcome is in the event A OR B if the outcome is in A, is in B, or is in both A and B. The Or Event An outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B. Tree Diagram the useful visual representation of a sample space and events in the form of a “tree” with branches marked by possible outcomes together with associated probabilities (frequencies, relative frequencies) This content is available for free at http://textbooke
quity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 Venn Diagram the visual representation of a sample space and events in the form of circles or ovals showing their CHAPTER 3 \| PROBABILITY TOPICS 199 intersections CHAPTER REVIEW 3.1 Terminology In this module we learned the basic terminology of probability. The set of all possible outcomes of an experiment is called the sample space. Events are subsets of the sample space, and they are assigned a probability that is a number between zero and one, inclusive. 3.2 Independent and Mutually Exclusive Events Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent. In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other. 3.3 Two Basic Rules of Probability The multiplication rule and the addition rule are used for computing the probability of A and B, as well as the probability of A or B for two given events A, B defined on the sample space. In sampling with replacement each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events A and B are mutually exclusive events when they do not have any outcomes in common. 3.4 Contingency Tables There are several tools you can use to help organize and sort data when calculating probabilities. Contingency tables help display data and are particularly useful when calculating probabilites that have multiple dependent variables. 3.5 Tree and Venn Diagrams A tree diagram use branches to show the different outcomes of experiments and makes complex probability questions easy to visualize. A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. A Venn diagram is especially helpful for visualizing the
OR event, the AND event, and the complement of an event and for understanding conditional probabilities. FORMULA REVIEW 3.1 Terminology A and B are events P(S) = 1 where S is the sample space 0 ≤ P(A) ≤ 1 P(A\|B) = P(AANDB) P(B) 3.2 Independent and Mutually Exclusive Events PRACTICE 3.1 Terminology If A and B are independent, P(A AND B) = P(A)P(B), P(A\|B) = P(A) and P(B\|A) = P(B). If A and B are mutually exclusive, P(A OR B) = P(A) + P(B) and P(A AND B) = 0. 3.3 Two Basic Rules of Probability The multiplication rule: P(A AND B) = P(A\|B)P(B) The addition rule: P(A OR B) = P(A) + P(B) - P(A AND B) 200 CHAPTER 3 \| PROBABILITY TOPICS 1. In a particular college class, there are male and female students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events for parts a through j. (Note that you cannot find numerical answers here. You were not given enough information to find any probability values yet; concentrate on understanding the symbols.) • Let F be the event that a student is female. • Let M be the event that a student is male. • Let S be the event that a student has short hair. • Let L be the event that a student has long hair. a. The probability that a student does not have long hair. b. The probability that a student is male or has short hair. c. The probability that a student is a female and has long hair. d. The probability that a student is male, given that the student has long hair. e. The probability that a student has long hair, given that the student is male. f. Of all the female students, the probability that a student has short hair. g. Of all students with long hair, the probability that a student is female. h. The probability that a student is female or has long hair. i. The probability that a randomly selected student is a male student with short hair. j. The probability that a student is female. Use the following
information to answer the next four exercises. A box is filled with several party favors. It contains 12 hats, 15 noisemakers, ten finger traps, and five bags of confetti. Let H = the event of getting a hat. Let N = the event of getting a noisemaker. Let F = the event of getting a finger trap. Let C = the event of getting a bag of confetti. 2. Find P(H). 3. Find P(N). 4. Find P(F). 5. Find P(C). Use the following information to answer the next six exercises. A jar of 150 jelly beans contains 22 red jelly beans, 38 yellow, 20 green, 28 purple, 26 blue, and the rest are orange. Let B = the event of getting a blue jelly bean Let G = the event of getting a green jelly bean. Let O = the event of getting an orange jelly bean. Let P = the event of getting a purple jelly bean. Let R = the event of getting a red jelly bean. Let Y = the event of getting a yellow jelly bean. 6. Find P(B). 7. Find P(G). 8. Find P(P). 9. Find P(R). 10. Find P(Y). 11. Find P(O). Use the following information to answer the next six exercises. There are 23 countries in North America, 12 countries in South America, 47 countries in Europe, 44 countries in Asia, 54 countries in Africa, and 14 in Oceania (Pacific Ocean region). Let A = the event that a country is in Asia. Let E = the event that a country is in Europe. Let F = the event that a country is in Africa. Let N = the event that a country is in North America. Let O = the event that a country is in Oceania. Let S = the event that a country is in South America. 12. Find P(A). 13. Find P(E). This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 201 14. Find P(F). 15. Find P(N). 16. Find P(O). 17. Find P(S). 18. What is the probability of drawing a red card in a standard deck of 52 cards? 19. What is the
probability of drawing a club in a standard deck of 52 cards? 20. What is the probability of rolling an even number of dots with a fair, six-sided die numbered one through six? 21. What is the probability of rolling a prime number of dots with a fair, six-sided die numbered one through six? Use the following information to answer the next two exercises. You see a game at a local fair. You have to throw a dart at a color wheel. Each section on the color wheel is equal in area. Figure 3.10 Let B = the event of landing on blue. Let R = the event of landing on red. Let G = the event of landing on green. Let Y = the event of landing on yellow. 22. If you land on Y, you get the biggest prize. Find P(Y). 23. If you land on red, you don’t get a prize. What is P(R)? Use the following information to answer the next ten exercises. On a baseball team, there are infielders and outfielders. Some players are great hitters, and some players are not great hitters. Let I = the event that a player in an infielder. Let O = the event that a player is an outfielder. Let H = the event that a player is a great hitter. Let N = the event that a player is not a great hitter. 24. Write the symbols for the probability that a player is not an outfielder. 25. Write the symbols for the probability that a player is an outfielder or is a great hitter. 26. Write the symbols for the probability that a player is an infielder and is not a great hitter. 27. Write the symbols for the probability that a player is a great hitter, given that the player is an infielder. 28. Write the symbols for the probability that a player is an infielder, given that the player is a great hitter. 29. Write the symbols for the probability that of all the outfielders, a player is not a great hitter. 30. Write the symbols for the probability that of all the great hitters, a player is an outfielder. 31. Write the symbols for the probability that a player is an infielder or is not a great hitter. 202 CHAPTER 3 \| PROBABILITY TOPICS 32. Write the symbols for the probability that a player is an outfielder and is a great hitter. 33. Write the symbols for the probability that a player is an infielder. 34. What is the word for the
set of all possible outcomes? 35. What is conditional probability? 36. A shelf holds 12 books. Eight are fiction and the rest are nonfiction. Each is a different book with a unique title. The fiction books are numbered one to eight. The nonfiction books are numbered one to four. Randomly select one book Let F = event that book is fiction Let N = event that book is nonfiction What is the sample space? 37. What is the sum of the probabilities of an event and its complement? Use the following information to answer the next two exercises. You are rolling a fair, six-sided number cube. Let E = the event that it lands on an even number. Let M = the event that it lands on a multiple of three. 38. What does P(E\|M) mean in words? 39. What does P(E OR M) mean in words? 3.2 Independent and Mutually Exclusive Events 40. E and F are mutually exclusive events. P(E) = 0.4; P(F) = 0.5. Find P(E∣F). 41. J and K are independent events. P(J\|K) = 0.3. Find P(J). 42. U and V are mutually exclusive events. P(U) = 0.26; P(V) = 0.37. Find: a. P(U AND V) = b. P(U\|V) = c. P(U OR V) = 43. Q and R are independent events. P(Q) = 0.4 and P(Q AND R) = 0.1. Find P(R). 3.3 Two Basic Rules of Probability Use the following information to answer the next ten exercises. Forty-eight percent of all Californians registered voters prefer life in prison without parole over the death penalty for a person convicted of first degree murder. Among Latino California registered voters, 55% prefer life in prison without parole over the death penalty for a person convicted of first degree murder. 37.6% of all Californians are Latino. In this problem, let: • C = Californians (registered voters) preferring life in prison without parole over the death penalty for a person convicted of first degree murder. • L = Latino Californians Suppose that one Californian is randomly selected. 44. Find P(C). 45. Find P(L). 46. Find P(C\|L). 47. In words, what is C\|
L? 48. Find P(L AND C). 49. In words, what is L AND C? 50. Are L and C independent events? Show why or why not. 51. Find P(L OR C). 52. In words, what is L OR C? 53. Are L and C mutually exclusive events? Show why or why not. 3.4 Contingency Tables Use the following information to answer the next four exercises. Table 3.15 shows a random sample of musicians and how they learned to play their instruments. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 203 Gender Self-taught Studied in School Private Instruction Total Female Male Total 12 19 31 Table 3.15 38 24 62 22 15 37 72 58 130 54. Find P(musician is a female). 55. Find P(musician is a male AND had private instruction). 56. Find P(musician is a female OR is self taught). 57. Are the events “being a female musician” and “learning music in school” mutually exclusive events? 3.5 Tree and Venn Diagrams 58. The probability that a man develops some form of cancer in his lifetime is 0.4567. The probability that a man has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51. Let: C = a man develops cancer in his lifetime; P = man has at least one false positive. Construct a tree diagram of the situation. BRINGING IT TOGETHER: PRACTICE Use the following information to answer the next seven exercises. An article in the New England Journal of Medicine, reported about a study of smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans, and 7,650 Whites. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 Whites. Of the people smoking 21 to
30 cigarettes per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 Whites. Of the people smoking at least 31 cigarettes per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 Whites. 59. Complete the table using the data provided. Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day. Smoking Level African American Native Hawaiian Latino Japanese Americans White TOTALS 1–10 11–20 21–30 31+ TOTALS Table 3.16 Smoking Levels by Ethnicity 60. Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day. 61. Find the probability that the person was Latino. 62. In words, explain what it means to pick one person from the study who is “Japanese American AND smokes 21 to 30 cigarettes per day.” Also, find the probability. 63. In words, explain what it means to pick one person from the study who is “Japanese American OR smokes 21 to 30 cigarettes per day.” Also, find the probability. 64. In words, explain what it means to pick one person from the study who is “Japanese American GIVEN that person smokes 21 to 30 cigarettes per day.” Also, find the probability. 65. Prove that smoking level/day and ethnicity are dependent events. 204 CHAPTER 3 \| PROBABILITY TOPICS HOMEWORK 3.1 Terminology 66. Figure 3.11 The graph in Figure 3.11 displays the sample sizes and percentages of people in different age and gender groups who were polled concerning their approval of Mayor Ford’s actions in office. The total number in the sample of all the age groups is 1,045. a. Define three events in the graph. b. Describe in words what the entry 40 means. c. Describe in words the complement of the entry in question 2. d. Describe in words what the entry 30 means. e. Out of the males and females, what percent are males? f. Out of the females, what percent disapprove of Mayor Ford? g. Out of all the age groups, what percent approve of Mayor Ford? h. Find P(Approve\|Male). i. Out of the age groups,
what percent are more than 44 years old? j. Find P(Approve\|Age < 35). 67. Explain what is wrong with the following statements. Use complete sentences. a. If there is a 60% chance of rain on Saturday and a 70% chance of rain on Sunday, then there is a 130% chance of rain over the weekend. b. The probability that a baseball player hits a home run is greater than the probability that he gets a successful hit. 3.2 Independent and Mutually Exclusive Events Use the following information to answer the next 12 exercises. The graph shown is based on more than 170,000 interviews done by Gallup that took place from January through December 2012. The sample consists of employed Americans 18 years of age or older. The Emotional Health Index Scores are the sample space. We randomly sample one Emotional Health Index Score. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 205 Figure 3.12 68. Find the probability that an Emotional Health Index Score is 82.7. 69. Find the probability that an Emotional Health Index Score is 81.0. 70. Find the probability that an Emotional Health Index Score is more than 81? 71. Find the probability that an Emotional Health Index Score is between 80.5 and 82? 72. If we know an Emotional Health Index Score is 81.5 or more, what is the probability that it is 82.7? 73. What is the probability that an Emotional Health Index Score is 80.7 or 82.7? 74. What is the probability that an Emotional Health Index Score is less than 80.2 given that it is already less than 81. 75. What occupation has the highest emotional index score? 76. What occupation has the lowest emotional index score? 77. What is the range of the data? 78. Compute the average EHIS. 79. If all occupations are equally likely for a certain individual, what is the probability that he or she will have an occupation with lower than average EHIS? 3.3 Two Basic Rules of Probability 80. On February 28, 2013, a Field Poll Survey reported that 61% of California registered voters approved of allowing two people of the same gender to marry and have regular marriage laws apply to them. Among 18 to 39 year olds (
California registered voters), the approval rating was 78%. Six in ten California registered voters said that the upcoming Supreme Court’s ruling about the constitutionality of California’s Proposition 8 was either very or somewhat important to them. Out of those CA registered voters who support same-sex marriage, 75% say the ruling is important to them. In this problem, let: • C = California registered voters who support same-sex marriage. • B = California registered voters who say the Supreme Court’s ruling about the constitutionality of California’s Proposition 8 is very or somewhat important to them • A = California registered voters who are 18 to 39 years old. a. Find P(C). b. Find P(B). c. Find P(C\|A). d. Find P(B\|C). e. In words, what is C\|A? 206 CHAPTER 3 \| PROBABILITY TOPICS In words, what is B\|C? f. g. Find P(C AND B). h. i. Find P(C OR B). j. Are C and B mutually exclusive events? Show why or why not. In words, what is C AND B? 81. After Rob Ford, the mayor of Toronto, announced his plans to cut budget costs in late 2011, the Forum Research polled 1,046 people to measure the mayor’s popularity. Everyone polled expressed either approval or disapproval. These are the results their poll produced: • • • In early 2011, 60 percent of the population approved of Mayor Ford’s actions in office. In mid-2011, 57 percent of the population approved of his actions. In late 2011, the percentage of popular approval was measured at 42 percent. a. What is the sample size for this study? b. What proportion in the poll disapproved of Mayor Ford, according to the results from late 2011? c. How many people polled responded that they approved of Mayor Ford in late 2011? d. What is the probability that a person supported Mayor Ford, based on the data collected in mid-2011? e. What is the probability that a person supported Mayor Ford, based on the data collected in early 2011? Use the following information to answer the next three exercises. The casino game, roulette, allows the gambler to bet on the probability of a ball, which spins in the roulette wheel, landing on a particular color, number, or range of numbers. The table used to place bets contains of 38 numbers, and each
number is assigned to a color and a range. Figure 3.13 (credit: film8ker/wikibooks) 82. a. List the sample space of the 38 possible outcomes in roulette. b. You bet on red. Find P(red). c. You bet on -1st 12- (1st Dozen). Find P(-1st 12-). d. You bet on an even number. Find P(even number). e. f. Find two mutually exclusive events. g. Are the events Even and 1st Dozen independent? Is getting an odd number the complement of getting an even number? Why? 83. Compute the probability of winning the following types of bets: a. Betting on two lines that touch each other on the table as in 1-2-3-4-5-6 b. Betting on three numbers in a line, as in 1-2-3 c. Betting on one number d. Betting on four numbers that touch each other to form a square, as in 10-11-13-14 e. Betting on two numbers that touch each other on the table, as in 10-11 or 10-13 f. Betting on 0-00-1-2-3 g. Betting on 0-1-2; or 0-00-2; or 00-2-3 84. Compute the probability of winning the following types of bets: a. Betting on a color b. Betting on one of the dozen groups c. Betting on the range of numbers from 1 to 18 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 d. Betting on the range of numbers 19–36 e. Betting on one of the columns f. Betting on an even or odd number (excluding zero) 85. Suppose that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card. CHAPTER 3 \| PROBABILITY TOPICS 207 • G = card drawn is green • E = card drawn is even-numbered a. List the sample space. b. P(G) = _____ c. P(G\|
E) = _____ d. P(G AND E) = _____ e. P(G OR E) = _____ f. Are G and E mutually exclusive? Justify your answer numerically. 86. Roll two fair dice. Each die has six faces. a. List the sample space. b. Let A be the event that either a three or four is rolled first, followed by an even number. Find P(A). c. Let B be the event that the sum of the two rolls is at most seven. Find P(B). d. e. Are A and B mutually exclusive events? Explain your answer in one to three complete sentences, including In words, explain what “P(A\|B)” represents. Find P(A\|B). numerical justification. f. Are A and B independent events? Explain your answer in one to three complete sentences, including numerical justification. 87. A special deck of cards has ten cards. Four are green, three are blue, and three are red. When a card is picked, its color of it is recorded. An experiment consists of first picking a card and then tossing a coin. a. List the sample space. b. Let A be the event that a blue card is picked first, followed by landing a head on the coin toss. Find P(A). c. Let B be the event that a red or green is picked, followed by landing a head on the coin toss. Are the events A and B mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification. d. Let C be the event that a red or blue is picked, followed by landing a head on the coin toss. Are the events A and C mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification. 88. An experiment consists of first rolling a die and then tossing a coin. a. List the sample space. b. Let A be the event that either a three or a four is rolled first, followed by landing a head on the coin toss. Find P(A). c. Let B be the event that the first and second tosses land on heads. Are the events A and B mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification. 89. An experiment consists of tossing a nickel, a dime, and a quarter. Of interest is the side the coin lands on. a. List the sample space. b. Let A be the event that
there are at least two tails. Find P(A). c. Let B be the event that the first and second tosses land on heads. Are the events A and B mutually exclusive? Explain your answer in one to three complete sentences, including justification. 90. Consider the following scenario: Let P(C) = 0.4. Let P(D) = 0.5. Let P(C\|D) = 0.6. a. Find P(C AND D). b. Are C and D mutually exclusive? Why or why not? c. Are C and D independent events? Why or why not? d. Find P(C OR D). e. Find P(D\|C). 91. Y and Z are independent events. a. Rewrite the basic Addition Rule P(Y OR Z) = P(Y) + P(Z) - P(Y AND Z) using the information that Y and Z are independent events. b. Use the rewritten rule to find P(Z) if P(Y OR Z) = 0.71 and P(Y) = 0.42. 92. G and H are mutually exclusive events. P(G) = 0.5 P(H) = 0.3 a. Explain why the following statement MUST be false: P(H\|G) = 0.4. b. Find P(H OR G). c. Are G and H independent or dependent events? Explain in a complete sentence. 208 CHAPTER 3 \| PROBABILITY TOPICS 93. Approximately 281,000,000 people over age five live in the United States. Of these people, 55,000,000 speak a language other than English at home. Of those who speak another language at home, 62.3% speak Spanish. Let: E = speaks English at home; E′ = speaks another language at home; S = speaks Spanish; Finish each probability statement by matching the correct answer. Probability Statements Answers a. P(E′) = b. P(E) = c. P(S and E′) = d. P(S\|E′) = Table 3.17 i. 0.8043 ii. 0.623 iii. 0.1957 iv. 0.1219 94. 1994, the U.S. government held a lottery to issue 55,000 Green Cards (permits for non-citizens to work legally in the U.S.). Renate Deutsch, from Germany, was
one of approximately 6.5 million people who entered this lottery. Let G = won green card. a. What was Renate’s chance of winning a Green Card? Write your answer as a probability statement. b. In the summer of 1994, Renate received a letter stating she was one of 110,000 finalists chosen. Once the finalists were chosen, assuming that each finalist had an equal chance to win, what was Renate’s chance of winning a Green Card? Write your answer as a conditional probability statement. Let F = was a finalist. c. Are G and F independent or dependent events? Justify your answer numerically and also explain why. d. Are G and F mutually exclusive events? Justify your answer numerically and explain why. 95. Three professors at George Washington University did an experiment to determine if economists are more selfish than other people. They dropped 64 stamped, addressed envelopes with $10 cash in different classrooms on the George Washington campus. 44% were returned overall. From the economics classes 56% of the envelopes were returned. From the business, psychology, and history classes 31% were returned. Let: R = money returned; E = economics classes; O = other classes a. Write a probability statement for the overall percent of money returned. b. Write a probability statement for the percent of money returned out of the economics classes. c. Write a probability statement for the percent of money returned out of the other classes. d. e. Based upon this study, do you think that economists are more selfish than other people? Explain why or why not. Is money being returned independent of the class? Justify your answer numerically and explain it. Include numbers to justify your answer. 96. The following table of data obtained from www.baseball-almanac.com shows hit information for four players. Suppose that one hit from the table is randomly selected. Name Single Double Triple Home Run Total Hits Babe Ruth 1,517 Jackie Robinson 1,054 Ty Cobb Hank Aaron Total Table 3.18 3,603 2,294 8,471 506 273 174 624 1,577 136 54 295 98 583 714 137 114 755 2,873 1,518 4,189 3,771 1,720 12,351 Are "the hit being made by Hank Aaron" and "the hit being a double" independent events? a. Yes, because P(hit by Hank Aaron\|hit is a double) = P(
hit by Hank Aaron) b. No, because P(hit by Hank Aaron\|hit is a double) ≠ P(hit is a double) c. No, because P(hit is by Hank Aaron\|hit is a double) ≠ P(hit by Hank Aaron) d. Yes, because P(hit is by Hank Aaron\|hit is a double) = P(hit is a double) This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 209 97. United Blood Services is a blood bank that serves more than 500 hospitals in 18 states. According to their website, a person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any bloodtype. Their data show that 43% of people have type O blood and 15% of people have Rh- factor; 52% of people have type O or Rh- factor. a. Find the probability that a person has both type O blood and the Rh- factor. b. Find the probability that a person does NOT have both type O blood and the Rh- factor. 98. At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 32% of courses have a research paper and a final exam. Let F be the event that a course has a final exam. Let R be the event that a course requires a research paper. a. Find the probability that a course has a final exam or a research project. b. Find the probability that a course has NEITHER of these two requirements. 99. In a box of assorted cookies, 36% contain chocolate and 12% contain nuts. Of those, 8% contain both chocolate and nuts. Sean is allergic to both chocolate and nuts. a. Find the probability that a cookie contains chocolate or nuts (he can't eat it). b. Find the probability that a cookie does not contain chocolate or nuts (he can eat it). 100. A college finds that 10% of students have taken a distance learning class and that 40% of students are part time students. Of the part time students, 20% have taken a distance learning class. Let D = event that a student takes a distance learning class and E = event that a student is a part time student a. Find P(D AND E). b. Find P
(E\|D). c. Find P(D OR E). d. Using an appropriate test, show whether D and E are independent. e. Using an appropriate test, show whether D and E are mutually exclusive. 3.4 Contingency Tables Use the information in the Table 3.19 to answer the next eight exercises. The table shows the political party affiliation of each of 67 members of the US Senate in June 2012, and when they are up for reelection. Up for reelection: Democratic Party Republican Party Other Total November 2014 November 2016 20 10 13 24 0 0 Total Table 3.19 101. What is the probability that a randomly selected senator has an “Other” affiliation? 102. What is the probability that a randomly selected senator is up for reelection in November 2016? 103. What is the probability that a randomly selected senator is a Democrat and up for reelection in November 2016? 104. What is the probability that a randomly selected senator is a Republican or is up for reelection in November 2014? 105. Suppose that a member of the US Senate is randomly selected. Given that the randomly selected senator is up for reelection in November 2016, what is the probability that this senator is a Democrat? 106. Suppose that a member of the US Senate is randomly selected. What is the probability that the senator is up for reelection in November 2014, knowing that this senator is a Republican? 107. The events “Republican” and “Up for reelection in 2016” are ________ independent. a. mutually exclusive. b. c. both mutually exclusive and independent. d. neither mutually exclusive nor independent. 108. The events “Other” and “Up for reelection in November 2016” are ________ independent. a. mutually exclusive. b. c. both mutually exclusive and independent. d. neither mutually exclusive nor independent. 109. Table 3.20 gives the number of suicides estimated in the U.S. for a recent year by age, race (black or white), and sex. We are interested in possible relationships between age, race, and sex. We will let suicide victims be our population. 210 CHAPTER 3 \| PROBABILITY TOPICS Race and Sex 1–14 15–24 25–64 over 64 TOTALS white, male 210 3,360 13,610 white, female black, male black, female 80 10 0 580 460 40 3,380 1,060 270 22,050 4,930 1,670 330 all others TOTALS Table 3.20
310 4,650 18,780 29,760 Do not include "all others" for parts f and g. a. Fill in the column for the suicides for individuals over age 64. b. Fill in the row for all other races. c. Find the probability that a randomly selected individual was a white male. d. Find the probability that a randomly selected individual was a black female. e. Find the probability that a randomly selected individual was black f. Find the probability that a randomly selected individual was male. g. Out of the individuals over age 64, find the probability that a randomly selected individual was a black or white male. Use the following information to answer the next two exercises. The table of data obtained from www.baseball-almanac.com shows hit information for four well known baseball players. Suppose that one hit from the table is randomly selected. NAME Single Double Triple Home Run TOTAL HITS Babe Ruth 1,517 Jackie Robinson 1,054 Ty Cobb Hank Aaron TOTAL Table 3.21 3,603 2,294 8,471 506 273 174 624 1,577 136 54 295 98 583 714 137 114 755 2,873 1,518 4,189 3,771 1,720 12,351 110. Find P(hit was made by Babe Ruth). a. b. c. d. 1518 2873 2873 12351 583 12351 4189 12351 111. Find P(hit was made by Ty Cobb\|The hit was a Home Run). a. b. c. d. 4189 12351 114 1720 1720 4189 114 12351 112. Table 3.22 identifies a group of children by one of four hair colors, and by type of hair. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 211 Hair Type Brown Blond Black Red Totals 20 80 Wavy Straight Totals Table 3.22 15 3 12 43 215 15 20 a. Complete the table. b. What is the probability that a randomly selected child will have wavy hair? c. What is the probability that a randomly selected child will have either brown or blond hair? d. What is the probability that a randomly selected child will have wavy brown hair? e. What is the probability that a randomly selected child will have red hair, given that he
or she has straight hair? f. g. If B is the event of a child having brown hair, find the probability of the complement of B. In words, what does the complement of B represent? 113. In a previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data were compiled into the following table. Shirt# ≤ 210 211–250 251–290 > 290 1–33 21 34–66 66–99 6 6 Table 3.23 5 18 12 0 7 22 0 4 5 For the following, suppose that you randomly select one player from the 49ers or Cowboys. a. Find the probability that his shirt number is from 1 to 33. b. Find the probability that he weighs at most 210 pounds. c. Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210 pounds. d. Find the probability that his shirt number is from 1 to 33 OR he weighs at most 210 pounds. e. Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs at most 210 pounds. 3.5 Tree and Venn Diagrams Use the following information to answer the next two exercises. This tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing three red (R), four yellow (Y) and five blue (B) beads. For the coin, P(H) = 2 3 where H is heads and T is tails. and P(T) = 1 3 212 CHAPTER 3 \| PROBABILITY TOPICS Figure 3.14 114. Find P(tossing a Head on the coin AND a Red bead) a. b. c. d. 2 3 5 15 6 36 5 36 b. a. 115. Find P(Blue bead). 15 36 10 36 10 12 6 36 d. c. 116. A box of cookies contains three chocolate and seven butter cookies. Miguel randomly selects a cookie and eats it. Then he randomly selects another cookie and eats it. (How many cookies did he take?) a. Draw the tree that represents the possibilities for the cookie selections. Write the probabilities along each branch of the tree. b. Are the probabilities for the flavor of the SECOND cookie that Miguel selects independent of his first selection? Explain. c. For each complete path through the tree, write the event it represents and find the probabilities. d. Let S be the event that both
cookies selected were the same flavor. Find P(S). e. Let T be the event that the cookies selected were different flavors. Find P(T) by two different methods: by using the complement rule and by using the branches of the tree. Your answers should be the same with both methods. f. Let U be the event that the second cookie selected is a butter cookie. Find P(U). BRINGING IT TOGETHER: HOMEWORK 117. A previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data are compiled into Table 3.24. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 213 Shirt# ≤ 210 211–250 251–290 290≤ 1–33 21 34–66 66–99 6 6 Table 3.24 5 18 12 0 7 22 0 4 5 For the following, suppose that you randomly select one player from the 49ers or Cowboys. If having a shirt number from one to 33 and weighing at most 210 pounds were independent events, then what should be true about P(Shirt# 1–33\|≤ 210 pounds)? 118. The probability that a male develops some form of cancer in his lifetime is 0.4567. The probability that a male has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51. Some of the following questions do not have enough information for you to answer them. Write “not enough information” for those answers. Let C = a man develops cancer in his lifetime and P = man has at least one false positive. a. P(C) = ______ b. P(P\|C) = ______ c. P(P\|C') = ______ d. If a test comes up positive, based upon numerical values, can you assume that man has cancer? Justify numerically and explain why or why not. 119. Given events G and H: P(G) = 0.43; P(H) = 0.26; P(H AND G) = 0.14 a. Find P(H OR G). b. Find the probability of the complement of event (H AND G). c. Find the probability
of the complement of event (H OR G). 120. Given events J and K: P(J) = 0.18; P(K) = 0.37; P(J OR K) = 0.45 a. Find P(J AND K). b. Find the probability of the complement of event (J AND K). c. Find the probability of the complement of event (J AND K). Use the following information to answer the next two exercises. Suppose that you have eight cards. Five are green and three are yellow. The cards are well shuffled. 121. Suppose that you randomly draw two cards, one at a time, with replacement. Let G1 = first card is green Let G2 = second card is green a. Draw a tree diagram of the situation. b. Find P(G1 AND G2). c. Find P(at least one green). d. Find P(G2\|G1). e. Are G2 and G1 independent events? Explain why or why not. 122. Suppose that you randomly draw two cards, one at a time, without replacement. G1 = first card is green G2 = second card is green a. Draw a tree diagram of the situation. b. Find P(G1 AND G2). c. Find P(at least one green). d. Find P(G2\|G1). e. Are G2 and G1 independent events? Explain why or why not. Use the following information to answer the next two exercises. The percent of licensed U.S. drivers (from a recent year) that are female is 48.60. Of the females, 5.03% are age 19 and under; 81.36% are age 20–64; 13.61% are age 65 or over. Of the licensed U.S. male drivers, 5.04% are age 19 and under; 81.43% are age 20–64; 13.53% are age 65 or over. 123. Complete the following. a. Construct a table or a tree diagram of the situation. b. Find P(driver is female). c. Find P(driver is age 65 or over\|driver is female). d. Find P(driver is age 65 or over AND female). e. f. Find P(driver is age 65 or over). In words, explain the difference between the probabilities in part c and part d. 214 CHAPTER 3 \| PROBABILITY TOPICS g. Are
being age 65 or over and being female mutually exclusive events? How do you know? 124. Suppose that 10,000 U.S. licensed drivers are randomly selected. a. How many would you expect to be male? b. Using the table or tree diagram, construct a contingency table of gender versus age group. c. Using the contingency table, find the probability that out of the age 20–64 group, a randomly selected driver is female. 125. Approximately 86.5% of Americans commute to work by car, truck, or van. Out of that group, 84.6% drive alone and 15.4% drive in a carpool. Approximately 3.9% walk to work and approximately 5.3% take public transportation. a. Construct a table or a tree diagram of the situation. Include a branch for all other modes of transportation to work. b. Assuming that the walkers walk alone, what percent of all commuters travel alone to work? c. Suppose that 1,000 workers are randomly selected. How many would you expect to travel alone to work? d. Suppose that 1,000 workers are randomly selected. How many would you expect to drive in a carpool? 126. When the Euro coin was introduced in 2002, two math professors had their statistics students test whether the Belgian one Euro coin was a fair coin. They spun the coin rather than tossing it and found that out of 250 spins, 140 showed a head (event H) while 110 showed a tail (event T). On that basis, they claimed that it is not a fair coin. a. Based on the given data, find P(H) and P(T). b. Use a tree to find the probabilities of each possible outcome for the experiment of tossing the coin twice. c. Use the tree to find the probability of obtaining exactly one head in two tosses of the coin. d. Use the tree to find the probability of obtaining at least one head. 127. Use the following information to answer the next two exercises. The following are real data from Santa Clara County, CA. As of a certain time, there had been a total of 3,059 documented cases of AIDS in the county. They were grouped into the following categories: Homosexual/Bisexual IV Drug User* Heterosexual Contact Other Totals Female 0 Male 2,146 Totals ____ 70 463 ____ 136 60 ____ Table 3.25 * includes homosexual/bisexual IV drug users 49 135 ____ ____ ____
____ Suppose a person with AIDS in Santa Clara County is randomly selected. a. Find P(Person is female). b. Find P(Person has a risk factor heterosexual contact). c. Find P(Person is female OR has a risk factor of IV drug user). d. Find P(Person is female AND has a risk factor of homosexual/bisexual). e. Find P(Person is male AND has a risk factor of IV drug user). f. Find P(Person is female GIVEN person got the disease from heterosexual contact). g. Construct a Venn diagram. Make one group females and the other group heterosexual contact. 128. Answer these questions using probability rules. Do NOT use the contingency table. Three thousand fifty-nine cases of AIDS had been reported in Santa Clara County, CA, through a certain date. Those cases will be our population. Of those cases, 6.4% obtained the disease through heterosexual contact and 7.4% are female. Out of the females with the disease, 53.3% got the disease from heterosexual contact. a. Find P(Person is female). b. Find P(Person obtained the disease through heterosexual contact). c. Find P(Person is female GIVEN person got the disease from heterosexual contact) d. Construct a Venn diagram representing this situation. Make one group females and the other group heterosexual contact. Fill in all values as probabilities. REFERENCES 3.1 Terminology “Countries List by Continent.” Worldatlas, 2013. Available online at http://www.worldatlas.com/cntycont.htm (accessed May 2, 2013). 3.2 Independent and Mutually Exclusive Events This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 215 Lopez, Shane, Preety Sidhu. “U.S. Teachers Love Their Lives, but Struggle in the Workplace.” Gallup Wellbeing, 2013. http://www.gallup.com/poll/161516/teachers-love-lives-struggle-workplace.aspx (accessed May 2, 2013). Data from Gallup. Available online at www.gallup.com/ (accessed May 2, 2013). 3.3 Two Basic Rules of Probability DiCamillo, Mark, Mervin
Field. “The File Poll.” Field Research Corporation. Available online at http://www.field.com/ fieldpollonline/subscribers/Rls2443.pdf (accessed May 2, 2013). Rider, David, “Ford support plummeting, poll suggests,” The Star, September 14, 2011. Available online at http://www.thestar.com/news/gta/2011/09/14/ford_support_plummeting_poll_suggests.html (accessed May 2, 2013). “Mayor’s Approval Down.” News Release by Forum Research Inc. Available online at http://www.forumresearch.com/ Releases/74209_TO_Issues_forms/News _Mayoral_Approval_%28Forum_Research%29%2820130320%29.pdf (accessed May 2, 2013). Archives/News “Roulette.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Roulette (accessed May 2, 2013). Shin, Hyon B., Robert A. Kominski. “Language Use in the United States: 2007.” United States Census Bureau. Available online at http://www.census.gov/hhes/socdemo/language/data/acs/ACS-12.pdf (accessed May 2, 2013). Data from the Baseball-Almanac, 2013. Available online at www.baseball-almanac.com (accessed May 2, 2013). Data from U.S. Census Bureau. Data from the Wall Street Journal. Data from The Roper Center: Public Opinion Archives at http://www.ropercenter.uconn.edu/ (accessed May 2, 2013). the University of Connecticut. Available online at Data from Field Research Corporation. Available online at www.field.com/fieldpollonline (accessed May 2,2 013). 3.4 Contingency Tables “Blood Types.” American Red Cross, 2013. Available online at http://www.redcrossblood.org/learn-about-blood/bloodtypes (accessed May 3, 2013). Data from the National Center for Health Statistics, part of the United States Department of Health and Human Services. Data from United States Senate. Available online at www.senate.gov (accessed May 2, 2013). Haiman, Christopher
A., Daniel O. Stram, Lynn R. Wilkens, Malcom C. Pike, Laurence N. Kolonel, Brien E. Henderson, and Loīc Le Marchand. “Ethnic and Racial Differences in the Smoking-Related Risk of Lung Cancer.” The New England Journal of Medicine, 2013. Available online at http://www.nejm.org/doi/full/10.1056/NEJMoa033250 (accessed May 2, 2013). “Human Blood Types.” Unite Blood Services, 2011. Available online at http://www.unitedbloodservices.org/ learnMore.aspx (accessed May 2, 2013). Samuel, T. M. “Strange Facts about RH Negative Blood.” eHow Health, 2013. Available online at http://www.ehow.com/ facts_5552003_strange-rh-negative-blood.html (accessed May 2, 2013). “United States: Uniform Crime Report – State Statistics from 1960–2011.” The Disaster Center. Available online at http://www.disastercenter.com/crime/ (accessed May 2, 2013). 3.5 Tree and Venn Diagrams Data from Clara County Public H.D. Data from the American Cancer Society. Data from The Data and Story Library, 1996. Available online at http://lib.stat.cmu.edu/DASL/ (accessed May 2, 2013). Data from the Federal Highway Administration, part of the United States Department of Transportation. Data from the United States Census Bureau, part of the United States Department of Commerce. Data from USA Today. “Environment.” The World Bank, 2013. Available online at http://data.worldbank.org/topic/environment (accessed May 2, 2013). “Search for Datasets.” Roper Center: Public Opinion Archives, University of Connecticut., 2013. Available online at http://www.ropercenter.uconn.edu/data_access/data/search_for_datasets.html (accessed May 2, 2013). 216 CHAPTER 3 \| PROBABILITY TOPICS SOLUTIONS 1 a. P(L′) = P(S) b. P(M OR S) c. P(F AND L) d. P(M\|L) e. P(L\|M) f. P(S\|
F) g. P(F\|L) h. P(F OR L) i. P(M AND S) j. P(F) 3 P(N) = 15 42 = 5 14 = 0.36 5 P(C) = 5 42 = 0.12 7 P(G) = 20 150 = 2 15 = 0.13 9 P(R) = 22 150 = 11 75 = 0.15 11 P(O) = 150 - 22 - 38 - 20 - 28 - 26 150 = 16 150 = 8 75 = 0.11 13 P(E) = 47 194 = 0.24 15 P(N) = 23 194 = 0.12 17 P(S) = 12 194 = 6 97 = 0.06 19 13 52 = 1 4 = 0.25 21 3 6 = 1 2 = 0.5 23 P(R) = 4 8 = 0.5 25 P(O OR H) 27 P(H\|I) 29 P(N\|O) 31 P(I OR N) 33 P(I) 35 The likelihood that an event will occur given that another event has already occurred. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 217 37 1 39 the probability of landing on an even number or a multiple of three 41 P(J) = 0.3 43 P(Q AND R) = P(Q)P(R) 0.1 = (0.4)P(R) P(R) = 0.25 45 0.376 47 C\|L means, given the person chosen is a Latino Californian, the person is a registered voter who prefers life in prison without parole for a person convicted of first degree murder. 49 L AND C is the event that the person chosen is a Latino California registered voter who prefers life without parole over the death penalty for a person convicted of first degree murder. 51 0.6492 53 No, because P(L AND C) does not equal 0. 55 P(musician is a male AND had private instruction) = 15 130 = 3 26 = 0.12 57 P(being a female musician AND learning music in school) = 38 130 = 19 65 = 0.29 P(being a female musician)P(learning music in school) = ⎛ �
�� 72 130 ⎛ ⎞ ⎝ ⎠ 62 130 ⎞ ⎠ = 4, 464 16, 900 = 1, 116 4, 225 = 0.26 No, they are not independent because P(being a female musician AND learning music in school) is not equal to P(being a female musician)P(learning music in school). 58 Figure 3.15 60 35,065 100,450 62 To pick one person from the study who is Japanese American AND smokes 21 to 30 cigarettes per day means that the person has to meet both criteria: both Japanese American and smokes 21 to 30 cigarettes. The sample space should include everyone in the study. The probability is 4,715 100,450. 218 CHAPTER 3 \| PROBABILITY TOPICS 64 To pick one person from the study who is Japanese American given that person smokes 21-30 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 21-30 cigarettes per day. The probability is 4715 15,273. 67 a. You can't calculate the joint probability knowing the probability of both events occurring, which is not in the information given; the probabilities should be multiplied, not added; and probability is never greater than 100% b. A home run by definition is a successful hit, so he has to have at least as many successful hits as home runs. 69 0 71 0.3571 73 0.2142 75 Physician (83.7) 77 83.7 − 79.6 = 4.1 79 P(Occupation < 81.3) = 0.5 81 a. The Forum Research surveyed 1,046 Torontonians. b. 58% c. 42% of 1,046 = 439 (rounding to the nearest integer) d. 0.57 e. 0.60. 83 a. P(Betting on two line that touch each other on the table) = 6 38 b. P(Betting on three numbers in a line) = 3 38 c. P(Bettting on one number) = 1 38 d. P(Betting on four number that touch each other to form a square) = 4 38 e. P(Betting on two number that touch each other on the table ) = 2 38 f. P(Betting on 0-00-1-2-3) = 5 38 g. P(Betting on 0-1-2;
or 0-00-2; or 00-2-3) = 3 38 85 a. {G1, G2, G3, G4, G5, Y1, Y2, Y3} b. c. d. 5 8 2 3 2 8 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 219 e. 6 8 f. No, because P(G AND E) does not equal 0. 87 NOTE The coin toss is independent of the card picked first. a. {(G,H) (G,T) (B,H) (B,T) (R,H) (R,T)} b. P(A) = P(blue)P(head) = ⎛ ⎝ ⎞ ⎠ 3 10 ⎞ ⎠ ⎛ ⎝ 1 2 = 3 20 c. Yes, A and B are mutually exclusive because they cannot happen at the same time; you cannot pick a card that is both blue and also (red or green). P(A AND B) = 0 d. No, A and C are not mutually exclusive because they can occur at the same time. In fact, C includes all of the outcomes of A; if the card chosen is blue it is also (red or blue). P(A AND C) = P(A) = 3 20 89 a. S = {(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)} b. 4 8 c. Yes, because if A has occurred, it is impossible to obtain two tails. In other words, P(A AND B) = 0. 91 a. If Y and Z are independent, then P(Y AND Z) = P(Y)P(Z), so P(Y OR Z) = P(Y) + P(Z) - P(Y)P(Z). b. 0.5 93 iii; i; iv; ii 95 a. P(R) = 0.44 b. P(R\|E) = 0.56 c. P(R\|O) = 0.31 d. No, whether the money is returned is not independent of which class
the money was placed in. There are several ways to justify this mathematically, but one is that the money placed in economics classes is not returned at the same overall rate; P(R\|E) ≠ P(R). e. No, this study definitely does not support that notion; in fact, it suggests the opposite. The money placed in the economics classrooms was returned at a higher rate than the money place in all classes collectively; P(R\|E) > P(R). 97 a. P(type O OR Rh-) = P(type O) + P(Rh-) - P(type O AND Rh-) 0.52 = 0.43 + 0.15 - P(type O AND Rh-); solve to find P(type O AND Rh-) = 0.06 6% of people have type O, Rh- blood b. P(NOT(type O AND Rh-)) = 1 - P(type O AND Rh-) = 1 - 0.06 = 0.94 94% of people do not have type O, Rh- blood 99 a. Let C = be the event that the cookie contains chocolate. Let N = the event that the cookie contains nuts. 220 CHAPTER 3 \| PROBABILITY TOPICS b. P(C OR N) = P(C) + P(N) - P(C AND N) = 0.36 + 0.12 - 0.08 = 0.40 c. P(NEITHER chocolate NOR nuts) = 1 - P(C OR N) = 1 - 0.40 = 0.60 Race and Sex 1–14 15–24 25–64 over 64 TOTALS white, male 210 3,360 13,610 4,870 22,050 white, female black, male black, female 80 10 0 580 460 40 3,380 1,060 270 4,930 1,670 330 890 140 20 100 310 4,650 18,780 6,020 29,760 all others TOTALS Table 3.26 Race and Sex 1–14 15–24 25–64 over 64 TOTALS white, male 210 3,360 13,610 4,870 22,050 white, female black, male black, female all others TOTALS Table 3.27 80 10 0 10 580 460 40 210 3,380 1,060 270 460 890 140 20 100 4,930 1,670 330 780 310 4,650 18,780 6,020 29,760 101 0
103 10 67 105 10 34 107 d 109 a. b. c. d. e. f. g. 22,050 29,760 330 29,760 2,000 29,760 23,720 29,760 5,010 6,020 111 b 113 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 221 a. b. c. d. e. 26 106 33 106 21 106 ⎛ ⎝ 26 106 ⎞ ⎠ + ⎛ ⎝ ⎞ ⎠ 33 106 - ⎛ ⎝ ⎞ ⎠ 21 106 = ⎛ ⎝ ⎞ ⎠ 38 106 21 33 115 a 118 a. P(C) = 0.4567 b. not enough information c. not enough information d. No, because over half (0.51) of men have at least one false positive text 120 a. P(J OR K) = P(J) + P(K) − P(J AND K); 0.45 = 0.18 + 0.37 - P(J AND K); solve to find P(J AND K) = 0.10 b. P(NOT (J AND K)) = 1 - P(J AND K) = 1 - 0.10 = 0.90 c. P(NOT (J OR K)) = 1 - P(J OR K) = 1 - 0.45 = 0.55 121 Figure 3.16 a. b. P(GG = 25 64 c. P(at least one green) = P(GG) + P(GY) + P(YG) = 25 64 + 15 64 + 15 64 = 55 64 d. P(G\|G) = 5 8 222 CHAPTER 3 \| PROBABILITY TOPICS e. Yes, they are independent because the first card is placed back in the bag before the second card is drawn; the composition of cards in the bag remains the same from draw one to draw two. 123 a. b. P(F) = 0.486 c. P(>64\|F) = 0.1361 <20 20–64 >64 Totals Female 0.0244 0.3954 0.0661 0.486 Male 0.0
259 0.4186 0.0695 0.514 Totals 0.0503 0.8140 0.1356 1 Table 3.28 d. P(>64 and F) = P(F) P(>64\|F) = (0.486)(0.1361) = 0.0661 e. P(>64\|F) is the percentage of female drivers who are 65 or older and P(>64 and F) is the percentage of drivers who are female and 65 or older. f. P(>64) = P(>64 and F) + P(>64 and M) = 0.1356 g. No, being female and 65 or older are not mutually exclusive because they can occur at the same time P(>64 and F) = 0.0661. 125 a. Car, Truck or Van Walk Public Transportation Other Totals Alone 0.7318 Not Alone 0.1332 Totals 0.8650 Table 3.29 0.0390 0.0530 0.0430 1 b. If we assume that all walkers are alone and that none from the other two groups travel alone (which is a big assumption) we have: P(Alone) = 0.7318 + 0.0390 = 0.7708. c. Make the same assumptions as in (b) we have: (0.7708)(1,000) = 771 d. (0.1332)(1,000) = 133 127 The completed contingency table is as follows: Homosexual/Bisexual IV Drug User* Heterosexual Contact Other Totals Female 0 Male 2,146 Totals 2,146 70 463 533 136 60 196 Table 3.30 * includes homosexual/bisexual IV drug users 49 135 184 255 2,804 3,059 a. 255 3059 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 3 \| PROBABILITY TOPICS 223 b. c. 196 3059 718 3059 d. 0 e. f. 463 3059 136 196 g. Figure 3.17 224 CHAPTER 3 \| PROBABILITY TOPICS This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562
/1.16 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 225 4 \| DISCRETE RANDOM VARIABLES Figure 4.1 You can use probability and discrete random variables to calculate the likelihood of lightening striking the ground five times during a half-hour thunderstorm. (Credit: Leszek Leszczynski) Introduction Chapter Objectives By the end of this chapter, the student should be able to: • Recognize and understand discrete probability distribution functions, in general. • Calculate and interpret expected values. • Recognize the binomial probability distribution and apply it appropriately. • Recognize the Poisson probability distribution and apply it appropriately. • Recognize the geometric probability distribution and apply it appropriately. • Recognize the hypergeometric probability distribution and apply it appropriately. • Classify discrete word problems by their distributions. A student takes a ten-question, true-false quiz. Because the student had such a busy schedule, he or she could not study and guesses randomly at each answer. What is the probability of the student passing the test with at least a 70%? Small companies might be interested in the number of long-distance phone calls their employees make during the peak time of the day. Suppose the average is 20 calls. What is the probability that the employees make more than 20 long-distance phone calls during the peak time? 226 CHAPTER 4 \| DISCRETE RANDOM VARIABLES These two examples illustrate two different types of probability problems involving discrete random variables. Recall that discrete data are data that you can count. A random variable describes the outcomes of a statistical experiment in words. The values of a random variable can vary with each repetition of an experiment. Random Variable Notation Upper case letters such as X or Y denote a random variable. Lower case letters like x or y denote the value of a random variable. If X is a random variable, then X is written in words, and x is given as a number. For example, let X = the number of heads you get when you toss three fair coins. The sample space for the toss of three fair coins is TTT; THH; HTH; HHT; HTT; THT; TTH; HHH. Then, x = 0, 1, 2, 3. X is in words and x is a number. Notice that for this example, the x values are countable outcomes. Because you can count the possible values that X can take on and the outcomes are random (
the x values 0, 1, 2, 3), X is a discrete random variable. Toss a coin ten times and record the number of heads. After all members of the class have completed the experiment (tossed a coin ten times and counted the number of heads), fill in Table 4.1. Let X = the number of heads in ten tosses of the coin. x Frequency of x Relative Frequency of x Table 4.1 a. Which value(s) of x occurred most frequently? b. If you tossed the coin 1,000 times, what values could x take on? Which value(s) of x do you think would occur most frequently? c. What does the relative frequency column sum to? 4.1 \| Probability Distribution Function (PDF) for a Discrete Random Variable A discrete probability distribution function has two characteristics: 1. Each probability is between zero and one, inclusive. 2. The sum of the probabilities is one. Example 4.1 A child psychologist is interested in the number of times a newborn baby's crying wakes its mother after midnight. For a random sample of 50 mothers, the following information was obtained. Let X = the number of times per week a newborn baby's crying wakes its mother after midnight. For this example, x = 0, 1, 2, 3, 4, 5. P(x) = probability that X takes on a value x. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 227 x P(x) 0 1 2 3 4 5 P(x = 0) = 2 50 P(x = 1) = 11 50 P(x = 2) = 23 50 P(x = 3) = 9 50 P(x = 4) = 4 50 P(x = 5) = 1 50 Table 4.2 X takes on the values 0, 1, 2, 3, 4, 5. This is a discrete PDF because: a. Each P(x) is between zero and one, inclusive. b. The sum of the probabilities is one, that is, 2 50 + 11 50 + 23 50 + 9 50 + 4 50 + 1 50 = 1 4.1 A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift
. For a random sample of 50 patients, the following information was obtained. Let X = the number of times a patient rings the nurse during a 12-hour shift. For this exercise, x = 0, 1, 2, 3, 4, 5. P(x) = the probability that X takes on value x. Why is this a discrete probability distribution function (two reasons)? X P(x) 0 P(x = 0) = 4 50 1 P(x = 1) = 8 50 2 P(x = 2) = 16 50 3 P(x = 3) = 14 50 4 P(x = 4) = 6 50 5 P(x = 5) = 2 50 Table 4.3 228 CHAPTER 4 \| DISCRETE RANDOM VARIABLES Example 4.2 Suppose Nancy has classes three days a week. She attends classes three days a week 80% of the time, two days 15% of the time, one day 4% of the time, and no days 1% of the time. Suppose one week is randomly selected. a. Let X = the number of days Nancy ____________________. Solution 4.2 a. Let X = the number of days Nancy attends class per week. b. X takes on what values? Solution 4.2 b. 0, 1, 2, and 3 c. Suppose one week is randomly chosen. Construct a probability distribution table (called a PDF table) like the one in Example 4.1. The table should have two columns labeled x and P(x). What does the P(x) column sum to? Solution 4.2 c. x P(x) 0 1 2 3 0.01 0.04 0.15 0.80 Table 4.4 4.2 Jeremiah has basketball practice two days a week. Ninety percent of the time, he attends both practices. Eight percent of the time, he attends one practice. Two percent of the time, he does not attend either practice. What is X and what values does it take on? 4.2 \| Mean or Expected Value and Standard Deviation The expected value is often referred to as the "long-term" average or mean. This means that over the long term of doing an experiment over and over, you would expect this average. You toss a coin and record the result. What is the probability that the result is heads? If you flip a coin two times, does probability tell you that these flips will result in one heads and one
tail? You might toss a fair coin ten times and record nine heads. As you learned in Section 3., probability does not describe the short-term results of an experiment. It gives information about what can be expected in the long term. To demonstrate this, Karl Pearson once tossed a fair coin 24,000 times! He recorded the results of each toss, obtaining heads 12,012 times. In his experiment, Pearson illustrated the Law of Large Numbers. The Law of Large Numbers states that, as the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency approaches zero (the theoretical probability and the relative frequency get closer and closer together). When evaluating the long-term results of statistical experiments, we often want to know the “average” outcome. This “long-term average” is known as the mean or expected value of the This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 experiment and is denoted by the Greek letter μ. In other words, after conducting many trials of an experiment, you would expect this average value. CHAPTER 4 \| DISCRETE RANDOM VARIABLES 229 NOTE To find the expected value or long term average, μ, simply multiply each value of the random variable by its probability and add the products. Example 4.3 A men's soccer team plays soccer zero, one, or two days a week. The probability that they play zero days is 0.2, the probability that they play one day is 0.5, and the probability that they play two days is 0.3. Find the long-term average or expected value, μ, of the number of days per week the men's soccer team plays soccer. To do the problem, first let the random variable X = the number of days the men's soccer team plays soccer per week. X takes on the values 0, 1, 2. Construct a PDF table adding a column xP(x). In this column, you will multiply each x value by its probability. x P(x) xP(x) 0 1 2 0.2 0.5 0.3 (0)(0.2) = 0 (1)(0.5) = 0.5 (2)(0.3) = 0.6 Table 4.5 Expected Value Table This table is called
an expected value table. The table helps you calculate the expected value or long-term average. Add the last column xP(x) to find the long term average or expected value: (0)(0.2) + (1)(0.5) + (2)(0.3) = 0 + 0.5 + 0.6 = 1.1. The expected value is 1.1. The men's soccer team would, on the average, expect to play soccer 1.1 days per week. The number 1.1 is the long-term average or expected value if the men's soccer team plays soccer week after week after week. We say μ = 1.1. Example 4.4 Find the expected value of the number of times a newborn baby's crying wakes its mother after midnight. The expected value is the expected number of times per week a newborn baby's crying wakes its mother after midnight. Calculate the standard deviation of the variable as well. x P(x) xP(x) (x – μ)2 ⋅⋅ P(x) 0 1 P(x = 0) = 2 50 (0) ⎛ ⎝ ⎞ ⎠ 2 50 = 0 (0 – 2.1)2 ⋅ 0.04 = 0.1764 P(x = 1) = ⎛ ⎝ ⎞ ⎠ 11 50 (1) ⎛ ⎝ ⎞ ⎠ 11 50 = 11 50 (1 – 2.1)2 ⋅ 0.22 = 0.2662 Table 4.6 You expect a newborn to wake its mother after midnight 2.1 times per week, on the average. 230 CHAPTER 4 \| DISCRETE RANDOM VARIABLES x P(x) x*P(x) (x – μ)2 ⋅⋅ P(x) 2 3 4 5 P(x = 2) = 23 50 (2) ⎛ ⎝ ⎞ ⎠ 23 50 = 46 50 (2 – 2.1)2 ⋅ 0.46 = 0.0046 P(x = 3) = 27 50 (3) ⎛ ⎝ ⎞ ⎠ 9 50 = 27 50 (3 – 2.1)2 ⋅ 0.18 = 0.1458 P(x = 4) = 4
50 (4) ⎛ ⎝ ⎞ ⎠ 4 50 = 16 50 (4 – 2.1)2 ⋅ 0.08 = 0.2888 P(x = 5) = 1 50 (5) ⎛ ⎝ ⎞ ⎠ 1 50 = 5 50 (5 – 2.1)2 ⋅ 0.02 = 0.1682 Table 4.6 You expect a newborn to wake its mother after midnight 2.1 times per week, on the average. Add the values in the third column of the table to find the expected value of X: μ = Expected Value = 105 50 = 2.1 Use μ to complete the table. The fourth column of this table will provide the values you need to calculate the standard deviation. For each value x, multiply the square of its deviation by its probability. (Each deviation has the format x – μ). Add the values in the fourth column of the table: 0.1764 + 0.2662 + 0.0046 + 0.1458 + 0.2888 + 0.1682 = 1.05 The standard deviation of X is the square root of this sum: σ = 1.05 ≈ 1.0247 4.4 A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. What is the expected value? x P(x) 0 1 2 3 4 5 P(x = 0) = 4 50 P(x = 1) = 8 50 P(x = 2) = 16 50 P(x = 3) = 14 50 P(x = 4) = 6 50 P(x = 5) = 2 50 Table 4.7 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 231 Example 4.5 Suppose you play a game of chance in which five numbers are chosen from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. A computer randomly selects five numbers from zero to nine with replacement. You pay $2 to play and could profit $100,000 if you match all five numbers in order (you get
your $2 back plus $100,000). Over the long term, what is your expected profit of playing the game? To do this problem, set up an expected value table for the amount of money you can profit. Let X = the amount of money you profit. The values of x are not 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Since you are interested in your profit (or loss), the values of x are 100,000 dollars and −2 dollars. To win, you must get all five numbers correct, in order. The probability of choosing one correct number is 1 10 because there are ten numbers. You may choose a number more than once. The probability of choosing all five numbers correctly and in order is ⎛ ⎝ 1 10 ⎞ ⎛ ⎝ ⎠ 1 10 ⎞ ⎛ ⎝ ⎠ 1 10 ⎞ ⎛ ⎝ ⎠ 1 10 ⎞ ⎛ ⎝ ⎠ 1 10 ⎞ ⎠ = (1)(10−5) = 0.00001. Therefore, the probability of winning is 0.00001 and the probability of losing is The expected value table is as follows: 1 − 0.00001 = 0.99999. x P(x) x*P(x) Loss –2 0.99999 (–2)(0.99999) = –1.99998 Profit 100,000 0.00001 (100000)(0.00001) = 1 Table 4.8 Αdd the last column. –1.99998 + 1 = –0.99998 Since –0.99998 is about –1, you would, on average, expect to lose approximately $1 for each game you play. However, each time you play, you either lose $2 or profit $100,000. The $1 is the average or expected LOSS per game after playing this game over and over. 4.5 You are playing a game of chance in which four cards are drawn from a standard deck of 52 cards. You guess the suit of each card before it is drawn. The cards are replaced in the deck on each draw. You pay $1 to play. If you guess the right suit every time, you get your money back and $256. What is your expected profit of playing the game over the long term? Example 4
.6 Suppose you play a game with a biased coin. You play each game by tossing the coin once. P(heads) = 2 3 and. If you toss a head, you pay $6. If you toss a tail, you win $10. If you play this game many times, P(tails) = 1 3 will you come out ahead? a. Define a random variable X. Solution 4.6 a. X = amount of profit 232 CHAPTER 4 \| DISCRETE RANDOM VARIABLES b. Complete the following expected value table. Solution 4.6 b. x ____ ____ WIN 10 1 3 LOSE ____ ____ ____ –12 3 Table 4.9 x P(x) xP(x) 1 3 2 3 WIN 10 LOSE –6 Table 4.10 10 3 –12 3 c. What is the expected value, μ? Do you come out ahead? Solution 4.6 c. Add the last column of the table. The expected value μ = –2 3 you play the game so you do not come out ahead.. You lose, on average, about 67 cents each time 4.6 Suppose you play a game with a spinner. You play each game by spinning the spinner once. P(red) = 2 5, P(blue) = 2 5, and P(green) = 1 5. If you land on red, you pay $10. If you land on blue, you don't pay or win anything. If you land on green, you win $10. Complete the following expected value table. –20 5 x P(x) Red Blue 2 5 20 Table 4.11 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 Like data, probability distributions have standard deviations. To calculate the standard deviation (σ) of a probability distribution, find each deviation from its expected value, square it, multiply it by its probability, add the products, and take the square root. To understand how to do the calculation, look at the table for the number of days per week a men's soccer team plays soccer. To find the standard deviation, add the entries in the column labeled (x – μ)2P(x) and take the square root. CHAPTER 4 \| DISCRETE RANDOM VARIABLES 233 x P(
x) x*P(x) (x – μ)2P(x) 0 1 2 0.2 0.5 0.3 (0)(0.2) = 0 (0 – 1.1)2(0.2) = 0.242 (1)(0.5) = 0.5 (1 – 1.1)2(0.5) = 0.005 (2)(0.3) = 0.6 (2 – 1.1)2(0.3) = 0.243 Table 4.12 Add the last column in the table. 0.242 + 0.005 + 0.243 = 0.490. The standard deviation is the square root of 0.49, or σ = 0.49 = 0.7 Generally for probability distributions, we use a calculator or a computer to calculate μ and σ to reduce roundoff error. For some probability distributions, there are short-cut formulas for calculating μ and σ. Example 4.7 Toss a fair, six-sided die twice. Let X = the number of faces that show an even number. Construct a table like Table 4.11 and calculate the mean μ and standard deviation σ of X. Solution 4.7 Tossing one fair six-sided die twice has the same sample space as tossing two fair six-sided dice. The sample space has 36 outcomes: (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) Table 4.13 Use the sample space to complete the following table: x P(x) xP(x) (x – μ)2 ⋅⋅ P(x) 0 1
2 9 36 18 36 9 36 0 18 36 18 36 (0 – 1)2 ⋅ 9 36 = 9 36 (1 – 1)2 ⋅ 18 36 = 0 (1 – 1)2 ⋅ 9 36 = 9 36 Table 4.14 Calculating μ and σ. 234 CHAPTER 4 \| DISCRETE RANDOM VARIABLES Add the values in the third column to find the expected value: μ = 36 36 = 1. Use this value to complete the fourth column. Add the values in the fourth column and take the square root of the sum: σ = 18 36 ≈ 0.7071. Example 4.8 On May 11, 2013 at 9:30 PM, the probability that moderate seismic activity (one moderate earthquake) would occur in the next 48 hours in Iran was about 21.42%. Suppose you make a bet that a moderate earthquake will occur in Iran during this period. If you win the bet, you win $50. If you lose the bet, you pay $20. Let X = the amount of profit from a bet. P(win) = P(one moderate earthquake will occur) = 21.42% P(loss) = P(one moderate earthquake will not occur) = 100% – 21.42% If you bet many times, will you come out ahead? Explain your answer in a complete sentence using numbers. What is the standard deviation of X? Construct a table similar to Table 4.12 and Table 4.12 to help you answer these questions. Solution 4.8 x P(x) x(Px) (x – μ)2P(x) win 50 0.2142 10.71 [50 – (–5.006)]2(0.2142) = 648.0964 loss –20 0.7858 –15.716 [–20 – (–5.006)]2(0.7858) = 176.6636 Table 4.15 Mean = Expected Value = 10.71 + (–15.716) = –5.006. If you make this bet many times under the same conditions, your long term outcome will be an average loss of $5.01 per bet. Standard Deviation = 648.0964 + 176.6636 ≈ 28.7186 4.8 On May 11, 2013 at 9:30 PM, the probability that moderate seismic activity (one moderate earthquake) would occur in
the next 48 hours in Japan was about 1.08%. As in Example 4.8, you bet that a moderate earthquake will occur in Japan during this period. If you win the bet, you win $100. If you lose the bet, you pay $10. Let X = the amount of profit from a bet. Find the mean and standard deviation of X. Some of the more common discrete probability functions are binomial, geometric, hypergeometric, and Poisson. Most elementary courses do not cover the geometric, hypergeometric, and Poisson. Your instructor will let you know if he or she wishes to cover these distributions. A probability distribution function is a pattern. You try to fit a probability problem into a pattern or distribution in order to perform the necessary calculations. These distributions are tools to make solving probability problems easier. Each distribution has its own special characteristics. Learning the characteristics enables you to distinguish among the different distributions. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 235 4.3 \| Binomial Distribution There are three characteristics of a binomial experiment. 1. There are a fixed number of trials. Think of trials as repetitions of an experiment. The letter n denotes the number of trials. 2. There are only two possible outcomes, called "success" and "failure," for each trial. The letter p denotes the probability of a success on one trial, and q denotes the probability of a failure on one trial. p + q = 1. 3. The n trials are independent and are repeated using identical conditions. Because the n trials are independent, the outcome of one trial does not help in predicting the outcome of another trial. Another way of saying this is that for each individual trial, the probability, p, of a success and probability, q, of a failure remain the same. For example, randomly guessing at a true-false statistics question has only two outcomes. If a success is guessing correctly, then a failure is guessing incorrectly. Suppose Joe always guesses correctly on any statistics true-false question with probability p = 0.6. Then, q = 0.4. This means that for every true-false statistics question Joe answers, his probability of success (p = 0.6) and his probability of failure (q = 0.4) remain the same.
The outcomes of a binomial experiment fit a binomial probability distribution. The random variable X = the number of successes obtained in the n independent trials. The mean, μ, and variance, σ2, for the binomial probability distribution are μ = np and σ2 = npq. The standard deviation, σ, is then σ = npq. Any experiment that has characteristics two and three and where n = 1 is called a Bernoulli Trial (named after Jacob Bernoulli who, in the late 1600s, studied them extensively). A binomial experiment takes place when the number of successes is counted in one or more Bernoulli Trials. Example 4.9 At ABC College, the withdrawal rate from an elementary physics course is 30% for any given term. This implies that, for any given term, 70% of the students stay in the class for the entire term. A "success" could be defined as an individual who withdrew. The random variable X = the number of students who withdraw from the randomly selected elementary physics class. 4.9 The state health board is concerned about the amount of fruit available in school lunches. Forty-eight percent of schools in the state offer fruit in their lunches every day. This implies that 52% do not. What would a "success" be in this case? Example 4.10 Suppose you play a game that you can only either win or lose. The probability that you win any game is 55%, and the probability that you lose is 45%. Each game you play is independent. If you play the game 20 times, write the function that describes the probability that you win 15 of the 20 times. Here, if you define X as the number of wins, then X takes on the values 0, 1, 2, 3,..., 20. The probability of a success is p = 0.55. The probability of a failure is q = 0.45. The number of trials is n = 20. The probability question can be stated mathematically as P(x = 15). 4.10 A trainer is teaching a dolphin to do tricks. The probability that the dolphin successfully performs the trick is 35%, and the probability that the dolphin does not successfully perform the trick is 65%. Out of 20 attempts, you want to find the probability that the dolphin succeeds 12 times. State the probability question mathematically. 236 CHAPTER 4 \| DISCRETE RANDOM VARIABLES Example 4.11 A fair coin is flipped 15
times. Each flip is independent. What is the probability of getting more than ten heads? Let X = the number of heads in 15 flips of the fair coin. X takes on the values 0, 1, 2, 3,..., 15. Since the coin is fair, p = 0.5 and q = 0.5. The number of trials is n = 15. State the probability question mathematically. Solution 4.11 P(x > 10) 4.11 A fair, six-sided die is rolled ten times. Each roll is independent. You want to find the probability of rolling a one more than three times. State the probability question mathematically. Example 4.12 Approximately 70% of statistics students do their homework in time for it to be collected and graded. Each student does homework independently. In a statistics class of 50 students, what is the probability that at least 40 will do their homework on time? Students are selected randomly. a. This is a binomial problem because there is only a success or a __________, there are a fixed number of trials, and the probability of a success is 0.70 for each trial. Solution 4.12 a. failure b. If we are interested in the number of students who do their homework on time, then how do we define X? Solution 4.12 b. X = the number of statistics students who do their homework on time c. What values does x take on? Solution 4.12 c. 0, 1, 2, …, 50 d. What is a "failure," in words? Solution 4.12 d. Failure is defined as a student who does not complete his or her homework on time. The probability of a success is p = 0.70. The number of trials is n = 50. e. If p + q = 1, then what is q? Solution 4.12 e. q = 0.30 f. The words "at least" translate as what kind of inequality for the probability question P(x ____ 40). Solution 4.12 f. greater than or equal to (≥) The probability question is P(x ≥ 40). This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 237 4.12 Sixty-five percent of people pass the state driver�
�s exam on the first try. A group of 50 individuals who have taken the driver’s exam is randomly selected. Give two reasons why this is a binomial problem. Notation for the Binomial: B = Binomial Probability Distribution Function X ~ B(n, p) Read this as "X is a random variable with a binomial distribution." The parameters are n and p; n = number of trials, p = probability of a success on each trial. Example 4.13 It has been stated that about 41% of adult workers have a high school diploma but do not pursue any further education. If 20 adult workers are randomly selected, find the probability that at most 12 of them have a high school diploma but do not pursue any further education. How many adult workers do you expect to have a high school diploma but do not pursue any further education? Let X = the number of workers who have a high school diploma but do not pursue any further education. X takes on the values 0, 1, 2,..., 20 where n = 20, p = 0.41, and q = 1 – 0.41 = 0.59. X ~ B(20, 0.41) Find P(x ≤ 12). P(x ≤ 12) = 0.9738. (calculator or computer) Go into 2nd DISTR. The syntax for the instructions are as follows: To calculate (x = value): binompdf(n, p, number) if "number" is left out, the result is the binomial probability table. To calculate P(x ≤ value): binomcdf(n, p, number) if "number" is left out, the result is the cumulative binomial probability table. For this problem: After you are in 2nd DISTR, arrow down to binomcdf. Press ENTER. Enter 20,0.41,12). The result is P(x ≤ 12) = 0.9738. NOTE If you want binomcdf(20,0.41,12). to find P(x = 12), use the pdf (binompdf). If you want to find P(x > 12), use 1 - The probability that at most 12 workers have a high school diploma but do not pursue any further education is 0.9738. The graph of X ~ B(20, 0.41) is as follows: 238 CHAPTER 4 \| DISCRETE RANDOM VARIABLES Figure 4
.2 The y-axis contains the probability of x, where X = the number of workers who have only a high school diploma. The number of adult workers that you expect to have a high school diploma but not pursue any further education is the mean, μ = np = (20)(0.41) = 8.2. The formula for the variance is σ2 = npq. The standard deviation is σ = npq. σ = (20)(0.41)(0.59) = 2.20. 4.13 About 32% of students participate in a community volunteer program outside of school. If 30 students are selected at random, find the probability that at most 14 of them participate in a community volunteer program outside of school. Use the TI-83+ or TI-84 calculator to find the answer. Example 4.14 In the 2013 Jerry’s Artarama art supplies catalog, there are 560 pages. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let X = the number of pages that feature signature artists. a. What values does x take on? b. What is the probability distribution? Find the following probabilities: i. ii. the probability that two pages feature signature artists the probability that at most six pages feature signature artists iii. the probability that more than three pages feature signature artists. c. Using the formulas, calculate the (i) mean and (ii) standard deviation. Solution 4.14 a. x = 0, 1, 2, 3, 4, 5, 6, 7, 8 b. X ~ B ⎛ ⎝100, 8 560 ⎞ ⎠ This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 239 i. P(x = 2) = binompdf ⎛ ⎝100, 8 560, 2 ⎞ ⎠ = 0.2466 ii. P(x ≤ 6) = binomcdf ⎛ ⎝100, 8 560, 6 ⎞ ⎠ = 0.9994 iii. P(x > 3) = 1 – P(x ≤ 3) = 1 – binomcdf ⎛ ⎝100, 8 560, 3 ⎞ ⎠ =
1 – 0.9443 = 0.0557 c. i. Mean = np = (100) ⎛ ⎝ ⎞ ⎠ 8 560 = 800 560 ≈ 1.4286 ⎛ ii. Standard Deviation = npq = (100) ⎝ 8 560 ⎞ ⎛ ⎝ ⎠ ⎞ ⎠ 552 560 ≈ 1.1867 4.14 According to a Gallup poll, 60% of American adults prefer saving over spending. Let X = the number of American adults out of a random sample of 50 who prefer saving to spending. a. What is the probability distribution for X? b. Use your calculator to find the following probabilities: i. ii. the probability that 25 adults in the sample prefer saving over spending the probability that at most 20 adults prefer saving iii. the probability that more than 30 adults prefer saving c. Using the formulas, calculate the (i) mean and (ii) standard deviation of X. Example 4.15 The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Suppose we randomly sample 200 people. Let X = the number of people who will develop pancreatic cancer. a. What is the probability distribution for X? b. Using the formulas, calculate the (i) mean and (ii) standard deviation of X. c. Use your calculator to find the probability that at most eight people develop pancreatic cancer d. Is it more likely that five or six people will develop pancreatic cancer? Justify your answer numerically. Solution 4.15 a. X ∼ B(200, 0.0128) b. i. Mean = np = 200(0.0128) = 2.56 ii. Standard Deviation = npq = (200)(0.0128)(0.9853) ≈ 1.5897 c. Using the TI-83, 83+, 84 calculator with instructions as provided in Example 4.13: P(x ≤ 8) = binomcdf(200, 0.0128, 8) = 0.9988 d. P(x = 5) = binompdf(200, 0.0128, 5) = 0.0707 P(x = 6) = binompdf(200, 0.0128, 6) = 0.0298 So P(x = 5) > P(x = 6); it is more likely that five people
will develop cancer than six. 240 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 4.15 During the 2013 regular NBA season, DeAndre Jordan of the Los Angeles Clippers had the highest field goal completion rate in the league. DeAndre scored with 61.3% of his shots. Suppose you choose a random sample of 80 shots made by DeAndre during the 2013 season. Let X = the number of shots that scored points. a. What is the probability distribution for X? b. Using the formulas, calculate the (i) mean and (ii) standard deviation of X. c. Use your calculator to find the probability that DeAndre scored with 60 of these shots. d. Find the probability that DeAndre scored with more than 50 of these shots. Example 4.16 The following example illustrates a problem that is not binomial. It violates the condition of independence. ABC College has a student advisory committee made up of ten staff members and six students. The committee wishes to choose a chairperson and a recorder. What is the probability that the chairperson and recorder are both students? The names of all committee members are put into a box, and two names are drawn without replacement. The first name drawn determines the chairperson and the second name the recorder. There are two trials. However, the trials are not independent because the outcome of the first trial affects the outcome of the second trial. The probability of a student on the first draw is 6 16. The probability of a student on the second draw is 5 15, when the first draw selects a student. The probability is 6 15, when the first draw selects a staff member. The probability of drawing a student's name changes for each of the trials and, therefore, violates the condition of independence. 4.16 A lacrosse team is selecting a captain. The names of all the seniors are put into a hat, and the first three that are drawn will be the captains. The names are not replaced once they are drawn (one person cannot be two captains). You want to see if the captains all play the same position. State whether this is binomial or not and state why. 4.4 \| Geometric Distribution There are three main characteristics of a geometric experiment. 1. There are one or more Bernoulli trials with all failures except the last one, which is a success. In other words, you keep repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bullsey
e until you hit the bullseye. The first time you hit the bullseye is a "success" so you stop throwing the dart. It might take six tries until you hit the bullseye. You can think of the trials as failure, failure, failure, failure, failure, success, STOP. 2. In theory, the number of trials could go on forever. There must be at least one trial. 3. The probability, p, of a success and the probability, q, of a failure is the same for each trial. p + q = 1 and q = 1 − p. For example, the probability of rolling a three when you throw one fair die is 1 6. This is true no matter how many times you roll the die. Suppose you want to know the probability of getting the first three on the fifth roll. On rolls one through four, you do not get a face with a three. The probability for each of the rolls is q = 5 6, the probability of a failure. The probability of getting a three on the fifth roll is ⎛ ⎝.0804 X = the number of independent trials until the first success. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 241 Example 4.17 You play a game of chance that you can either win or lose (there are no other possibilities) until you lose. Your probability of losing is p = 0.57. What is the probability that it takes five games until you lose? Let X = the number of games you play until you lose (includes the losing game). Then X takes on the values 1, 2, 3,... (could go on indefinitely). The probability question is P(x = 5). 4.17 You throw darts at a board until you hit the center area. Your probability of hitting the center area is p = 0.17. You want to find the probability that it takes eight throws until you hit the center. What values does X take on? Example 4.18 A safety engineer feels that 35% of all industrial accidents in her plant are caused by failure of employees to follow instructions. She decides to look at the accident reports (selected randomly and replaced in the pile after reading) until she finds one that shows an accident caused by failure of employees to follow instructions
. On average, how many reports would the safety engineer expect to look at until she finds a report showing an accident caused by employee failure to follow instructions? What is the probability that the safety engineer will have to examine at least three reports until she finds a report showing an accident caused by employee failure to follow instructions? Let X = the number of accidents the safety engineer must examine until she finds a report showing an accident caused by employee failure to follow instructions. X takes on the values 1, 2, 3,.... The first question asks you to find the expected value or the mean. The second question asks you to find P(x ≥ 3). ("At least" translates to a "greater than or equal to" symbol). 4.18 An instructor feels that 15% of students get below a C on their final exam. She decides to look at final exams (selected randomly and replaced in the pile after reading) until she finds one that shows a grade below a C. We want to know the probability that the instructor will have to examine at least ten exams until she finds one with a grade below a C. What is the probability question stated mathematically? Example 4.19 Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says he or she lives within five miles of you. What is the probability that you need to contact four people? This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays the same each time you ask a student if he or she lives within five miles of you. There is no definite number of trials (number of times you ask a student). a. Let X = the number of ____________ you must ask ____________ one says yes. Solution 4.19 a. Let X = the number of students you must ask until one says yes. b. What values does X take on? Solution 4.19 242 CHAPTER 4 \| DISCRETE RANDOM VARIABLES b. 1, 2, 3, …, (total number of students) c. What are p and q? Solution 4.19 c. p = 0.55; q = 0.45 d. The probability question is P(_______). Solution 4.19 d. P(x = 4) 4.19 You need
to find a store that carries a special printer ink. You know that of the stores that carry printer ink, 10% of them carry the special ink. You randomly call each store until one has the ink you need. What are p and q? Notation for the Geometric: G = Geometric Probability Distribution Function X ~ G(p) Read this as "X is a random variable with a geometric distribution." The parameter is p; p = the probability of a success for each trial. Example 4.20 Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective? Let X = the number of computer components tested until the first defect is found. X takes on the values 1, 2, 3,... where p = 0.02. X ~ G(0.02) Find P(x = 7). P(x = 7) = 0.0177. To find the probability that x = 7, • Enter 2nd, DISTR • Scroll down and select geometpdf( • Press ENTER • Enter 0.02, 7); press ENTER to see the result: P(x = 7) = 0.0177 To find the probability that x ≤ 7, follow the same instructions EXCEPT select E:geometcdf(as the distribution function. The probability that the seventh component is the first defect is 0.0177. The graph of X ~ G(0.02) is: This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 243 Figure 4.3 The y-axis contains the probability of x, where X = the number of computer components tested. The number of components that you would expect to test until you find the first defective one is the mean, µ = 50. The formula for the mean is μ = 1 p = 1 0.02 = 50 The formula for the variance is σ2 = ⎛ ⎝.02 ⎞ ⎛ ⎝ ⎠ 1 0.02 − 1 ⎞ ⎠ = 2,450 The standard deviation is.02 ⎛ ⎞ ⎝ �
�� 1 0.02 − 1 ⎞ ⎠ = 49.5 4.20 The probability of a defective steel rod is 0.01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84 calculator to find the answer. Example 4.21 The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Let X = the number of people you ask until one says he or she has pancreatic cancer. Then X is a discrete random variable with a geometric distribution: X ~ G ⎛ ⎝ or X ~ G(0.0128). ⎞ ⎠ 1 78 a. What is the probability of that you ask ten people before one says he or she has pancreatic cancer? b. What is the probability that you must ask 20 people? c. Find the (i) mean and (ii) standard deviation of X. Solution 4.21 a. P(x = 10) = geometpdf(0.0128, 10) = 0.0114 b. P(x = 20) = geometpdf(0.0128, 20) = 0.01 c. i. Mean = μ = 1 p = 1 0.0128 = 78 244 CHAPTER 4 \| DISCRETE RANDOM VARIABLES ii. Standard Deviation = σ = 1 − p p2 = 1 − 0.0128 0.01282 ≈ 77.6234 4.21 The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate for women in Afghanistan is 12%. Let X = the number of Afghani women you ask until one says that she is literate. a. What is the probability distribution of X? b. What is the probability that you ask five women before one says she is literate? c. What is the probability that you must ask ten women? d. Find the (i) mean and (ii) standard deviation of X. 4.5 \| Hypergeometric Distribution There are five characteristics of a hypergeometric experiment. 1. You take samples from two groups. 2. You are concerned with a group of interest, called the first group. 3. You sample without replacement from the combined groups. For example, you want to choose a softball team from a combined group of 11 men and 13 women.
The team consists of ten players. 4. Each pick is not independent, since sampling is without replacement. In the softball example, the probability of picking a woman first is 13 24. The probability of picking a man second is 11 23 if a woman was picked first. It is 10 23 if a man was picked first. The probability of the second pick depends on what happened in the first pick. 5. You are not dealing with Bernoulli Trials. The outcomes of a hypergeometric experiment fit a hypergeometric probability distribution. The random variable X = the number of items from the group of interest. Example 4.22 A candy dish contains 100 jelly beans and 80 gumdrops. Fifty candies are picked at random. What is the probability that 35 of the 50 are gumdrops? The two groups are jelly beans and gumdrops. Since the probability question asks for the probability of picking gumdrops, the group of interest (first group) is gumdrops. The size of the group of interest (first group) is 80. The size of the second group is 100. The size of the sample is 50 (jelly beans or gumdrops). Let X = the number of gumdrops in the sample of 50. X takes on the values x = 0, 1, 2,..., 50. What is the probability statement written mathematically? Solution 4.22 P(x = 35) 4.22 A bag contains letter tiles. Forty-four of the tiles are vowels, and 56 are consonants. Seven tiles are picked at random. You want to know the probability that four of the seven tiles are vowels. What is the group of interest, the size of the group of interest, and the size of the sample? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 245 Example 4.23 Suppose a shipment of 100 DVD players is known to have ten defective players. An inspector randomly chooses 12 for inspection. He is interested in determining the probability that, among the 12 players, at most two are defective. The two groups are the 90 non-defective DVD players and the 10 defective DVD players. The group of interest (first group) is the defective group because the probability question asks for the probability of at most two defective DVD players. The size of the sample is 12
DVD players. (They may be non-defective or defective.) Let X = the number of defective DVD players in the sample of 12. X takes on the values 0, 1, 2,..., 10. X may not take on the values 11 or 12. The sample size is 12, but there are only 10 defective DVD players. Write the probability statement mathematically. Solution 4.23 P(x ≤ 2) 4.23 A gross of eggs contains 144 eggs. A particular gross is known to have 12 cracked eggs. An inspector randomly chooses 15 for inspection. She wants to know the probability that, among the 15, at most three are cracked. What is X, and what values does it take on? Example 4.24 You are president of an on-campus special events organization. You need a committee of seven students to plan a special birthday party for the president of the college. Your organization consists of 18 women and 15 men. You are interested in the number of men on your committee. If the members of the committee are randomly selected, what is the probability that your committee has more than four men? This is a hypergeometric problem because you are choosing your committee from two groups (men and women). a. Are you choosing with or without replacement? Solution 4.24 a. without b. What is the group of interest? Solution 4.24 b. the men c. How many are in the group of interest? Solution 4.24 c. 15 men d. How many are in the other group? Solution 4.24 d. 18 women e. Let X = _________ on the committee. What values does X take on? Solution 4.24 e. Let X = the number of men on the committee. x = 0, 1, 2, …, 7. 246 CHAPTER 4 \| DISCRETE RANDOM VARIABLES f. The probability question is P(_______). Solution 4.24 f. P(x > 4) 4.24 A palette has 200 milk cartons. Of the 200 cartons, it is known that ten of them have leaked and cannot be sold. A stock clerk randomly chooses 18 for inspection. He wants to know the probability that among the 18, no more than two are leaking. Give five reasons why this is a hypergeometric problem. Notation for the Hypergeometric: H = Hypergeometric Probability Distribution Function X ~ H(r, b, n) Read this as "X is a random variable
with a hypergeometric distribution." The parameters are r, b, and n; r = the size of the group of interest (first group), b = the size of the second group, n = the size of the chosen sample. Example 4.25 A school site committee is to be chosen randomly from six men and five women. If the committee consists of four members chosen randomly, what is the probability that two of them are men? How many men do you expect to be on the committee? Let X = the number of men on the committee of four. The men are the group of interest (first group). X takes on the values 0, 1, 2, 3, 4, where r = 6, b = 5, and n = 4. X ~ H(6, 5, 4) Find P(x = 2). P(x = 2) = 0.4545 (calculator or computer) NOTE Currently, the TI-83+ and TI-84 do not have hypergeometric probability functions. There are a number of computer packages, including Microsoft Excel, that do. The probability that there are two men on the committee is about 0.45. The graph of X ~ H(6, 5, 4) is: This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 247 Figure 4.4 The y-axis contains the probability of X, where X = the number of men on the committee. You would expect m = 2.18 (about two) men on the committee. The formula for the mean is µ = nr r + b = (4)(6) 6 + 5 = 2.18 4.25 An intramural basketball team is to be chosen randomly from 15 boys and 12 girls. The team has ten slots. You want to know the probability that eight of the players will be boys. What is the group of interest and the sample? 4.6 \| Poisson Distribution There are two main characteristics of a Poisson experiment. 1. The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event. For example, a book editor might be interested in the number of words spelled incorrectly in a particular
book. It might be that, on the average, there are five words spelled incorrectly in 100 pages. The interval is the 100 pages. 2. The Poisson distribution may be used to approximate the binomial if the probability of success is "small" (such as 0.01) and the number of trials is "large" (such as 1,000). You will verify the relationship in the homework exercises. n is the number of trials, and p is the probability of a "success." The random variable X = the number of occurrences in the interval of interest. Example 4.26 The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12. Of interest is the number of loaves of bread put on the shelf in five minutes. The time interval of interest is five minutes. What is the probability that the number of loaves, selected randomly, put on the shelf in five minutes is three? Let X = the number of loaves of bread put on the shelf in five minutes. If the average number of loaves put on the shelf in 30 minutes (half-hour) is 12, then the average number of loaves put on the shelf in five minutes is ⎛ ⎝ (12) = 2 loaves of bread. ⎞ ⎠ 5 30 248 CHAPTER 4 \| DISCRETE RANDOM VARIABLES The probability question asks you to find P(x = 3). 4.26 The average number of fish caught in an hour is eight. Of interest is the number of fish caught in 15 minutes. The time interval of interest is 15 minutes. What is the average number of fish caught in 15 minutes? Example 4.27 A bank expects to receive six bad checks per day, on average. What is the probability of the bank getting fewer than five bad checks on any given day? Of interest is the number of checks the bank receives in one day, so the time interval of interest is one day. Let X = the number of bad checks the bank receives in one day. If the bank expects to receive six bad checks per day then the average is six checks per day. Write a mathematical statement for the probability question. Solution 4.27 P(x < 5) 4.27 An electronics store expects to have ten returns per day on average. The manager wants to know the probability of the store getting fewer than eight returns on any given day. State the probability question mathematically. Example 4.28
You notice that a news reporter says "uh," on average, two times per broadcast. What is the probability that the news reporter says "uh" more than two times per broadcast. This is a Poisson problem because you are interested in knowing the number of times the news reporter says "uh" during a broadcast. a. What is the interval of interest? Solution 4.28 a. one broadcast b. What is the average number of times the news reporter says "uh" during one broadcast? Solution 4.28 b. 2 c. Let X = ____________. What values does X take on? Solution 4.28 c. Let X = the number of times the news reporter says "uh" during one broadcast. x = 0, 1, 2, 3,... d. The probability question is P(______). Solution 4.28 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 d. P(x > 2) CHAPTER 4 \| DISCRETE RANDOM VARIABLES 249 4.28 An emergency room at a particular hospital gets an average of five patients per hour. A doctor wants to know the probability that the ER gets more than five patients per hour. Give the reason why this would be a Poisson distribution. Notation for the Poisson: P = Poisson Probability Distribution Function X ~ P(μ) Read this as "X is a random variable with a Poisson distribution." The parameter is μ (or λ); μ (or λ) = the mean for the interval of interest. Example 4.29 Leah's answering machine receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than one call in the next 15 minutes? Let X = the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or 1 4 hour.) x = 0, 1, 2, 3,... If Leah receives, on the average, six telephone calls in two hours, and there are eight 15 minute intervals in two hours, then Leah receives ⎞ ⎠ ⎛ ⎝ 1 8 (6) = 0.75 calls in 15 minutes, on average. So, μ = 0.75 for this problem. X ~ P(0.75) Find P(x > 1). P(x
> 1) = 0.1734 (calculator or computer) • Press 1 – and then press 2nd DISTR. • Arrow down to poissoncdf. Press ENTER. • Enter (.75,1). • The result is P(x > 1) = 0.1734. NOTE The TI calculators use λ (lambda) for the mean. The probability that Leah receives more than one telephone call in the next 15 minutes is about 0.1734: P(x > 1) = 1 − poissoncdf(0.75, 1). The graph of X ~ P(0.75) is: 250 CHAPTER 4 \| DISCRETE RANDOM VARIABLES Figure 4.5 The y-axis contains the probability of x where X = the number of calls in 15 minutes. 4.29 A customer service center receives about ten emails every half-hour. What is the probability that the customer service center receives more than four emails in the next six minutes? Use the TI-83+ or TI-84 calculator to find the answer. Example 4.30 According to Baydin, an email management company, an email user gets, on average, 147 emails per day. Let X = the number of emails an email user receives per day. The discrete random variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson distribution: X ~ P(147). The mean is 147 emails. a. What is the probability that an email user receives exactly 160 emails per day? b. What is the probability that an email user receives at most 160 emails per day? c. What is the standard deviation? Solution 4.30 a. P(x = 160) = poissonpdf(147, 160) ≈ 0.0180 b. P(x ≤ 160) = poissoncdf(147, 160) ≈ 0.8666 c. Standard Deviation = σ = µ = 147 ≈ 12.1244 4.30 According to a recent poll by the Pew Internet Project, girls between the ages of 14 and 17 send an average of 187 text messages each day. Let X = the number of texts that a girl aged 14 to 17 sends per day. The discrete random variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson distribution: X ~ P(187). The mean is 187 text messages. a. What is
the probability that a teen girl sends exactly 175 texts per day? b. What is the probability that a teen girl sends at most 150 texts per day? c. What is the standard deviation? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 251 Example 4.31 Text message users receive or send an average of 41.5 text messages per day. a. How many text messages does a text message user receive or send per hour? b. What is the probability that a text message user receives or sends two messages per hour? c. What is the probability that a text message user receives or sends more than two messages per hour? Solution 4.31 a. Let X = the number of texts that a user sends or receives in one hour. The average number of texts received per hour is 41.5 24 ≈ 1.7292. b. X ~ P(1.7292), so P(x = 2) = poissonpdf(1.7292, 2) ≈ 0.2653 c. P(x > 2) = 1 – P(x ≤ 2) = 1 – poissoncdf(1.7292, 2) ≈ 1 – 0.7495 = 0.2505 4.31 Atlanta’s Hartsfield-Jackson International Airport is the busiest airport in the world. On average there are 2,500 arrivals and departures each day. a. How many airplanes arrive and depart the airport per hour? b. What is the probability that there are exactly 100 arrivals and departures in one hour? c. What is the probability that there are at most 100 arrivals and departures in one hour? Example 4.32 On May 13, 2013, starting at 4:30 PM, the probability of low seismic activity for the next 48 hours in Alaska was reported as about 1.02%. Use this information for the next 200 days to find the probability that there will be low seismic activity in ten of the next 200 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close? Solution 4.32 Let X = the number of days with low seismic activity. Using the binomial distribution: • P(x = 10) = binompdf(200,.0102, 10) ≈ 0.0000
39 Using the Poisson distribution: • Calculate μ = np = 200(0.0102) ≈ 2.04 • P(x = 10) = poissonpdf(2.04, 10) ≈ 0.000045 We expect the approximation to be good because n is large (greater than 20) and p is small (less than 0.05). The results are close—both probabilities reported are almost 0. 4.32 On May 13, 2013, starting at 4:30 PM, the probability of moderate seismic activity for the next 48 hours in the Kuril Islands off the coast of Japan was reported at about 1.43%. Use this information for the next 100 days to find the probability that there will be low seismic activity in five of the next 100 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close? 252 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 4.1 Discrete Distribution (Playing Card Experiment) Class Time: Names: Student Learning Outcomes • The student will compare empirical data and a theoretical distribution to determine if an everyday experiment fits a discrete distribution. • The student will demonstrate an understanding of long-term probabilities. Supplies • One full deck of playing cards Procedure The experimental procedure is to pick one card from a deck of shuffled cards. 1. The theoretical probability of picking a diamond from a deck is _________. 2. Shuffle a deck of cards. 3. Pick one card from it. 4. Record whether it was a diamond or not a diamond. 5. Put the card back and reshuffle. 6. Do this a total of ten times. 7. Record the number of diamonds picked. 8. Let X = number of diamonds. Theoretically, X ~ B(_____,_____) Organize the Data 1. Record the number of diamonds picked for your class in Table 4.16. Then calculate the relative frequency Frequency Relative Frequency __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ 10 __________ __________ Table 4.16 2. Calculate the following: This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content
/col11562/1.16 a. b. x¯ = ________ s = ________ 3. Construct a histogram of the empirical data. CHAPTER 4 \| DISCRETE RANDOM VARIABLES 253 Figure 4.6 Theoretical Distribution a. Build the theoretical PDF chart based on the distribution in the Procedure section. x P(x 10 Table 4.17 b. Calculate the following: a. μ = ____________ b. σ = ____________ c. Construct a histogram of the theoretical distribution. 254 CHAPTER 4 \| DISCRETE RANDOM VARIABLES Figure 4.7 Using the Data NOTE RF = relative frequency Use the table from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places. • P(x = 3) = _______________________ • P(1 < x < 4) = _______________________ • P(x ≥ 8) = _______________________ Use the data from the Organize the Data section to calculate the following answers. Round your answers to four decimal places. • RF(x = 3) = _______________________ • RF(1 < x < 4) = _______________________ • RF(x ≥ 8) = _______________________ Discussion Questions For questions 1 and 2, think about the shapes of the two graphs, the probabilities, the relative frequencies, the means, and the standard deviations. 1. Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical and empirical distributions. Use complete sentences. 2. Describe the three most significant differences between the graphs or distributions of the theoretical and empirical distributions. 3. Using your answers from questions 1 and 2, does it appear that the data fit the theoretical distribution? In complete sentences, explain why or why not. 4. Suppose that the experiment had been repeated 500 times. Would you expect Table 4.16 or Table 4.17 to change, and how would it change? Why? Why wouldn’t the other table change? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 255 4.2 Discrete Distribution (Lucky Dice Experiment) Class Time: Names: Student Learning Outcomes • The student will compare empirical data and a theoretical distribution to determine if a Tet gambling game fits a discrete
distribution. • The student will demonstrate an understanding of long-term probabilities. Supplies • one “Lucky Dice” game or three regular dice Procedure Round answers to relative frequency and probability problems to four decimal places. 1. The experimental procedure is to bet on one object. Then, roll three Lucky Dice and count the number of matches. The number of matches will decide your profit. 2. What is the theoretical probability of one die matching the object? 3. Choose one object to place a bet on. Roll the three Lucky Dice. Count the number of matches. 4. Let X = number of matches. Theoretically, X ~ B(______,______) 5. Let Y = profit per game. Organize the Data In Table 4.18, fill in the y value that corresponds to each x value. Next, record the number of matches picked for your class. Then, calculate the relative frequency. 1. Complete the table. y Frequency Relative Frequency x 0 1 2 3 Table 4.18 2. Calculate the following: a. b. c. d. x¯ = _______ sx = ________ y¯ = _______ sy = _______ 3. Explain what x¯ represents. 4. Explain what y¯ represents. 256 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 5. Based upon the experiment: a. What was the average profit per game? b. Did this represent an average win or loss per game? c. How do you know? Answer in complete sentences. 6. Construct a histogram of the empirical data. Figure 4.8 Theoretical Distribution Build the theoretical PDF chart for x and y based on the distribution from the Procedure section. y P(x) = P(y) x 0 1 2 3 Table 4.19 1. 2. Calculate the following: a. μx = _______ b. σx = _______ c. μx = _______ 3. Explain what μx represents. 4. Explain what μy represents. 5. Based upon theory: a. What was the expected profit per game? b. Did the expected profit represent an average win or loss per game? c. How do you know? Answer in complete sentences. 6. Construct a histogram of the theoretical distribution. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CH
APTER 4 \| DISCRETE RANDOM VARIABLES 257 Figure 4.9 Use the Data NOTE RF = relative frequency Use the data from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places. 1. P(x = 3) = _________________ 2. P(0 < x < 3) = _________________ 3. P(x ≥ 2) = _________________ Use the data from the Organize the Data section to calculate the following answers. Round your answers to four decimal places. 1. RF(x = 3) = _________________ 2. RF(0 < x < 3) = _________________ 3. RF(x ≥ 2) = _________________ Discussion Question For questions 1 and 2, consider the graphs, the probabilities, the relative frequencies, the means, and the standard deviations. 1. Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical and empirical distributions. Use complete sentences. 2. Describe the three most significant differences between the graphs or distributions of the theoretical and empirical distributions. 3. Thinking about your answers to questions 1 and 2, does it appear that the data fit the theoretical distribution? In complete sentences, explain why or why not. 4. Suppose that the experiment had been repeated 500 times. Would you expect Table 4.18 or Table 4.19 to change, and how would it change? Why? Why wouldn’t the other table change? 258 CHAPTER 4 \| DISCRETE RANDOM VARIABLES KEY TERMS Bernoulli Trials an experiment with the following characteristics: 1. There are only two possible outcomes called “success” and “failure” for each trial. 2. The probability p of a success is the same for any trial (so the probability q = 1 − p of a failure is the same for any trial). Binomial Experiment a statistical experiment that satisfies the following three conditions: 1. There are a fixed number of trials, n. 2. There are only two possible outcomes, called "success" and, "failure," for each trial. The letter p denotes the probability of a success on one trial, and q denotes the probability of a failure on one trial. 3. The n trials are independent and are repeated using identical conditions. Binomial Probability Distribution a discrete random variable (RV) that arises from Bernoulli trials; there are a fixed number, n, of independent trials. �
�Independent” means that the result of any trial (for example, trial one) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV X is defined as the number of successes in n trials. The notation is: X ~ B(n, p). The mean is μ = np and the standard deviation is σ = npq. The probability of exactly x successes in n trials is n P(X = x) = ⎛ x ⎝ ⎞ ⎠ pxqn − x. Expected Value expected arithmetic average when an experiment is repeated many times; also called the mean. Notations: μ. For a discrete random variable (RV) with probability distribution function P(x),the definition can also be written in the form μ = ∑ xP(x). Geometric Distribution a discrete random variable (RV) that arises from the Bernoulli trials; the trials are repeated until the first success. The geometric variable X is defined as the number of trials until the first success. Notation: X ~ G(p). The mean is μ = 1 p and the standard deviation is ⎞ ⎠. The probability of exactly x failures before the first success is given by the formula: P(X = x) = p(1 – p)x – 1. Geometric Experiment a statistical experiment with the following properties: 1. There are one or more Bernoulli trials with all failures except the last one, which is a success. 2. In theory, the number of trials could go on forever. There must be at least one trial. 3. The probability, p, of a success and the probability, q, of a failure do not change from trial to trial. Hypergeometric Experiment a statistical experiment with the following properties: 1. You take samples from two groups. 2. You are concerned with a group of interest, called the first group. 3. You sample without replacement from the combined groups. 4. Each pick is not independent, since sampling is without replacement. 5. You are not dealing with Bernoulli Trials. Hypergeometric Probability a discrete random variable (RV) that is characterized by: 1. A fixed number of trials. 2. The probability of success is not the same from trial to trial. We sample from two groups of items when we are interested in only one group. X is defined as the number of
successes out of the total number of items chosen. Notation: X ~ H(r, b, n), where r = the number of items in the group of interest, b = the number of items in the group not of interest, and n = the number of items chosen. Mean of a Probability Distribution the long-term average of many trials of a statistical experiment Mean a number that measures the central tendency; a common name for mean is ‘average.’ The term ‘mean’ is a shortened form of ‘arithmetic mean.’ By definition, the mean for a sample (detonated by x¯ ) is This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 x¯ = Sum of all values in the sample Number of values in the sample Sum of all values in the population Number of values in the population. CHAPTER 4 \| DISCRETE RANDOM VARIABLES 259 and the mean for a population (denoted by μ) is μ = Poisson Probability Distribution a discrete random variable (RV) that counts the number of times a certain event will occur in a specific interval; characteristics of the variable: • The probability that the event occurs in a given interval is the same for all intervals. • The events occur with a known mean and independently of the time since the last event. The distribution is defined by the mean μ of the event in the interval. Notation: X ~ P(μ). The mean is μ = np. The µ x standard deviation is σ = µ. The probability of having exactly x successes in r trials is P(X = x ) = (e−µ x! The Poisson distribution is often used to approximate the binomial distribution, when n is “large” and p is “small” (a general rule is that n should be greater than or equal to 20 and p should be less than or equal to 0.05). ). Probability Distribution Function (PDF) a mathematical description of a discrete random variable (RV), given either in the form of an equation (formula) or in the form of a table listing all the possible outcomes of an experiment and the probability associated with each outcome. Random Variable (RV) a characteristic of interest in a population being studied; common notation for variables are upper case Latin letters X
, Y, Z,...; common notation for a specific value from the domain (set of all possible values of a variable) are lower case Latin letters x, y, and z. For example, if X is the number of children in a family, then x represents a specific integer 0, 1, 2, 3,.... Variables in statistics differ from variables in intermediate algebra in the two following ways. • The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if X = hair color then the domain is {black, blond, gray, green, orange}. • We can tell what specific value x the random variable X takes only after performing the experiment. Standard Deviation of a Probability Distribution a number that measures how far the outcomes of a statistical experiment are from the mean of the distribution The Law of Large Numbers As the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency probability approaches zero. CHAPTER REVIEW 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable The characteristics of a probability distribution function (PDF) for a discrete random variable are as follows: 1. Each probability is between zero and one, inclusive (inclusive means to include zero and one). 2. The sum of the probabilities is one. 4.2 Mean or Expected Value and Standard Deviation The expected value, or mean, of a discrete random variable predicts the long-term results of a statistical experiment that has been repeated many times. The standard deviation of a probability distribution is used to measure the variability of possible outcomes. 4.3 Binomial Distribution A statistical experiment can be classified as a binomial experiment if the following conditions are met: 1. There are a fixed number of trials, n. 2. There are only two possible outcomes, called "success" and, "failure" for each trial. The letter p denotes the probability of a success on one trial and q denotes the probability of a failure on one trial. 3. The n trials are independent and are repeated using identical conditions. 260 CHAPTER 4 \| DISCRETE RANDOM VARIABLES The outcomes of a binomial experiment fit a binomial probability distribution. The random variable X = the number of successes obtained in the n independent trials. The mean of X can be calculated using the formula μ = np, and the standard deviation is given by the formula σ = npq. 4.
4 Geometric Distribution There are three characteristics of a geometric experiment: 1. There are one or more Bernoulli trials with all failures except the last one, which is a success. 2. In theory, the number of trials could go on forever. There must be at least one trial. 3. The probability, p, of a success and the probability, q, of a failure are the same for each trial. In a geometric experiment, define the discrete random variable X as the number of independent trials until the first success. We say that X has a geometric distribution and write X ~ G(p) where p is the probability of success in a single trial. The mean of the geometric distribution X ~ G(p) is μ = 1 − p p2 =.5 Hypergeometric Distribution A hypergeometric experiment is a statistical experiment with the following properties: 1. You take samples from two groups. 2. You are concerned with a group of interest, called the first group. 3. You sample without replacement from the combined groups. 4. Each pick is not independent, since sampling is without replacement. 5. You are not dealing with Bernoulli Trials. The outcomes of a hypergeometric experiment fit a hypergeometric probability distribution. The random variable X = the number of items from the group of interest. The distribution of X is denoted X ~ H(r, b, n), where r = the size of the group of interest (first group), b = the size of the second group, and n = the size of the chosen sample. It follows that n ≤ r + b. The mean of X is μ = nr r + b and the standard deviation is σ = rbn(r + b − n) (r + b)2 (r + b − 1). 4.6 Poisson Distribution A Poisson probability distribution of a discrete random variable gives the probability of a number of events occurring in a fixed interval of time or space, if these events happen at a known average rate and independently of the time since the last event. The Poisson distribution may be used to approximate the binomial, if the probability of success is "small" (less than or equal to 0.05) and the number of trials is "large" (greater than or equal to 20). FORMULA REVIEW 4.2 Mean or Expected Value and Standard Deviation Mean or Expected Value: µ = ∑ x ∈ X xP(x) Standard
Deviation: σ = ∑ x ∈ X (x − µ)2 P(x) X = the number of successes in n independent trials n = the number of independent trials X takes on the values x = 0, 1, 2, 3,..., n p = the probability of a success for any trial q = the probability of a failure for any trial.3 Binomial Distribution The mean of X is μ = np. The standard deviation of X is σ = X ~ B(n, p) means that the discrete random variable X has a binomial probability distribution with n trials and probability of success p. npq. 4.4 Geometric Distribution This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 X ~ G(p) means that the discrete random variable X has a geometric probability distribution with probability of success in a single trial p. X = the number of independent trials until the first success X takes on the values x = 1, 2, 3,... up to the size of the group of interest. (The minimum value for X may be larger than zero in some instances.) n ≤ r + b The mean of X is given by the formula μ = nr r + b and the CHAPTER 4 \| DISCRETE RANDOM VARIABLES 261 p = the probability of a success for any trial standard deviation is = rbn(r + b − n) (r + b)2(r + b − 1). q = the probability of a failure for any trial The mean is μ = 1 p. The standard deviation is σ = 1 – p p2 =.5 Hypergeometric Distribution X ~ H(r, b, n) means that the discrete random variable X has a hypergeometric probability distribution with r = the size of the group of interest (first group), b = the size of the second group, and n = the size of the chosen sample. X = the number of items from the group of interest that are in the chosen sample, and X may take on the values x = 0, 1,..., PRACTICE 4.6 Poisson Distribution X ~ P(μ) means that X has a Poisson probability distribution where X = the number of occurrences in the interval of interest. X takes on the values x = 0, 1, 2, 3,
... The mean μ is typically given. The variance is σ2 = μ, and the standard deviation is σ = µ. When P(μ) is used to approximate a binomial distribution, μ = np where n represents the number of independent trials and p represents the probability of success in a single trial. 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable Use the following information to answer the next five exercises: A company wants to evaluate its attrition rate, in other words, how long new hires stay with the company. Over the years, they have established the following probability distribution. Let X = the number of years a new hire will stay with the company. Let P(x) = the probability that a new hire will stay with the company x years. 1. Complete Table 4.20 using the data provided. x P(x) 0 1 2 3 4 5 6 0.12 0.18 0.30 0.15 0.10 0.05 Table 4.20 2. P(x = 4) = _______ 3. P(x ≥ 5) = _______ 4. On average, how long would you expect a new hire to stay with the company? 5. What does the column “P(x)” sum to? Use the following information to answer the next six exercises: A baker is deciding how many batches of muffins to make to 262 CHAPTER 4 \| DISCRETE RANDOM VARIABLES sell in his bakery. He wants to make enough to sell every one and no fewer. Through observation, the baker has established a probability distribution. x P(x) 1 2 3 4 0.15 0.35 0.40 0.10 Table 4.21 6. Define the random variable X. 7. What is the probability the baker will sell more than one batch? P(x > 1) = _______ 8. What is the probability the baker will sell exactly one batch? P(x = 1) = _______ 9. On average, how many batches should the baker make? Use the following information to answer the next four exercises: Ellen has music practice three days a week. She practices for all of the three days 85% of the time, two days 8% of the time, one day 4% of the time, and no days 3% of the time. One week is selected at random. 10. Define the random variable X. 11. Construct a probability distribution table for the
data. 12. We know that for a probability distribution function to be discrete, it must have two characteristics. One is that the sum of the probabilities is one. What is the other characteristic? Use the following information to answer the next five exercises: Javier volunteers in community events each month. He does not do more than five events in a month. He attends exactly five events 35% of the time, four events 25% of the time, three events 20% of the time, two events 10% of the time, one event 5% of the time, and no events 5% of the time. 13. Define the random variable X. 14. What values does x take on? 15. Construct a PDF table. 16. Find the probability that Javier volunteers for less than three events each month. P(x < 3) = _______ 17. Find the probability that Javier volunteers for at least one event each month. P(x > 0) = _______ 4.2 Mean or Expected Value and Standard Deviation 18. Complete the expected value table. x P(x) xP(x) 0 1 2 3 0.2 0.2 0.4 0.2 Table 4.22 19. Find the expected value from the expected value table. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 \| DISCRETE RANDOM VARIABLES 263 20. Find the standard deviation. x P(x) xP(x) 2 4 6 8 0.1 0.3 0.4 0.2 2(0.1) = 0.2 4(0.3) = 1.2 6(0.4) = 2.4 8(0.2) = 1.6 Table 4.23 x P(x) x*P(x) (x – μ)2P(x) 2(0.1) = 0.2 (2–5.4)2(0.1) = 1.156 4(0.3) = 1.2 (4–5.4)2(0.3) = 0.588 6(0.4) = 2.4 (6–5.4)2(0.4) = 0.144 8(0.2) = 1.6 (8–5.4

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