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 ∩ B ) = P ( B ) P ( A | B ) . We can think of the intersection symbol as substituting for the word "and". This rule may also be written as: P ( A | B ) = P ( A ∩ B ) P ( B ) This equation is read as 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 ∩ B ) = P ( A | B ) P ( B ) becomes P ( A ∩ B ) = P ( A ) ( B ) because the P ( A | B ) = P ( A ) if A and B are independent. One easy way to remember the multiplication rule is that the word "and" means that the event has to satisfy two conditions. For example the name drawn from the class roster is to be both a woman and a sophomore. It is harder to satisfy two conditions than only one and of course when we multiply fractions the result is always smaller. This reflects the increasing difficulty of satisfying two conditions. The Addition Rule If A and B are defined on a sample space, then: P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) . We can think of the union symbol substituting for the word "or". The reason we subtract the intersection of A and B is to keep from double counting elements that are in both A and B . If A and B are mutually exclusive , then P ( A ∩ B ) = 0 . Then P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) becomes P ( A ∪ B ) = P ( A ) + P ( B ) . 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 ∩ 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 ∪ 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. Try It Anna has to buy a new car. She has two choices, car A and car B. Anna can afford only one car. The probability that Anna will buy car A is P ( A ) = 0 . 25 , and the probability that Anna will buy car B is P ( B ) = 0 . 65 . Find: P ( A A N D B ) P ( A O R B ) P ( A A N D B ) = 0 because it is given that Anna can afford only one car. Since P ( A A N D B ) = 0 , then P ( A O R B ) = P ( A ) + P ( B ) = 0 . 25 + 0 . 65 = 0 . 9 . Note that the probability that Anna does not a car must be 0.1. 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 | that he made the first goal is 0.90. What is the probability that he makes both goals? What is the probability that Carlos makes either the first goal or the second goal? Are A and B independent? Are A and B mutually exclusive? a. The problem is asking you to find P ( A ∩ B ) = P ( B ∩ A ) . Since P ( B | A ) = 0.90: P ( B ∩ A ) = P ( B | A ) P ( A ) = (0.90)(0.65) = 0.585 Carlos makes the first and second goals with probability 0.585. b. The problem is asking you to find P ( A ∪ B ). P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) = 0.65 + 0.65 - 0.585 = 0.715 Carlos makes either the first goal or the second goal with probability 0.715. c. No, they are not, because P ( B ∩ A ) = 0.585. P ( B ) P ( A ) = (0.65)(0.65) = 0.423 0.423 ≠ 0.585 = P ( B ∩ A ) So, P ( B ∩ A ) is not equal to P ( B ) P ( A ). d. No, they are not because P ( A ∩ B ) = 0.585. To be mutually exclusive, P ( A ∩ B ) must equal zero. Try It 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? P ( D | C ) = 0.85 P ( C ∩ D ) = P ( D ∩ C ) P ( D ∩ C ) = P ( D | C ) P ( C ) = (0.85)(0.75) = 0.6375 Helen makes the first and second free throws with probability 0.6375. A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven 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. What is the probability that the member is a novice swimmer? What is the probability that the member practices four times a week? What is the probability that the member is an advanced swimmer and practices four times a week? 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? Are being a novice swimmer and practicing four times a week independent events? Why or why not? 28 150 80 150 40 150 P (advanced ∩ intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time. No, these are not independent events. P (novice ∩ practices four times per week) = 0.0667 P (novice) P (practices four times per week) = 0.0996 0.0667 ≠ 0.0996 Try It 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? P = 200 − 140 − 40 200 = 20 200 = 0.1 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 | that she enrolls in speech class is 0.25. Let: M = math class, S = speech class, M | S = math given speech What is the probability that Felicity enrolls in math and speech? Find P ( M ∩ S ) = P ( M | S ) P ( S ). What is the probability that Felicity enrolls in math or speech classes? Find P ( M ∪ S ) = P ( M ) + P ( S ) - P ( M ∩ S ). Are M and S independent? Is P ( M | S ) = P ( M )? Are M and S mutually exclusive? Is P ( M ∩ S ) = 0? a. 0.1625, b. 0.6875, c. No, d. No Try It 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. Find P ( B ∩ D ). Find P ( B ∪ D ). P ( B ∩ D ) = P ( D | B ) P ( B ) = (0.5)(0.4) = 0.20. P ( B ∪ D ) = P ( B ) + P ( D ) − P ( B ∩ D ) = 0.40 + 0.30 − 0.20 = 0.50 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. What is the probability that the woman develops breast cancer? What is the probability that woman tests negative? Given that the woman has breast cancer, what is the probability that she tests negative? What is the probability that the woman has breast cancer AND tests negative? What is the probability that the woman has breast cancer or tests negative? Are having breast cancer and testing negative independent events? Are having breast cancer and testing negative mutually exclusive? P ( B ) = 0.143; P ( N ) = 0.85 P ( N | B ) = 0.02 P ( B ∩ N ) = P ( B ) P ( N | B ) = (0.143)(0.02) = 0.0029 P ( B ∪ N ) = P ( B ) + P ( N ) - P ( B ∩ N ) = 0.143 + 0.85 - 0.0029 = 0.9901 No. P ( N ) = 0.85; P ( N | B ) = 0.02. So, P ( N | B ) does not equal P ( N ). No. P ( B ∩ N ) = 0.0029. For B and N to be mutually exclusive, P ( B ∩ N ) must be zero. Try It 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? Let A = student is a senior going to college. Let B = student plays sports. P ( B ) = 140 200 P ( B | A ) = 50 140 P ( A ∩ B ) = P ( B | A ) P ( A ) P ( A ∩ B ) = ( 140 200 ) ( 50 140 ) = 1 4 Refer to the information in . P = tests positive. Given that a woman develops breast cancer, what is the probability that she tests positive. Find P ( P | B ) = 1 - P ( N | B ). What is the probability that a woman develops breast cancer and tests positive. Find P ( B ∩ P ) = P ( P | B ) P ( B ). What is the probability that a woman does not develop breast cancer. Find P ( B′ ) = 1 - P ( B ). What is the probability that a woman tests positive for breast cancer. Find P ( P ) = 1 - P ( N ). a. 0.98; b. 0.1401; c. 0.857; d. 0.15 Try It 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. Find P ( B′ ). Find P ( D ∩ B ). Find P ( B | D ). Find P ( D ∩ B′ ). Find P ( D | B′ ). P ( B′ ) = 0.60 P ( D ∩ B ) = P ( D | B ) P ( B ) = 0.20 P ( B | D ) = P ( B ∩ D ) P ( D ) = ( 0.20 ) (0 .30) = 0.66 P ( D ∩ B′ ) = P ( D ) - P ( D ∩ B ) = 0.30 - 0.20 = 0.10 P ( D | B′ ) = P ( D ∩ B′ ) P ( B′ ) = ( P ( D ) - P ( D ∩ B ))(0.60) = (0.10)(0.60) = 0.06 References 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/forms/News Archives/News Releases/74209_TO_Issues_-_Mayoral_Approval_%28Forum_Research%29%2820130320%29.pdf (accessed May 2, 2013). “Roulette.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Roulette (accessed May 2, 2013). Data from U.S. Census Bureau. www.census.gov/library/stories/2022/12/languages-we-speak-in-united-states.html (accessed June 19, 2023). 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 the University of Connecticut. Available online at http://www.ropercenter.uconn.edu/ (accessed May 2, 2013). Data from Field Research Corporation. Available online at www.field.com/fieldpollonline (accessed May 2,2 013). Chapter Review 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. Formula Review The multiplication rule: P ( A ∩ B ) = P ( A | B ) P ( B ) The addition rule: P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) Use the following information to answer the next ten exercises. A local restaurant knows that the probability that a customer will order a pizza is 87%. The restaurant also knows that the probability that a customer will order a salad is 32%. Of the customers who order pizzas, 55% of them also order a salad In this problem, let: Z = event that a customer orders a pizza S = event that a customer orders a salad Suppose that one customer is randomly selected. Find P ( Z ). Find P ( Z ) : P ( Z ) = 0.87 Find P ( S ). Find P ( S ) : P ( S ) = 0.32 Find P ( S | Z ). Find P ( S | Z ) : P ( S | Z ) = 0.55 In words, what is S | Z ? In words, what is S | Z ? : S | Z means, given that the customer has ordered pizza, the person also orders a salad. Find P ( Z ∩ S ) . Find P ( Z A N D S ) : P ( Z A N D S ) = 0.4785 In words, what is Z ∩ S ? In words, what is Z A N D S ? : Z A N D S represents the event that a customer orders a pizza and salad. Are P and S independent events? Show why or why not. Are P and S independent events? Show why or why not. No, because P ( S ) does not equal P ( S | Z ). Find P ( Z ∪ S ) . Find P ( Z O R S ) : P ( Z O R S ) = 0.7115 In words, what is Z ∪ S ? In words, what is Z O R S ? Z O R S represents the event that a customer orders a pizza or salad. Are Z and S mutually exclusive events? Show why or why not. Are Z and S mutually exclusive events? Show why or why not. No, because P ( S and Z ) does not equal 0. Homework Prior to the 2015 Supreme Court decision legalizing same-sex marriage nationwide, a 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. Find P ( C ). Find P ( B ). Find P ( C | A ). Find P ( B | C ). In words, what is C | A ? In words, what is B | C ? Find P ( C ∩ B ). In words, what is C ∩ B ? Find P ( C ∪ B ). Are C and B mutually exclusive events? Show why or why not. A survey was conducted in a large city to measure the popularity of that city’s mayor. The survey was repeated every year for three years. The survey polled 1,046 people to measure the mayor’s popularity. Everyone polled expressed either approval or disapproval. These are the results the poll produced: In Year 1, 60% of the population approved of the mayor’s actions in office. In Year 2, 57% of the population approved of his actions. In Year 3, the percentage of popular approval was measured at 42%. What is the sample size for this study? What proportion in the poll disapproved of the mayor, according to the results from Year 3? How many people polled responded that they approved the mayor based on results from Year 3? What is the probability that a person supported the mayor, based on the data collected in Year 2? What is the probability that a person supported the mayor, based on the data collected in Year 1? The Forum Research surveyed 1,046 Torontonians. 58% 42% of 1,046 = 439 (rounding to the nearest integer) 0.57 0.60. 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. (credit: film8ker/wikibooks) List the sample space of the 38 possible outcomes in roulette. You bet on red. Find P (red). You bet on -1st 12- (1st Dozen). Find P (-1st 12-). You bet on an even number. Find P (even number). Is getting an odd number the complement of getting an even number? Why? Find two mutually exclusive events. Are the events Even and 1st Dozen independent? Compute the probability of winning the following types of bets: Betting on two lines that touch each other on the table as in 1-2-3-4-5-6 Betting on three numbers in a line, as in 1-2-3 Betting on one number Betting on four numbers that touch each other to form a square, as in 10-11-13-14 Betting on two numbers that touch each other on the table, as in 10-11 or 10-13 Betting on 0-00-1-2-3 Betting on 0-1-2; or 0-00-2; or 00-2-3 P (Betting on two line that touch each other on the table) = 6 38 P (Betting on three numbers in a line) = 3 38 P (Betting on one number) = 1 38 P (Betting on four number that touch each other to form a square) = 4 38 P (Betting on two number that touch each other on the table ) = 2 38 P (Betting on 0-00-1-2-3) = 5 38 P (Betting on 0-1-2; or 0-00-2; or 00-2-3) = 3 38 Compute the probability of winning the following types of bets: Betting on a color Betting on one of the dozen groups Betting on the range of numbers from 1 to 18 Betting on the range of numbers 19–36 Betting on one of the columns Betting on an even or odd number (excluding zero) 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. G = card drawn is green E = card drawn is even-numbered List the sample space. P ( G ) = _____ P ( G | E ) = _____ P ( G ∩ E ) = _____ P ( G ∪ E ) = _____ Are G and E mutually exclusive? Justify your answer numerically. { G 1, G 2, G 3, G 4, G 5, Y 1, Y 2, Y 3} 5 8 2 3 2 8 6 8 No, because P ( G ∩ E ) does not equal 0. Roll two fair dice separately. Each die has six faces. List the sample space. Let A be the event that either a three or four is rolled first, followed by an even number. Find P ( A ). Let B be the event that the sum of the two rolls is at most seven. Find P ( B ). In words, explain what “ P ( A | B )” represents. Find P ( A | B ). Are A and B mutually exclusive events? Explain your answer in one to three complete sentences, including numerical justification. Are A and B independent events? Explain your answer in one to three complete sentences, including numerical justification. 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. List the sample space. Let A be the event that a blue card is picked first, followed by landing a head on the coin toss. Find P ( A ). 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. 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. NOTE The coin toss is independent of the card picked first. {( G , H ) ( G , T ) ( B , H ) ( B , T ) ( R , H ) ( R , T )} P ( A ) = P (blue) P (head) = ( 3 10 ) ( 1 2 ) = 3 20 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 ∩ B ) = 0 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 ∩ C ) = P ( A ) = 3 20 An experiment consists of first rolling a die and then tossing a coin. List the sample space. 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 ). 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. An experiment consists of tossing a nickel, a dime, and a quarter. Of interest is the side the coin lands on. List the sample space. Let A be the event that there are at least two tails. Find P ( A ). 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. S = {( HHH ), ( HHT ), ( HTH ), ( HTT ), ( THH ), ( THT ), ( TTH ), ( TTT )} 4 8 Yes, because if A has occurred, it is impossible to obtain two tails. In other words, P ( A ∩ B ) = 0. Consider the following scenario: Let P ( C ) = 0.4. Let P ( D ) = 0.5. Let P ( C | D ) = 0.6. Find P ( C ∩ D ). Are C and D mutually exclusive? Why or why not? Are C and D independent events? Why or why not? Find P ( C ∪ D ). Find P ( D | C ). Y and Z are independent events. Rewrite the basic Addition Rule P ( Y ∪ Z ) = P ( Y ) + P ( Z ) - P ( Y ∩ Z ) using the information that Y and Z are independent events. Use the rewritten rule to find P ( Z ) if P ( Y ∪ Z ) = 0.71 and P ( Y ) = 0.42. If Y and Z are independent, then P ( Y ∩ Z ) = P ( Y ) P ( Z ), so P ( Y ∪ Z ) = P ( Y ) + P ( Z ) - P ( Y ) P ( Z ). 0.5 G and H are mutually exclusive events. P ( G ) = 0.5 P ( H ) = 0.3 Explain why the following statement MUST be false: P ( H | G ) = 0.4. Find P ( H ∪ G ). Are G and H independent or dependent events? Explain in a complete sentence. According to the 2019 U.S. Census, approximately 331,449,281 people live in the United States. Of these people, 67,800,000 speak a language other than English at home. Of those who speak another language at home, 61.6% 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 ′ ) = i. 0.7954 b. P ( E ) = ii. 0.616 c. P ( S ∩ E ′ ) = iii. 0.2046 d. P ( S | E ′ ) = iv. 0.1260 iii i iv ii 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. What was Renate’s chance of winning a Green Card? Write your answer as a probability statement. 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. Are G and F independent or dependent events? Justify your answer numerically and also explain why. Are G and F mutually exclusive events? Justify your answer numerically and explain why. 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 Write a probability statement for the overall percent of money returned. Write a probability statement for the percent of money returned out of the economics classes. Write a probability statement for the percent of money returned out of the other classes. Is money being returned independent of the class? Justify your answer numerically and explain it. Based upon this study, do you think that economists are more selfish than other people? Explain why or why not. Include numbers to justify your answer. P ( R ) = 0.44 P ( R | E ) = 0.56 P ( R | O ) = 0.31 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 ). 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 ). 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 1,517 506 136 714 2,873 Jackie Robinson 1,054 273 54 137 1,518 Ty Cobb 3,053 724 295 117 4,189 Hank Aaron 2,294 624 98 755 3,771 Total 7,918 2,127 583 1,723 12,351 Are "the hit being made by Hank Aaron" and "the hit being a double" independent events? Yes, because P (hit by Hank Aaron | hit is a double) = P (hit by Hank Aaron) No, because P (hit by Hank Aaron | hit is a double) ≠ P (hit is a double) No, because P (hit is by Hank Aaron | hit is a double) ≠ P (hit by Hank Aaron) Yes, because P (hit is by Hank Aaron | hit is a double) = P (hit is a double) According to the American Red Cross, 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 45% of people have type O blood and 7% of people have Rh- factor; 38% of people have type O or Rh+ factor. Find the probability that a person has both type O blood and the Rh- factor. Find the probability that a person does NOT have both type O blood and the Rh- factor. P (type O OR Rh-) = P (type O) + P (Rh-) – P (type O AND Rh-) 0.38 = 0.45 + 0.07 – P (type O AND Rh-); Solve to find P (type O AND Rh-) = 0.14. 14% of people have type O, Rh- blood. P (NOT(type O AND Rh-)) = 1 – P (type O AND Rh-) = 1 – 0.14 = 0.86 86% of people do not have type O, Rh- blood. 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. Find the probability that a course has a final exam or a research project. Find the probability that a course has NEITHER of these two requirements. In a box of assorted cookies, 36% contain chocolate and 12% contain nuts. In the box, 8% contain both chocolate and nuts. Sean is allergic to both chocolate and nuts. Find the probability that a cookie contains chocolate or nuts (he can't eat it). Find the probability that a cookie does not contain chocolate or nuts (he can eat it). Let C = be the event that the cookie contains chocolate. Let N = the event that the cookie contains nuts. P ( C ∪ N ) = P ( C ) + P ( N ) - P ( C ∩ N ) = 0.36 + 0.12 - 0.08 = 0.40 P (NEITHER chocolate NOR nuts) = 1 - P ( C ∪ N ) = 1 - 0.40 = 0.60 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 Find P ( D ∩ E ). Find P ( E | D ). Find P ( D ∪ E ). Using an appropriate test, show whether D and E are independent. Using an appropriate test, show whether D and E are mutually exclusive. 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: P ( A | B ) = P ( A ) P ( B | A ) = P ( B ) P ( A ∩ 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 ∩ B ) = 0.

Business


Contingency Tables and Probability Trees 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. 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 Total Uses cell phone while driving 25 280 305 Does not use cell phone while driving 45 405 450 Total 70 685 755 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. Find P (Driver is a cell phone user). Find P (Driver had no violation in the last year). Find P (Driver had no violation in the last year ∩ was a cell phone user). Find P (Driver is a cell phone user ∪ driver had no violation in the last year). Find P (Driver is a cell phone user | driver had a violation in the last year). Find P (Driver had no violation last year | driver was not a cell phone user) number of cell phone users total number in study = 305 755 number that had no violation total number in study = 685 755 280 755 ( 305 755 + 685 755 ) − 280 755 = 710 755 25 70 (The sample space is reduced to the number of drivers who had a violation.) 405 450 (The sample space is reduced to the number of drivers who were not cell phone users.) Try it 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 55 295 350 Does not stretch 231 219 450 Total 286 514 800 What is P (athlete stretches before exercising)? What is P (athlete stretches before exercising | no injury in the last year)? P (athlete stretches before exercising) = 350 800 = 0.4375 P (athlete stretches before exercising | no injury in the last year) = 295 514 = 0.5739 shows a random sample of 100 hikers and the areas of hiking they prefer. Hiking Area Preference Sex The coastline Near lakes and streams On mountain peaks Total Women 18 16 ___ 45 Men ___ ___ 14 55 Total ___ 41 ___ ___ a. Complete the table. a. Hiking Area Preference Sex The coastline Near lakes and streams On mountain peaks Total Women 18 16 11 45 Men 16 25 14 55 Total 34 41 25 100 b. Are the events "being a woman" and "preferring the coastline" independent events? Let F = being a woman and let C = preferring the coastline. Find P ( F ∩ C ) . 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. b. P ( F ∩ C ) = 18 100 = 0.18 P ( F ) P ( C ) = ( 45 100 ) ( 34 100 ) = (0.45)(0.34) = 0.153 P ( F ∩ C ) ≠ P ( F ) P ( C ), so the events F and C are not independent. c. Find the probability that a person is a man given that the person prefers hiking near lakes and streams. Let M = being a man, and let L = prefers hiking near lakes and streams. What word tells you this is a conditional? Fill in the blanks and calculate the probability: P (___ | ___) = ___. Is the sample space for this problem all 100 hikers? If not, what is it? c. The word 'given' tells you that this is a conditional. P ( M | L ) = 25 41 No, the sample space for this problem is the 41 hikers who prefer lakes and streams. d. Find the probability that a person is a woman or prefers hiking on mountain peaks. Let F = being a woman, and let P = prefers mountain peaks. Find P ( F ). Find P ( P ). Find P ( F ∩ P ) . Find P ( F ∪ P ) . d. P ( F ) = 45 100 P ( P ) = 25 100 P ( F ∩ P ) = 11 100 P ( F ∪ P ) = 45 100 + 25 100 - 11 100 = 59 100 Try It shows a random sample of 200 cyclists and the routes they prefer. Let O = older and H = hilly path. Age Group Lake Path Hilly Path Wooded Path Total Younger 45 38 27 110 Older 26 52 12 90 Total 71 90 39 200 Out of the older group, what is the probability that the cyclist prefers a hilly path? Are the events “being older” and “preferring the hilly path” independent events? P ( H | M ) = 52 90 = 0.5778 For M and H to be independent, show P ( H | M ) = P ( H ) P ( H | M ) = 0.5778, P ( H ) = 90 200 = 0.45 P ( H | M ) does not equal P ( H ) so M and H are NOT independent. 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 . It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is 1 3 . Door Choice Caught or not Door one Door two Door three Total Caught 1 15 1 12 1 6 ____ Not caught 4 15 3 12 1 6 ____ Total ____ ____ ____ 1 The first entry 1 15 = ( 1 5 ) ( 1 3 ) is P ( Door One ∩ Caught ) The entry 4 15 = ( 4 5 ) ( 1 3 ) is P ( Door One ∩ Not Caught ) Verify the remaining entries. a. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1. a. Door Choice Caught or not Door one Door two Door three Total Caught 1 15 1 12 1 6 19 60 Not caught 4 15 3 12 1 6 41 60 Total 5 15 4 12 2 6 1 b. What is the probability that Alissa does not catch Muddy? b. 41 60 c. What is the probability that Muddy chooses Door One ∪ Door Two given that Muddy is caught by Alissa? c. 9 19 Try It Fred's preference to drink coffee or iced tea depends on the season. When it is summer, Fred's preference to drink coffee is 1 4 , and his preference to drink iced tea is 3 4 . When it is rainy season, Fred's preference to drink coffee is 1 2 , and his preference to drink iced tea is 1 2 . When it is winter, Fred's preference to drink coffee is 1 5 , and his preference to drink iced tea is 1 5 . A day randomly selected in a year has a probability of 1 3 for each season. The event here is selection of a day randomly in a year. Season Summer Rainy Winter Total Coffee 1 12 1 6 4 15 __ Iced tea 1 4 1 6 1 15 __ Total __ __ __ 1 The first entry is 1 4 1 3 = 1 12 . The second entry is 3 4 1 3 = 1 4 . Verify the remaining entries. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1. What is the probability that Fred will drink iced tea? What is the probability that the day is in summer or rainy season given that Fred drinks iced tea? Season Summer Rainy Winter Total Coffee 1 12 1 6 4 15 31 60 Iced tea 1 4 1 6 1 15 29 60 Total 4 12 2 6 5 15 1 29 60 25 29 contains the number of crimes per 100,000 inhabitants in the United States over the span of several years. United States Crime Index Rates Per 100,000 Inhabitants Year Robbery Burglary Vandalism Vehicle Total 1 145.7 732.1 29.7 314.7 2 133.1 717.7 29.1 259.2 3 119.3 701 27.7 239.1 4 113.7 702.2 26.8 229.6 Total TOTAL each column and each row. Total data = 4,520.7 Find P (Year 2 AND Robbery). Find P (Year 3 AND Burglary). Find P (Year 3 OR Burglary). Find P (Year 4 | Vandalism). Find P (Vehicle | Year 1). a. 0.0294, b. 0.1551, c. 0.7165, d. 0.2365, e. 0.2575 Try It relates the weights and heights of a group of individuals participating in an observational study. Weight/height Tall Medium Short Totals Overweight 18 28 14 Typical Weight Range 20 51 28 Underweight 12 25 9 Totals Find the total for each row and column Find the probability that a randomly chosen individual from this group is Tall. Find the probability that a randomly chosen individual from this group is Overweight and Tall. Find the probability that a randomly chosen individual from this group is Tall given that the individual is Overweight. Find the probability that a randomly chosen individual from this group is Overweight given that the individual is Tall. Find the probability a randomly chosen individual from this group is Tall and Underweight. Are the events Overweight and Tall independent? Weight/height Tall Medium Short Totals Obese 18 28 14 60 Normal 20 51 28 99 Underweight 12 25 9 46 Totals 50 104 51 205 Row Totals: 60, 99, 46. Column totals: 50, 104, 51. P (Tall) = 50 205 = 0.244 P ( Obese ∩ Tall ) = 18 205 = 0.088 P (Tall | Obese) = 18 60 = 0.3 P (Obese | Tall) = 18 50 = 0.36 P ( Tall ∩ Underweight ) = 12 205 = 0.0585 No. P (Tall) does not equal P (Tall | Obese). Tree Diagrams Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams 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. 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. 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 R 1, R 2, and R 3 and each blue ball as B 1, B 2, B 3, B 4, B 5, B 6, B 7, and B 8. Then the nine RR outcomes can be written as: R 1 R 1 R 1 R 2 R 1 R 3 R 2 R 1 R 2 R 2 R 2 R 3 R 3 R 1 R 3 R 2 R 3 R 3 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: B 1 R 1, B 1 R 2, B 1 R 3, ... a. B 1 R 1 B 1 R 2 B 1 R 3 B 2 R 1 B 2 R 2 B 2 R 3 B 3 R 1 B 3 R 2 B 3 R 3 B 4 R 1 B 4 R 2 B 4 R 3 B 5 R 1 B 5 R 2 B 5 R 3 B 6 R 1 B 6 R 2 B 6 R 3 B 7 R 1 B 7 R 2 B 7 R 3 B 8 R 1 B 8 R 2 B 8 R 3 b. Using the tree diagram, calculate P ( RR ). b. P ( RR ) = ( 3 11 ) ( 3 11 ) = 9 121 c. Using the tree diagram, calculate P ( RB ∪ BR ) . c. P ( RB ∪ BR ) = ( 3 11 ) ( 8 11 ) + ( 8 11 ) ( 3 11 ) = 48 121 d. Using the tree diagram, calculate P ( R on 1st draw ∩ B on 2nd draw ) . d. P ( R on 1st draw ∩ B on 2nd draw ) = ( 3 11 ) ( 8 11 ) = 24 121 e. Using the tree diagram, calculate P ( R on 2nd draw | B on 1st draw). e. P ( R on 2nd draw | B on 1st draw) = P ( R on 2nd | B on 1st) = 24 88 = 3 11 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 ). f. P ( BB ) = 64 121 g. Using the tree diagram, calculate P ( B on the 2nd draw | R on the first draw). 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 . Try It 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 ). Total number of outcomes is 144 + 480 + 480 + 1600 = 2,704. P ( FF ) = 144 144 + 480 + 480 + 1,600 = 144 2 , 704 = 9 169 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, ( 3 11 ) ( 2 10 ) = 6 110 . 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 ) = ________ a. P ( RR ) = ( 3 11 ) ( 2 10 ) = 6 110 b. Fill in the blanks: P ( RB ∪ BR ) = ( 3 11 ) ( 8 10 ) + (___)(___) = 48 110 b. P ( RB ∪ BR ) = ( 3 11 ) ( 8 10 ) + ( 8 11 ) ( 3 10 ) = 48 110 c. P ( R on 2nd | B on 1st) = c. P ( R on 2nd | B on 1st) = 3 10 d. Fill in the blanks. P ( R on 1st ∩ B on 2nd ) = (___)(___) = 24 110 d. P ( R on 1st ∩ B on 2nd ) = ( 3 11 ) ( 8 10 ) = 24 110 e. Find P ( BB ). e. P ( BB ) = ( 8 11 ) ( 7 10 ) f. Find P ( B on 2nd | R on 1st). 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) Try It 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. Find P ( FN ∪ NF ) . Find P ( N | F ). Find P (at most one face card). Hint: "At most one face card" means zero or one face card. Find P (at least one face card). Hint: "At least one face card" means one or two face cards. P ( FN ∪ NF ) = 480 2,652 + 480 2,652 = 960 2,652 = 80 221 P ( N | F ) = 40 51 P (at most one face card) = (480 + 480 + 1,560) 2,652 = 2 , 520 2 , 652 P (at least one face card) = (132 + 480 + 480) 2,652 = 1,092 2,652 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. What is the probability that both kittens are tabby? a. ( 1 2 ) ( 1 2 ) b. ( 4 9 ) ( 4 9 ) c. ( 4 9 ) ( 3 8 ) d. ( 4 9 ) ( 5 9 ) What is the probability that one kitten of each coloring is selected? a. ( 4 9 ) ( 5 9 ) b. ( 4 9 ) ( 5 8 ) c. ( 4 9 ) ( 5 9 ) + ( 5 9 ) ( 4 9 ) d. ( 4 9 ) ( 5 8 ) + ( 5 9 ) ( 4 8 ) What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first? What is the probability of choosing two kittens of the same color? a. c, b. d, c. 4 8 , d. 32 72 Try It Suppose there are four red balls and three yellow balls in a box. Two balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected? ( 4 7 ) ( 3 6 ) + ( 3 7 ) ( 4 6 ) References “Blood Types.” American Red Cross, 2013. Available online at http://www.redcrossblood.org/learn-about-blood/blood-types (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, Malcolm 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). 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). 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 https://ropercenter.cornell.edu/?s=Search+for+Datasets (accessed February 6, 2019). Chapter Review 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 probabilities that have multiple dependent variables. A tree diagram use branches to show the different outcomes of experiments and makes complex probability questions easy to visualize. 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) 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.

Business


Venn Diagrams 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. Venn diagrams also help us to convert common English words into mathematical terms that help add precision. Venn diagrams are named for their inventor, John Venn, a mathematics professor at Cambridge and an Anglican minister. His main work was conducted during the late 1870's and gave rise to a whole branch of mathematics and a new way to approach issues of logic. We will develop the probability rules just covered using this powerful way to demonstrate the probability postulates including the Addition Rule, Multiplication Rule, Complement Rule, Independence, and Conditional Probability. 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 intersect B = A ∩ B = {6} and A union B = A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9}. . The Venn diagram is as follows: shows the most basic relationship among these numbers. First, the numbers are in groups called sets; set A and set B. Some number are in both sets; we say in set A ∩ in set B. The English word "and" means inclusive, meaning having the characteristics of both A and B, or in this case, being a part of both A and B. This condition is called the INTERSECTION of the two sets. All members that are part of both sets constitute the intersection of the two sets. The intersection is written as A ∩ B where ∩ is the mathematical symbol for intersection. The statement A ∩ B is read as "A intersect B." You can remember this by thinking of the intersection of two streets. There are also those numbers that form a group that, for membership, the number must be in either one or the other group. The number does not have to be in BOTH groups, but instead only in either one of the two. These numbers are called the UNION of the two sets and in this case they are the numbers 1-5 (from A exclusively), 7-9 (from set B exclusively) and also 6, which is in both sets A and B. The symbol for the UNION is ∪ , thus A ∪ B = numbers 1-9, but excludes number 10, 11, and 12. The values 10, 11, and 12 are part of the universe, but are not in either of the two sets. Translating the English word "AND" into the mathematical logic symbol ∩ , intersection, and the word "OR" into the mathematical symbol ∪ , union, provides a very precise way to discuss the issues of probability and logic. The general terminology for the three areas of the Venn diagram in is shown in . Try It 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 ∩ P = {blue} and C ∪ P = {green, blue, purple, red, yellow} . Draw a Venn diagram representing this situation. 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 ∩ B = {TT} . A ∪ 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: Try It 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 ∩ B = {3, 5} . A ∪ 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. 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. 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 Rh- factor. Let O = African American with Type O blood and R = African American with Rh- factor. P ( O ) = ___________ P ( R ) = ___________ P ( O ∩ R ) = ___________ P ( O ∪ R ) = ____________ 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. 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. Try It 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. 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. 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 ∩ PT ) = 0.05 the probability that the student belongs to a club given that the student works part time. P ( C | P T ) = P ( C ∩ P T ) P ( P T ) = 0.05 0.50 = 0.1 the probability that the student belongs to a club OR works part time. P ( C ∪ PT ) = P ( C ) + P ( PT ) - P ( C ∩ PT ) = 0.40 + 0.50 - 0.05 = 0.85 In order to solve we had to draw upon the concept of conditional probability from the previous section. There we used tree diagrams to track the changes in the probabilities, because the sample space changed as we drew without replacement. In short, conditional probability is the chance that something will happen given that some other event has already happened. Put another way, the probability that something will happen conditioned upon the situation that something else is also true. In the probability P ( C | PT ) is the conditional probability that the randomly drawn student is a member of the club, conditioned upon the fact that the student also is working part time. This allows us to see the relationship between Venn diagrams and the probability postulates. Try It 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. 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. Draw a Venn diagram representing the situation. Find the probability that the customer buys either a novel or a non-fiction book. In the Venn diagram, describe the overlapping area using a complete sentence. Suppose that some customers buy only compact disks. Draw an oval in your Venn diagram representing this event. a. and d. In the following Venn diagram below, the blue oval represent customers buying a novel, the red oval represents customer buying non-fiction, and the yellow oval customer who buy compact disks. b. P (novel or non-fiction) = P (Blue ∪ Red) = P (Blue) + P (Red) - P (Blue ∩ Red) = 0.6 + 0.4 - 0.2 = 0.8. c. The overlapping area of the blue oval and red oval represents the customers buying both a novel and a nonfiction book. A set of 20 German Shepherd dogs is observed. 12 are male, 8 are female, 10 have some brown coloring, and 5 have some white sections of fur. Answer the following using Venn Diagrams. Draw a Venn diagram simply showing the sets of male and female dogs. The Venn diagram below demonstrates the situation of mutually exclusive events where the outcomes are independent events. If a dog cannot be both male and female, then there is no intersection. Being male precludes being female and being female precludes being male: in this case, the characteristic gender is therefore mutually exclusive. A Venn diagram shows this as two sets with no intersection. The intersection is said to be the null set using the mathematical symbol ∅. Draw a second Venn diagram illustrating that 10 of the male dogs have brown coloring. The Venn diagram below shows the overlap between male and brown where the number 10 is placed in it. This represents Male ∩ Brown : both male and brown. This is the intersection of these two characteristics. To get the union of Male and Brown, then it is simply the two circled areas minus the overlap. In proper terms, Male ∪ Brown = Male + Brown − Male ∩ Brown will give us the number of dogs in the union of these two sets. If we did not subtract the intersection, we would have double counted some of the dogs. Now draw a situation depicting a scenario in which the non-shaded region represents "No white fur and female," or White fur′ ∩ Female. the prime above "fur" indicates "not white fur." The prime above a set means not in that set, e.g. A ′ means not A . Sometimes, the notation used is a line above the letter. For example, A ¯ = A ′ . The Addition Rule of Probability We met the addition rule earlier but without the help of Venn diagrams. Venn diagrams help visualize the counting process that is inherent in the calculation of probability. To restate the Addition Rule of Probability: P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) Remember that probability is simply the proportion of the objects we are interested in relative to the total number of objects. This is why we can see the usefulness of the Venn diagrams. shows how we can use Venn diagrams to count the number of dogs in the union of brown and male by reminding us to subtract the intersection of brown and male. We can see the effect of this directly on probabilities in the addition rule. Try It A set of 40 people are selected randomly. 24 are male, 16 are female, 21 like coffee, and 19 like tea. Answer the following using Venn diagrams. Draw a Venn diagram simply showing the sets of male and female. Draw a second Venn diagram illustrating 14 males liking coffee. Draw a situation with a region depicting female and not liking tea. IBS2e_UnFigure_03_Ex31_001 IBS2e_UnFigure_03_Ex31_002 IBS2e_UnFigure_03_Ex31_003 Let's sample 50 students who are in a statistics class. 20 are first-year students and 30 are sophomores. 15 students get a "B" in the course, and 5 students both get a "B" and are first-year students. Find the probability of selecting a student who either earns a "B" OR is a first-year student. We are translating the word OR to the mathematical symbol for the addition rule, which is the union of the two sets. We know that there are 50 students in our sample, so we know the denominator of our fraction to give us probability. We need only to find the number of students that meet the characteristics we are interested in, i.e. any first-year students and any student who earned a grade of "B." With the Addition Rule of probability, we can skip directly to probabilities. Let "A" = the number of first-year students, and let "B" = the grade of "B." Below we can see the process for using Venn diagrams to solve this. The P ( A ) = 20 50 = 0.40 , P ( B ) = 15 50 = 0.30 , and P ( A ∩ B ) = 5 50 = 0.10 . Therefore, P ( A ∪ B ) = 0.40 + 0.30 − 0.10 = 0.60 . If two events are mutually exclusive, then, like the example where we diagram the male and female dogs, the addition rule is simplified to just P ( A ∪ B ) = P ( A ) + P ( B ) − 0 . This is true because, as we saw earlier, the union of mutually exclusive events is the null set, ∅. The diagrams below demonstrate this. Try It A set of 40 random people is taken. 15 are male, and 25 are female. 18 people like coffee. 10 people are male and like coffee. Find the probability of selecting a person that likes coffee or is male. Let “A” = the number of males and let “B” = people who like coffee. The probabilities obtained are: P ( A ) = 15 40 = 0 . 375 , P ( B ) = 18 40 = 0 . 45 , P ( A ∩ B ) = 10 40 = 0 . 25 So, P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) = 0 . 375 + 0 . 45 - 0 . 25 = 0 . 575 . So the probability that a person likes coffee or is male is 0.575. The Multiplication Rule of Probability Restating the Multiplication Rule of Probability using the notation of Venn diagrams, we have: P ( A ∩ B ) = P ( A | B ) ⋅ P ( B ) The multiplication rule can be modified with a bit of algebra into the following conditional rule. Then Venn diagrams can then be used to demonstrate the process. The conditional rule: P ( A | B ) = P ( A ∩ B ) P ( B ) Using the same facts from above, find the probability that someone will earn a "B" if they are a "first-year students." P ( A | B ) = 0.10 0.30 = 1 3 The multiplication rule must also be altered if the two events are independent. Independent events are defined as a situation where the conditional probability is simply the probability of the event of interest. Formally, independence of events is defined as P ( A | B ) = P ( A ) or P ( B | A ) = P ( B ) . When flipping coins, the outcome of the second flip is independent of the outcome of the first flip; coins do not have memory. The Multiplication Rule of Probability for independent events thus becomes: P ( A ∩ B ) = P ( A ) ⋅ P ( B ) One easy way to remember this is to consider what we mean by the word "and." We see that the Multiplication Rule has translated the word "and" to the Venn notation for intersection. Therefore, the outcome must meet the two conditions of first-year and grade of "B" in the above example. It is harder, less probable, to meet two conditions than just one or some other one. We can attempt to see the logic of the Multiplication Rule of probability due to the fact that fractions multiplied times each other become smaller. The development of the Rules of Probability with the use of Venn diagrams can be shown to help as we wish to calculate probabilities from data arranged in a contingency table. is from a sample of 200 people who were asked how much education they completed. The columns represent the highest education they completed, and the rows separate the individuals by men and women. Less than high school grad High school grad Some college College grad Total Men 5 15 40 60 120 Women 8 12 30 30 80 Total 13 27 70 90 200 Now, we can use this table to answer probability questions. The following examples are designed to help understand the format above while connecting the knowledge to both Venn diagrams and the probability rules. What is the probability that a selected person both finished college and is a woman? This is a simple task of finding the value where the two characteristics intersect on the table, and then applying the postulate of probability, which states that the probability of an event is the proportion of outcomes that match the event in which we are interested as a proportion of all total possible outcomes. P ( College Grad ∩ Female ) = 30 200 = 0.15 What is the probability of selecting either a woman or someone who finished college? This task involves the use of the addition rule to solve for this probability. P ( College Grad ∪ Female ) = P ( F ) + P ( CG )− P ( F ∩ CG ) P ( College Grad ∪ Female ) = 80 200 + 90 200 − 30 200 = 140 200 = 0.70 What is the probability of selecting a high school graduate if we only select from the group of men? Here we must use the conditional probability rule (the modified multiplication rule) to solve for this probability. P ( HS Grad | Man = P ( HS Grad ∩ Man ) P ( Man ) = ( 15 200 ) ( 120 200 ) = 15 120 = 0.125 Can we conclude that the level of education attained by these 200 people is independent of the gender of the person? There are two ways to approach this test. The first method seeks to test if the intersection of two events equals the product of the events separately remembering that if two events are independent than P ( A )* P ( B ) = P ( A ∩ B ). For simplicity's sake, we can use calculated values from above. Does P(College Grad ∩ Woman) = P(CG) ⋅ P(F)? 30 200 ≠ 90 200 ⋅ 80 200 because 0.15 ≠ 0.18. Therefore, gender and education here are not independent. The second method is to test if the conditional probability of A given B is equal to the probability of A . Again for simplicity, we can use an already calculated value from above. Does P(HS Grad | Man) = P(HS Grad)? 15 120 ≠ 27 200 because 0.125 ≠ 0.135. Therefore, again gender and education here are not independent. Try It The table shows a spread of what 100 random people like to eat. The columns represent the fruit that they like to eat, and the rows separate the individuals by men and women. Grape Watermelon Apple Pomegranate Total Men 12 18 20 10 60 Women 13 7 15 5 40 Total 25 25 35 15 100 Answer the following questions: What is the probability that a selected person likes pomegranate and is a man? What is the probability of selecting either a man or someone who likes apple? What is the probability of selecting a person who likes watermelon if we select from the group of women? P ( P o m e g a n a t e ∩ m a l e ) = 10 100 = 0 . 1 P ( M a l e ∪ A p p l e ) = P ( M a l e ) + P ( A p p l e ) - P ( M a l e ∩ A p p l e ) = 60 100 + 35 100 - 20 100 = 0 . 75 P ( W a t e r m e l o n | F e m a l e ) = P ( W a t e r m e l o n ∩ F e m a l e ) P ( F e m a l e ) = 7 100 40 100 = 7 40 = 0 . 175 Chapter Review 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 or universe of the objects of interest together with circles or ovals. The circles or ovals represent groups of events called sets. A Venn diagram is especially helpful for visualizing the ∪ event, the ∩ event, and the complement of an event and for understanding conditional probabilities. A Venn diagram is especially helpful for visualizing an Intersection of two events, a Union of two events, or a Complement of one event. A system of Venn diagrams can also help to understand Conditional probabilities. Venn diagrams connect the brain and eyes by matching the literal arithmetic to a picture. It is important to note that more than one Venn diagram is needed to solve the probability rule formulas introduced in Section 3.3 . Use the following information to answer the next four exercises. shows a random sample of musicians and how they learned to play their instruments. Gender Self-taught Studied in school Private instruction Total Women 12 38 22 72 Men 19 24 15 58 Total 31 62 37 130 Find P (musician is a woman). Find P (musician is a man ∩ had private instruction). P (musician is a male ∩ had private instruction) = 15 130 = 3 26 = 0.12 Find P (musician is a woman ∪ is self taught). Are the events “being a woman musician” and “learning music in school” mutually exclusive events? The events are not mutually exclusive. It is possible to be a female musician who learned music in school. 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 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 Black people, 2,745 Native Hawaiian people, 12,831 Latino people, 8,378 Japanese American people, and 7,650 White people. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 Black people, 3,062 Native Hawaiian people, 4,932 Latino people, 10,680 Japanese American people, and 9,877 White people. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 Black people, 1,419 Native Hawaiian people, 1,406 Latino people, 4,715 Japanese American people, and 6,062 White people. Of the people smoking at least 31 cigarettes per day, there were 759 Black people, 788 Native Hawaiian people, 800 Latino people, 2,305 Japanese American people, and 3,970 White people. Complete the table using the data provided. Smoking Levels by Ethnicity Smoking level Black people Native Hawaiian Latino Japanese American White people TOTALS 1–10 11–20 21–30 31+ TOTALS Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day. 35,065 100,450 Find the probability that the person was Latino. 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. 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 . In words, explain what it means to pick one person from the study who is “Japanese American ∪ smokes 21 to 30 cigarettes per day.” Also, find the probability. In words, explain what it means to pick one person from the study who is “Japanese American | that person smokes 21 to 30 cigarettes per day.” Also, find the probability. 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 . Prove that smoking level/day and ethnicity are dependent events. 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. Suppose that you randomly draw two cards, one at a time, with replacement . Let G 1 = first card is green Let G 2 = second card is green Draw a tree diagram of the situation. Find P ( G 1 ∩ G 2 ). Find P (at least one green). Find P ( G 2 | G 1 ). Are G 2 and G 1 independent events? Explain why or why not. P ( GG ) = ( 5 8 ) ( 5 8 ) = 25 64 P (at least one green) = P ( GG ) + P ( GY ) + P ( YG ) = 25 64 + 15 64 + 15 64 = 55 64 P ( G | G ) = 5 8 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. Suppose that you randomly draw two cards, one at a time, without replacement . G 1 = first card is green G 2 = second card is green Draw a tree diagram of the situation. Find P ( G 1 ∩ G 2 ). Find P (at least one green). Find P ( G 2 | G 1 ). Are G 2 and G 1 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 women is 48.60. Of the women, 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. men drivers, 5.04% are age 19 and under; 81.43% are age 20–64; 13.53% are age 65 or over. Complete the following. Construct a table or a tree diagram of the situation. Find P (driver is a woman). Find P (driver is age 65 or over | driver is a woman). Find P (driver is age 65 or over ∩ a woman). In words, explain the difference between the probabilities in part c and part d. Find P (driver is age 65 or over). Are being age 65 or over and being a woman mutually exclusive events? How do you know? <20 20–64 >64 Totals Female 0.0244 0.3954 0.0661 0.486 Male 0.0259 0.4186 0.0695 0.514 Totals 0.0503 0.8140 0.1356 1 P ( F ) = 0.486 P (>64 | F ) = 0.1361 P (>64 and F ) = P ( F ) P (>64| F ) = (0.486)(0.1361) = 0.0661 P (>64 | F ) is the percentage of female drivers who are 65 or older and P (>64 ∩ F ) is the percentage of drivers who are female and 65 or older. P (> 64 ) = P (>64 ∩ F ) + P (>64 ∩ M ) = 0.1356 No, being female and 65 or older are not mutually exclusive because they can occur at the same time P(>64 ∩ F ) = 0.0661. Suppose that 10,000 U.S. licensed drivers are randomly selected. How many would you expect to be men? Using the table or tree diagram, construct a contingency table of gender versus age group. Using the contingency table, find the probability that out of the age 20–64 group, a randomly selected driver is a woman. 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. Construct a table or a tree diagram of the situation. Include a branch for all other modes of transportation to work. Assuming that the walkers walk alone, what percent of all commuters travel alone to work? Suppose that 1,000 workers are randomly selected. How many would you expect to travel alone to work? Suppose that 1,000 workers are randomly selected. How many would you expect to drive in a carpool? Car, truck or van Walk Public transportation Other Totals Alone 0.7318 Not alone 0.1332 Totals 0.8650 0.0390 0.0530 0.0430 1 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. Make the same assumptions as in (b) we have: (0.7708)(1,000) = 771 (0.1332)(1,000) = 133 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. Based on the given data, find P ( H ) and P ( T ). Use a tree to find the probabilities of each possible outcome for the experiment of tossing the coin twice. Use the tree to find the probability of obtaining exactly one head in two tosses of the coin. Use the tree to find the probability of obtaining at least one head. Homework Use the information in the to answer the next eight exercises. The table shows the political party affiliation for various members of the U.S. Senate during two separate years when they are up for reelection. Up for reelection: Democratic party Republican party Other Total Year A 20 13 0 Year B 10 24 0 Total What is the probability that a randomly selected senator has an “Other” affiliation? 0 What is the probability that a randomly selected senator is up for reelection in Year B? What is the probability that a randomly selected senator is a Democrat and up for reelection in Year B? 10 67 What is the probability that a randomly selected senator is a Republican or is up for reelection in Year A? Suppose that a member of the U.S. Senate is randomly selected. Given that the randomly selected senator is up for reelection in Year B, what is the probability that this senator is a Democrat? 10 34 Suppose that a member of the U.S. Senate is randomly selected. What is the probability that the senator is up for reelection in Year A, knowing that this senator is a Republican? The events “Republican” and “Up for reelection in Year B” are ________ mutually exclusive. independent. both mutually exclusive and independent. neither mutually exclusive nor independent. d The events “Other” and “Up for reelection in Year B” are ________ mutually exclusive. independent. both mutually exclusive and independent. neither mutually exclusive nor independent. gives the number of participants in the recent National Health Interview Survey who had been treated for cancer in the previous 12 months. The results are sorted 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. Race and sex 15–24 25–40 41–65 Over 65 TOTALS White, male 1,165 2,036 3,703 8,395 White, female 1,076 2,242 4,060 9,129 Black, male 142 194 384 824 Black, female 131 290 486 1,061 All others TOTALS 2,792 5,279 9,354 21,081 Do not include "all others" for parts f and g. Fill in the column for cancer treatment for individuals over age 65. Fill in the row for all other races. Find the probability that a randomly selected individual was a White male. Find the probability that a randomly selected individual was a Black female. Find the probability that a randomly selected individual was Black Find the probability that a randomly selected individual was male. Out of the individuals over age 65, find the probability that a randomly selected individual was a Black or White male. Race and sex 1–14 15–24 25–64 Over 64 TOTALS White, male 1,165 2,036 3,703 1,491 8,395 White, female 1,076 2,242 4,060 1,751 9,129 Black, male 142 194 384 104 824 Black, female 131 290 486 154 1,061 All others 156 TOTALS 2,792 5,279 9,354 3,656 21,081 Race and sex 1–14 15–24 25–64 Over 64 TOTALS White, male 1,165 2,036 3,703 1,491 8,395 White, female 1,076 2,242 4,060 1,751 9,129 Black, male 142 194 384 104 824 Black, female 131 290 486 154 1,061 All others 278 517 721 156 1672 TOTALS 2,792 5,279 9,354 3,656 21,081 8,395 21,081 ≈ 0.3982 1,061 21,081 ≈ 0.0503 1,885 21,081 ≈ 0.0894 9,219 19,409 ≈ 0.475 1,595 3,500 ≈ 0.456 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 1,517 506 136 714 2,873 Jackie Robinson 1,054 273 54 137 1,518 Ty Cobb 3,053 724 295 117 4,189 Hank Aaron 2,294 624 98 755 3,771 TOTAL 7,918 2,127 583 1,723 12,351 Find P (hit was made by Babe Ruth). 1518 2873 2873 12351 583 12351 4189 12351 Find P (hit was made by Ty Cobb | The hit was a Home Run). 4189 12351 117 1723 1723 4189 117 12351 b identifies a group of children by one of four hair colors, and by type of hair. Hair type Brown Blond Black Red Totals Wavy 20 15 3 43 Straight 80 15 12 Totals 20 215 Complete the table. What is the probability that a randomly selected child will have wavy hair? What is the probability that a randomly selected child will have either brown or blond hair? What is the probability that a randomly selected child will have wavy brown hair? What is the probability that a randomly selected child will have red hair, given that they have straight hair? 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? 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 5 0 0 34–66 6 18 7 4 66–99 6 12 22 5 For the following, suppose that you randomly select one player from the 49ers or Cowboys. Find the probability that his shirt number is from 1 to 33. Find the probability that he weighs at most 210 pounds. Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210 pounds. Find the probability that his shirt number is from 1 to 33 OR he weighs at most 210 pounds. Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs at most 210 pounds. 26 106 33 106 21 106 ( 26 106 ) + ( 33 106 ) - ( 21 106 ) = ( 38 106 ) 21 33 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 and P ( T ) = 1 3 where H is heads and T is tails. Find P (tossing a Head on the coin AND a Red bead) 2 3 5 15 6 36 5 36 Find P (Blue bead). 15 36 10 36 10 12 6 36 a 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?) Draw the tree that represents the possibilities for the cookie selections. Write the probabilities along each branch of the tree. Are the probabilities for the flavor of the SECOND cookie that Miguel selects independent of his first selection? Explain. For each complete path through the tree, write the event it represents and find the probabilities. Let S be the event that both cookies selected were the same flavor. Find P ( S ). 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. Let U be the event that the second cookie selected is a butter cookie. Find P ( U ). Venn Diagram the visual representation of a sample space and events in the form of circles or ovals showing their intersections

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Introduction You can use probability and discrete random variables to calculate the likelihood of lightning striking the ground five times during a half-hour thunderstorm. (credit: modification of work “CG lightning strike” by Axel Rouvin/ Flickr, CC BY 2.0) A student takes a ten-question, true-false quiz. Because the student had such a busy schedule, they could not study and guesses randomly at each answer. What is the probability of the student passing the test with at least a 70%? The manager of an auto dealership might be interested in the color preferences for new car buyers. Suppose on average the dealership sells 20 cars per month. What is the probability that a customer prefers red cars? These two examples illustrate two different types of probability problems involving discrete random variables. Recall that discrete data are data that you can count, that is, the random variable can only take on whole number values. 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, often called a trial. Random Variable Notation The upper case letter X denotes 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 as whole numbers that X can take on and the outcomes are random (the x values 0, 1, 2, 3), X is a discrete random variable. Probability Density Functions (PDF) for a Random Variable A probability density function or probability distribution function has two characteristics: Each probability is between zero and one, inclusive. The sum of the probabilities is one. A probability density function is a mathematical formula that calculates probabilities for specific types of events, what we have been calling experiments. There is a sort of magic to a probability density function (Pdf) partially because the same formula often describes very different types of events. For example, the binomial Pdf will calculate probabilities for flipping coins, yes/no questions on an exam, opinions of voters in an up or down opinion poll, indeed any binary event. Other probability density functions will provide probabilities for the time until a part will fail, when a customer will arrive at the turnpike booth, the number of telephone calls arriving at a central switchboard, the growth rate of a bacterium, and on and on. There are whole families of probability density functions that are used in a wide variety of applications, including medicine, business and finance, physics and engineering, among others. For our needs here we will concentrate on only a few probability density functions as we develop the tools of inferential statistics. Counting Formulas and the Combinational Formula To repeat, the probability of event A , P(A), is simply the number of ways the experiment will result in A, relative to the total number of possible outcomes of the experiment. As an equation this is: P ( A ) = number of ways to get A Total number of possible outcomes When we looked at the sample space for flipping 3 coins we could easily write the full sample space and thus could easily count the number of events that met our desired result, e.g. x = 1 , where X is the random variable defined as the number of heads. As we have larger numbers of items in the sample space, such as a full deck of 52 cards, the ability to write out the sample space becomes impossible. We see that probabilities are nothing more than counting the events in each group we are interested in and dividing by the number of elements in the universe, or sample space. This is easy enough if we are counting sophomores in a Stat class, but in more complicated cases listing all the possible outcomes may take a life time. There are, for example, 36 possible outcomes from throwing just two six-sided dice where the random variable is the sum of the number of spots on the up-facing sides. If there were four dice then the total number of possible outcomes would become 1,296. There are more than 2.5 MILLION possible 5 card poker hands in a standard deck of 52 cards. Obviously keeping track of all these possibilities and counting them to get at a single probability would be tedious at best. An alternative to listing the complete sample space and counting the number of elements we are interested in, is to skip the step of listing the sample space, and simply figuring out the number of elements in it and doing the appropriate division. If we are after a probability we really do not need to see each and every element in the sample space, we only need to know how many elements are there. Counting formulas were invented to do just this. They tell us the number of unordered subsets of a certain size that can be created from a set of unique elements. By unordered it is meant that, for example, when dealing cards, it does not matter if you got {ace, ace, ace, ace, king} or {king, ace, ace, ace, ace} or {ace, king, ace, ace, ace} and so on. Each of these subsets are the same because they each have 4 aces and one king. Combinational Formula n x = n C x = n ! x ! ( n - x ) ! This is the formula that tells the number of unique unordered subsets of size x that can be created from n unique elements. The formula is read “n combinatorial x”. Sometimes it is read as “n choose x." The exclamation point "!" is called a factorial and tells us to take all the numbers from 1 through the number before the ! and multiply them together thus 4! is 1·2·3·4=24. By definition 0! = 1. The formula is called the Combinatorial Formula. It is also called the Binomial Coefficient, for reasons that will be clear shortly. While this mathematical concept was understood long before 1653, Blaise Pascal is given major credit for his proof that he published in that year. Further, he developed a generalized method of calculating the values for combinatorials known to us as the Pascal Triangle. Pascal was one of the geniuses of an era of extraordinary intellectual advancement which included the work of Galileo, Rene Descartes, Isaac Newton, William Shakespeare and the refinement of the scientific method, the very rationale for the topic of this text. Let’s find the hard way the total number of combinations of the four aces in a deck of cards if we were going to take them two at a time. The sample space would be: S={Spade,Heart),(Spade, Diamond),(Spade,Club), (Diamond,Club),(Heart,Diamond),(Heart,Club)} There are 6 combinations; formally, six unique unordered subsets of size 2 that can be created from 4 unique elements. To use the combinatorial formula we would solve the formula as follows: 4 2 = 4 ! ( 4 - 2 ) ! 2 ! = 4 · 3 · 2 · 1 2 · 1 · 2 · 1 = 6 If we wanted to know the number of unique 5 card poker hands that could be created from a 52 card deck we simply compute: 52 5 where 52 is the total number of unique elements from which we are drawing and 5 is the size group we are putting them into. With the combinatorial formula we can count the number of elements in a sample space without having to write each one of them down, truly a lifetime's work for just the number of 5 card hands from a deck of 52 cards. We can now apply this tool to a very important probability density function, the hypergeometric distribution. Remember, a probability density function computes probabilities for us. We simply put the appropriate numbers in the formula and we get the probability of specific events. However, for these formulas to work they must be applied only to cases for which they were designed. Chapter Review The characteristics of a probability distribution or density function (PDF) are as follows: Each probability is between zero and one, inclusive ( inclusive means to include zero and one). The sum of the probabilities is one. 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. Complete using the data provided. x P ( x ) 0 0.12 1 0.18 2 0.30 3 0.15 4 5 0.10 6 0.05 x P ( x ) 0 0.12 1 0.18 2 0.30 3 0.15 4 0.10 5 0.10 6 0.05 P ( x = 4) = _______ P ( x ≥ 5) = _______ 0.10 + 0.05 = 0.15 On average, how long would you expect a new hire to stay with the company? What does the column “ P ( x )” sum to? 1 Use the following information to answer the next six exercises: A baker is deciding how many batches of muffins to make to 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 0.15 2 0.35 3 0.40 4 0.10 Define the random variable X . What is the probability the baker will sell more than one batch? P ( x > 1) = _______ 0.35 + 0.40 + 0.10 = 0.85 What is the probability the baker will sell exactly one batch? P ( x = 1) = _______ On average, how many batches should the baker make? 1(0.15) + 2(0.35) + 3(0.40) + 4(0.10) = 0.15 + 0.70 + 1.20 + 0.40 = 2.45 Use the following information to answer the next four exercises: Emery has music practice three days a week. They practice 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. Define the random variable X . Construct a probability distribution table for the data. x P ( x ) 0 0.03 1 0.04 2 0.08 3 0.85 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. Define the random variable X . Let X = the number of events Javier volunteers for each month. What values does x take on? Construct a PDF table. x P ( x ) 0 0.05 1 0.05 2 0.10 3 0.20 4 0.25 5 0.35 Find the probability that Javier volunteers for less than three events each month. P ( x < 3) = _______ Find the probability that Javier volunteers for at least one event each month. P ( x > 0) = _______ 1 – 0.05 = 0.95 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. 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.

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