Chapter 15 Probability Rules! 15-1 What s It About? Chapter 14 introduced students to basic probability concepts. Chapter 15 generalizes and expands the Addition and Multiplication Rules. We discuss conditional probability, using it to define independence and to find probabilities when events are not independent. We use Venn diagrams, two-way tables, and tree diagrams to help organize our thinking when facing more difficult issues, such as reverse conditioning. Comments Probability is a deceptively difficult topic. The arithmetic, especially with a calculator in hand, is pretty easy. The questions sound straightforward. The words and, or, not, of, if, independent, mutually exclusive, given that can be unexpectedly confusing. That s why we have separated the basic ideas in Chapter 14 from the more challenging issues found here. We return to each idea and look more deeply. The Addition Rule needs an adjustment if the events are not disjoint. When knowledge of other conditions changes the probability of an event, we say that the events are not independent, so we must also adjust the Multiplication Rule to use conditional probabilities. We ve tiptoed around independence ever since Chapter 3, and now it s finally time to pin that issue down. Finally, we face such issues as drawing without replacement, the difference between independent and disjoint events, and reversed conditioning (Bayes s Theorem). Expect this to be the most difficult of the probability chapters. It s useful to spend some time with examples translating from word problems to probability. A two-way table is a good place to start. If we have Sex as one variable and Political Party as the other (say Democratic, Republican, or Independent), most students have no difficulty finding the proportion of Females who are Democrats. Use this to introduce conditional probability. Show how a tree diagram can be useful if the conditional probabilities are given, while a two-way table or Venn Diagram might be better if the probabilities are known. Looking Ahead Independence will continue to be one of the most critical concepts in the course. It lies at the heart of combining random variables in Chapter 16 and the probability models we examine in Chapter 17. The Independence Assumption is fundamental to the sampling distribution models of Chapter 18 and throughout our look at inference the rest of the book! Class Do s The formulas should play a secondary role. The major issues are the concepts. Once those concepts are clear, formulas are just ways of implementing them. Well-understood examples and pictures can help with the concepts. Students who are not familiar with playing cards should be encouraged to obtain, or at least examine, a deck, regarding it as a tool for thinking about probabilities.
15-2 Part IV Randomness and Probability To find the probability of getting a red card or an ace from a deck, we can t add the 26 red cards and the 4 aces, because some aces are red. A Venn diagram shows that adding the probabilities of two such events counts their intersection twice. To find the probability of getting 2 aces in a row (when drawing without replacement), we cannot just multiply the probability of an ace by itself. The changing composition of the deck makes the probability of the second ace different. The probability of getting a second ace is different given that one ace is already missing. This introduces the idea of conditional probability. A tree diagram can help make this more clear. Emphasize the importance of the direction of conditioning. The probability that a red card is a heart (P(heart red) = 1/2) is not the same as the probability that a heart is red (P(red heart) = 1). The probability that a baseball game will be postponed if it rains (maybe) is not the same as the probability that if a game was postponed, the reason was rain (very likely). A two-way table is a good way to develop the formula for conditional probability when the joint probabilities are known. Students saw conditional probabilities in Chapter 3. The probability of an outcome on one variable given a certain result for the other variable is easily seen to be the number of ways that both happen out of the total number of ways the given event happened. Define independence in terms of conditional probability. Conceptually we know that independent events don t affect each other. More specifically, events are independent if the outcome on one of them does not change the probability of the other. That leads us directly to say that when given knowledge of A, the probability of B remains the same. Hence, in formula form, two events are independent if P(B A) = P(B). Make students interpret that requirement in context every time they face a question about independence. Continue to discuss the difference between independent events and disjoint events. This will be confusing for some students. Have ready examples. Ask them if events can be both disjoint and independent. We can hope that they will explain (with some coaxing, perhaps) that if the events are disjoint then they cannot both happen at the same time. That means that as soon as one of them happens the other cannot the probability of the other changes to 0. When the occurrence of one event changes the probability of the other, the events are not independent. Ever. Perhaps what is most confusing is the fact that while possible events that are disjoint cannot be independent, events that are not disjoint may or may not be independent. Baseball games are sometimes played in the rain, so baseball and rain are not mutually exclusive. But they re not independent either because games are less likely to be played when it rains. The most difficult probability questions students will encounter involve reverse conditioning. These are best approached with tree diagrams and common sense rather than formulas. Have students write a clear statement of the conditional probability first. Usually both the denominator and numerator can be found directly from the tree diagram. Bayes s Theorem per se is not very useful at this level. Finally, emphasize again the value of the Step-by-Step worked examples. Students can see how the Rules are implemented, how to use Venn diagrams, how to test for independence, and how to reverse the conditioning. They will find these examples a great help when they work on the exercises.
The Importance of What You Don t Say Chapter 15 Probability Rules! 15-3 Don t use P( A)i P(B) = P( A B) to define independence. Independence is based in conditional probability. Define and discuss it in that context. Always. The original Multiplication Rule works only if events are independent, but that s a byproduct of independence, not a definition or a test for independence. There is no insight or meaning to be found in that formula. Don t present Bayes s Formula. We show the formula in a footnote, and only reluctantly. Students need to be able to think their way through a probability question. Relying on formulas short-circuits thinking, allows one to do only those problems to which the formula applies, and even then only as long as one remembers the formula. This formula is way too scary and complex, and, more important, it produces no insight. A tree diagram is almost always better. Class Examples 1. If students are familiar with card games, a deck of cards makes a good frame of reference for many of the issues in this chapter. One card is drawn. What is the probability it is an ace or red? (General Addition Rule) Answer: Think The events ace and red are not disjoint. There are 2 red aces in the deck. The General Addition Rule can be used. Show P(ace or red) = P(ace) + P(red) P(ace and red) = 4 + 26 2 = 28 52 52 52 52 Tell The probability that a randomly selected card is an ace or red is 28 52. Two cards are drawn without replacement. What is the probability they are both aces? (General Multiplication Rule) Extend to the probability of getting 5 hearts in a row. Answer: Think Two cards are drawn without replacement. The two draws are not independent. The General Multiplication Rule can be used. Show P(both cards are aces) = P(ace)P(ace first draw was ace) = 4 52 () 3 51 Tell The probability that two randomly selected cards are both aces is P(5 hearts) = 13 52 () () 12 11 51 () 10 50 49 () 9 () 0.0005 48 ()= 12 12. 2652 I draw one card and look at it. I tell you it is red. What is the probability it is a heart? And what is the probability it is red, given that it is a heart? (Conditional probability) Answer: Think The card is selected at random. 2652 Show P(heart red) = P(heart and red) P(red) = 13 52 = 1 26 2 52 (half of the red cards are hearts) P(red heart) = P(red and heart) P(heart) = 13 52 = 1 (all of the hearts are red) 13 52 Tell If a randomly chosen card is a red, the probability that it is a heart is 0.5. If a randomly chosen card is a heart, it is certain to be red.
15-4 Part IV Randomness and Probability Are red card and spade independent? Mutually exclusive? Answer: Red card and spade are not independent events. There s a 25% chance a card is a spade, but if you know that you have a red card, you are certain that you don t have a spade. The events are mutually exclusive (disjoint), since there are no red cards that are also spades. Are red card and ace independent? Mutually exclusive? Answer: Red card and ace are independent events. The probability that a card is red is 0.5. The probability that a card chosen from the four aces is red is also 0.5. P(Red) = P(Red Ace) The events are not mutually exclusive (not disjoint), since there are two red aces. Are face card and king independent? Mutually exclusive? Answer: Face card and king are not independent. The probability of drawing a face card is 12. The probability that a face card is drawn from the four kings is 1. 52 P(face) P(face King) The events are not mutually exclusive (not disjoint), since all four kings are face cards. 2. Use one of the many 2-way tables in Chapter 3, or create one in class. By show of hands, count the number of males and females wearing jeans and not. (Collect your own data, or use the two-way table provided. Remember that your own data may or may not show evidence that gender and wearing jeans are independent.) Jeans Other Total M 12 5 17 F 8 11 19 Total 20 16 36 What is the probability that a male wears jeans? Answer: Think Assume that a male student is selected at random from the class. Consider only the M row. Show P(Jeans M) = 12 17 Tell The probability that a male wears jeans is 12 17. What is the probability that someone wearing jeans is a male? Answer: Think Assume that a student wearing jeans is selected at random from the class. Show P(M Jeans) = 12 20 Tell The probability that someone wearing jeans is male is 12 20. Are being male and wearing jeans disjoint? Answer: No, being male and wearing jeans are not disjoint events. There are 12 males who were wearing jeans.
Chapter 15 Probability Rules! 15-5 Are sex and attire independent? Have students explain in context what they need to know to ascertain independence. There are many ways to check for independence; write several such statements down, in words and formula. Answer: Sex and attire are not independent in this class, since P(Jeans) P(Jeans M). The overall probability of wearing jeans is 20, or 33%. The probability that a male is 36 wearing jeans is 12, or 71%. Males are more likely to be wearing jeans than students in 17 general. 3. In April 2003, Science magazine reported on a new com-puter-based test for ovarian cancer, clinical proteomics, that examines a blood sample for the presence of certain patterns of proteins. Ovarian cancer, though dangerous, is very rare, afflicting only 1 of every 5000 women. The test is highly sensitive, able to correctly detect the presence of ovarian cancer in 99.97% of women who have the disease. However, it is unlikely to be used as a screening test in the general population because the test gave false positives 5% of the time. Why are false positives such a big problem? Draw a tree diagram and determine the probability that a woman who tests positive using this method actually has ovarian cancer. Think We re given conditional probabilities, so a tree diagram is useful in this case. We are interested in the probability that a woman who tests positive actually has ovarian cancer. Show P(Ovarian Cancer positive) = P(Ovarian Cancer and positive) P(positive) (0.0002)(0.9997) = (0.0002)(0.9997) + (0.9998)(0.05) 0.00398 Tell The probability that a woman who tests positive using this method actually has ovarian cancer is only 0.4%. Since the cancer is so rare, the overwhelming majority of positive test results are false positives. Resources ActivStats Chapter 15 Conditional Probability and Independence. Simulation activities in this chapter generalize the Randomness Tool to generate random events with two kinds of outcomes that may or may not be independent.
15-6 Part IV Randomness and Probability Assignments Allow about 4 5 days for this chapter, including time to go over the last unit test and to let students continue working on their projects. Have them read the text over about three days while spiraling through 5 6 exercises a night. Questions on the homework will give you plenty to talk about in class. Three chapter quizzes are provided.
AP Statistics Quiz A Chapter 15 Name 1. According to the American Pet Products Manufacturers Association (APPMA) 2003-2004 National Pet Owners Survey, 39% of U.S. households own at least one dog and 34% of U.S. households own at least one cat. Assume that 60% of U.S. households own a cat or a dog. a. What is the probability that a randomly selected U.S. household owns neither a cat nor a dog? b. What is the probability that a randomly selected U.S. household owns both a cat and a dog? c. What is the probability that a randomly selected U.S. household owns a cat if the household has a dog? 2. A manufacturing firm orders computer chips from three different companies: 10% from Company A; 20% from Company B; and 70% from Company C. Some of the computer chips that are ordered are defective: 4% of chips from Company A are defective; 2% of chips from Company B are defective; and 0.5% of chips from Company C are defective. A worker at the manufacturing firm discovers that a randomly selected computer chip is defective. What is the probability that the computer chip came from Company B? Show your work. 15-7
3. A survey of an introductory statistics class in Autumn 2003 asked students whether or not they ate breakfast the morning of the survey. Results are as follows: Sex Breakfast Yes No Total Male 66 66 132 Female 125 74 199 Total 191 140 331 a. What is the probability that a randomly selected student is female? b. What is the probability that a randomly selected student ate breakfast? c. What is the probability that a randomly selected student is a female who ate breakfast d. What is the probability that a randomly selected student is female, given that the student ate breakfast? e. What is the probability that a randomly selected student ate breakfast, given that the student is female? f. Does it appear that whether or not a student ate breakfast is independent of the student s sex? Explain. 15-8
AP Statistics Quiz A Chapter 15 Key 1. According to the American Pet Products Manufacturers Association (APPMA) 2003-2004 National Pet Owners Survey, 39% of U.S. households own at least one dog and 34% of U.S. households own at least one cat. Assume that 60% of U.S. households own a cat or a dog. NOTE: solutions based on a Venn Diagram are acceptable. a. What is the probability that a randomly selected U.S. household owns neither a cat nor a dog? P neither cat nor dog = 1 P cat dog = 1 0.6 = 0.4 ( ) ( ) b. What is the probability that a randomly selected U.S. household owns both a cat and a dog? P cat dog = P cat + P dog P cat dog P ( ) ( ) ( ) ( ) 0.60 = 0.34 + 0.39 P ( cat dog) ( cat dog) = 0.13 c. What is the probability that a randomly selected U.S. household owns a cat if the household has a dog? P ( cat dog) 0.13 1 P ( cat dog) = = = P( dog) 0.39 3 2. A manufacturing firm orders computer chips from three different companies: 10% from Company A; 20% from Company B; and 70% from Company C. Some of the computer chips that are ordered are defective: 4% of chips from Company A are defective; 2% of chips from Company B are defective; and 0.5% of chips from Company C are defective. A worker at the manufacturing firm discovers that a randomly selected computer chip is defective. What is the probability that the computer chip came from Company B? Show your work. ( ) P( ) P B defective 0.0040 P ( B defective) = = = 0.3478 defective 0.0040 + 0.0040 + 0.0035 15-9
3. A survey of an introductory statistics class in Autumn 2003 asked students whether or not they ate breakfast the morning of the survey. Results are as follows: Breakfast Yes No Total Male 66 66 132 Sex Female 125 74 199 Total 191 140 331 a. What is the probability that a randomly selected student is female? 199 P ( female) = = 0.601 331 b. What is the probability that a randomly selected student ate breakfast? 191 P ( ate breakfast) = = 0.577 331 c. What is the probability that a randomly selected student is a female who ate breakfast? 125 P ( female ate breakfast) = = 0.378 331 d. What is the probability that a randomly selected student is female, given that the student ate breakfast? P(female breakfast) 125 P ( female breakfast) = = = 0.654 P(breakfast) 191 e. What is the probability that a randomly selected student ate breakfast, given that the student is female? P(breakfast female) 125 P ( breakfast female) = = = 0.628 P(female) 199 f. Does it appear that whether or not a student ate breakfast is independent of the student s sex? Explain. Since P( breakfast female) 0.628 0.577 P( breakfast) = =, eating breakfast is not independent of the sex of the student. Females seem to be a bit more likely to eat breakfast. 15-10
AP Statistics Quiz B Chapter 15 Name 1. A survey of families revealed that 58% of all families eat turkey at holiday meals, 44% eat ham, and 16% have both turkey and ham to eat at holiday meals. a. What is the probability that a family selected at random had neither turkey nor ham at their holiday meal? b. What is the probability that a family selected at random had only ham without having turkey at their holiday meal? c. What is the probability that a randomly selected family having turkey had ham at their holiday meal? d. Are having turkey and having ham disjoint events? Explain. 2. Many school administrators watch enrollment numbers for answers to questions parents ask. Some parents wondered if preferring a particular science course is related to the student s preference in foreign language. Students were surveyed to establish their preference in science and foreign language courses. Does it appear that preferences in science and foreign language are independent? Explain. Chemistry Physics Biology Total French 16 10 8 34 Spanish 35 23 44 102 Total 51 33 52 136 3. For purposes of making budget plans for staffing, a college reviewed student s year in school and area of study. Of the students, 22.5% are seniors, 25% are juniors, 25% are sophomores, and the rest are freshmen. Also, 40% of the seniors major in the area of humanities, as did 39% of the juniors, 40 % of the sophomores, and 36% of the freshmen. What is the probability that a randomly selected humanities major is a junior? Show your work. 15-11
AP Statistics Quiz B Chapter 15 Key 1. a.. P(neither ham nor turkey) = 1 P(ham turkey) = 1 [ P(ham) + P(turkey) P(ham turkey)] = 1 [0.44 + 0.58 0.16 = 1 0.86 = 0.14 Or, using the Venn diagram at the right, 14% b. c. P(ham only) = P(ham) P(ham turkey) = 0.44 0.16 = 0.28 Or, using Venn Diagram at the right, 28%. ( ) P ham turkey 0.16 P ( ham turkey) = = = 0.2759 P(turkey) 0.58 d. No, the events are not disjoint, since some families (16%) have both ham and turkey at their holiday meals. 2. Overall, 102 of 136, or 75%, preferred Spanish. 35 of 51, or 68.6%, of students in Chemistry had Spanish. 23 of 33, or 69.6%, of students in Physics had Spanish, and 44 of 52, or 84.6% of students in Biology had Spanish. Chemistry and Physics students were much less likely to take Spanish than Biology students. It appears that there is an association between preference in science and foreign language. 3. P(humanities major) = 0.09 + 0.0975 + 0.1 + 0.099 = 0.3865 P( junior humanities major) = P(J H) P(H) = 0.0975 0.3865 = 0.2523 15-12
AP Statistics Quiz C Chapter 15 Name 1. A survey of local car dealers revealed that 64% of all cars sold last month had CD players, 28% had alarm systems, and 22% had both CD players and alarm systems. a. What is the probability one of these cars selected at random had neither a CD player nor an alarm system? b. What is the probability that a car had a CD player unprotected by an alarm system? c. What is the probability a car with an alarm system had a CD player? d. Are having a CD player and an alarm system disjoint events? Explain. 2. All airline passengers must pass through security screenings, but some are subjected to additional searches as well. Some travelers who carry laptops wonder if that makes them more likely to be searched. Data for 420 passengers aboard a cross-country flight are summarized in the table shown. Does it appear that being subjected to an additional search is independent of carrying a laptop computer? Explain. Searched? Yes No Total Laptop 30 42 72 No laptop 145 203 348 Total 175 245 420 3. For purposes of making on-campus housing assignments, a college classifies its students as Priority A (seniors), Priority B (juniors), and Priority C (freshmen and sophomores). Of the students who choose to live on campus, 10% are seniors, 20% are juniors, and the rest are underclassmen. The most desirable dorm is the newly constructed Gold dorm, and 60% of the seniors elect to live there. 15% of the juniors also live there, along with only 5% of the freshmen and sophomores. What is the probability that a randomly selected resident of the Gold dorm is a senior? Show your work clearly on the back of this page. 15-13
AP Statistics Quiz C Chapter 15 Key 1. a. PCD ( A) = PCD ( ) + PA ( ) PCD ( A) = 0.64 + 0.28 0.22 = 0.70, so P(neither) = 1 0.70 = 0.30 b. PCD ( A C ) = 0.64 0.22 = 0.42 c. PCD ( A) 0.22 PCD ( A) = = 0.786 PA ( ) 0.28 d. No. PCD ( A) = 0.22. If they were disjoint, the probability of a car having both features would be 0. 2. Overall, 175 of 420 passengers, or 41.7% were searched. 30 of 72 people carrying laptops were searched, also 41.7%. Carrying a laptop does not change the probability that a passenger is searched, so the events are independent. 3. PA ( G) 0.060 0.060 PAG ( ) = = = = 0.48 PG ( ) 0.060 + 0.030 + 0.035 0.125 15-14