Section 5.7 Bayes’s Rule 5 - 1 5.7 BAYES’S RULE Objectives 1 Use the Rule of Total Probability 2 Use Bayes’s Rule to compute probabilities Now let’s look at probability experiments with sample spaces that will be divided into two or more mutually exclusive events. This is going to be a continuation of the study of conditional probability presented in Section 5.4. We will use the following notation: The symbol x means "and" and the symbol h means "or". In Other Words E ¨ F represents event E and F E ´ F represents event E or F Two rules for computing probabilities that will be used in this section are the Addition Rule for Disjoint Events and the General Multiplication Rule. We restate these rules using this notation. Addition Rule for Disjoint Events If E and F are disjoint (mutually exclusive) events, then P(E ´ F) = P(E) + P(F) In general, if E, F, G, . . . are disjoint (mutually exclusive) events, then P(E ´ F ´ G ´ g ) = P(E) + P(F) + P(G) + g General Multiplication Rule The probability that two events E and F both occur is P(E ¨ F) = P(E) # P(F E) 1 Use the Rule of Total Probability Let’s begin by looking at the following example. Example 1 Introduction to the Rule of Total Probability Problem Suppose that there are two urns. Urn I contains four black and three white balls. Urn II contains four black, seven white, and two red balls. Suppose that we roll a fair die. If the die is a one or a two, we randomly select a ball from urn I. However, if the die is a three, four, five, or six, we randomly select a ball from urn II. What is the probability that the ball is black? Approach We define the events as follows: U1: Urn I is chosen U2: Urn II is chosen B: Black ball is chosen W: White ball is chosen R: Red ball is chosen We answer the question using a tree diagram. Solution Figure 18 shows a tree diagram along with the probabilities corresponding to each branch. Copyright © 2013 Pearson Education, Inc. 7804_Ch05CD_pp01-10.indd 1 9/8/11 12:56 PM 5 - 2 Chapter 5 Probability Figure 18 1 3 2 3 P(B and U1) P(B U1) P(U1) P(B|U1) 13 47 4 7 B 3 7 W 4 13 B U1 U2 7 13 2 13 4 21 P(B and U2) P(B U2) P(U2) P(B|U2) W R 4 2 3 13 8 39 So P(B) = P[(B ¨ U1) ´ (B ¨ U2)] = P(B ¨ U1) + P(B ¨ U2) = Now Work Problem 29(a) 4 8 36 + = = 0.396 21 39 91 There is a 39.6% probability that the ball is black. Suppose that A1 and A2 are two nonempty sets. In addition, A1 and A2 are disjoint. Let S be the sample space such that the union of sets A1 and A2 is S. That is, A1 A2 A1 ¨ A2 = A1 ´ A2 = S Under these circumstances, we say that A1 and A2 form a partition of S. See Figure 19(a). For example, urn I and urn II partitioned the sample space into two disjoint events in Example 1. Figure 19 A1 A2 S A1 A2 S E A1 E A2 E (a) (b) If we define E to be any event in the sample space S, then we can write event E as E = (E ¨ A1) ´ (E ¨ A2) Figure 19(b) illustrates this. We can see that the events (E ¨ A1) and (E ¨ A2) are disjoint, so P((E ¨ A1) ¨ (E ¨ A2)) = 0. Therefore, we can express the probability of event E using the Addition Rule for Disjoint Events: P(E) = P(E ¨ A1) + P(E ¨ A2) Using the General Multiplication Rule, we obtain P(E) = P(A1) # P(E A1) + P(A2) # P(E A2) (1) We can use Formula (1) to find the probability of an event E when the sample space is partitioned into two sets, A1 and A2. Figure 20 shows a tree diagram that leads directly Figure 20 P(E|A1) P(A1) A1 EC P(E|A2) P(A2) E P(E) P(A1) P(E|A1) P(A2) P(E|A2) E A2 EC Copyright © 2013 Pearson Education, Inc. 7804_Ch05CD_pp01-10.indd 2 9/8/11 12:56 PM Section 5.7 Bayes’s Rule 5 - 3 to the result in Formula (1). It is probably better to construct a tree diagram like the one in Figure 20, rather than memorize Formula (1). Now let’s redo Example 1 using the results of Formula (1). Example 2 Using Formula (1) Problem Redo Example 1 using Formula (1). Approach We will define the events as we did in Example 1. We want the probability of selecting a black ball. This can occur by selecting urn I and then selecting a black ball or by selecting urn II and then selecting a black ball. With B representing the event “black ball selected,” we have P(B) = P(B ¨ U1) + P(B ¨ U2) = P(U1) # P(B U1) + P(U2) # P(B U2) where 1 3 4 P(B U1) = 7 P(U1) = 2 3 4 P(B U2) = 13 P(U2) = Solution Substituting, we obtain P(B) = P(U1) # P(B U1) + P(U2) # P(B U2) = Now Work Problem 29(b) 1#4 2# 4 4 8 36 + = + = = 0.396 3 7 3 13 21 39 91 This result agrees with the result in Example 1. What if we partition the sample space into three sets, A1, A2, and A3? Then we have S = A1 ´ A2 ´ A3 A1 ¨ A2 = A1 ¨ A3 = A2 ¨ A3 = A1 A2 A3 Now, some event E can be written E = (E ¨ A1) ´ (E ¨ A2) ´ (E ¨ A3) Figure 21 shows the event E. Figure 21 A1 A2 A3 S E A1 E A2 E A3 E Because the events E ¨ A1, E ¨ A2, and E ¨ A3 are all disjoint, the probability of event E is P(E) = P(E ¨ A1) + P(E ¨ A2) + P(E ¨ A3) = P(A1) # P(E A1) + P(A2) # P(E A2) + P(A3) # P(E A3) (2) The tree diagram in Figure 22 shows the results of Formula (2). Copyright © 2013 Pearson Education, Inc. 7804_Ch05CD_pp01-10.indd 3 9/8/11 12:57 PM 5 - 4 Chapter 5 Probability Figure 22 P(A1) P(A2) P(A3) Example 3 P(E|A1) E P(E|A2) EC E A1 A2 A3 P(E|A3) E E P(E) P(A1) P(E|A1) P(A2) P(E|A2) P(A3) P(E|A3) C EC Using Formula (2) Problem Suppose that a calculator manufacturer buys integrated circuits from three different suppliers: supplier I, supplier II, and supplier III. Based on past experience, it is known that 2% of circuits from supplier I are defective, 1% of circuits from supplier II are defective, and 1.6% of circuits from supplier III are defective. The calculator manufacturer buys 20% of its circuits from supplier I, 35% of its circuits from supplier II, and 45% of its circuits from supplier III. What is the probability that a randomly selected integrated circuit is found to be defective? Approach Let D represent the event that the integrated circuit is defective. Let S 1 , S 2 , and S 3 represent the events that the circuit came from suppliers I, II, or III, respectively. We have the following probabilities: P(S 1) = 0.2 P(S 2) = 0.35 P(S 3) = 0.45 P(D S 1) = 0.02 P(D S 2) = 0.01 P(D S 3) = 0.016 We wish to know the probability that the integrated circuit is defective. Since the sample space is partitioned into three disjoint events corresponding to suppliers I, II, and III, the probability of obtaining a defective circuit is the probability that the circuit is defective and comes from supplier I or defective and comes from supplier II or defective and comes from supplier III. P(D) = P((D ¨ S1) ´ (D ¨ S2) ´ (D ¨ S3)) These three events are disjoint, so we have P(D) = P(D ¨ S1) + P(D ¨ S2) + P(D ¨ S3) Using the General Multiplication Rule, this simplifies to P(D) = P(S 1) # P(D S 1) + P(S 2) # P(D S 2) + P(S 3) # P(D S 3) (3) Solution Substituting into (3), we obtain P(D) = 0.2(0.02) + 0.35(0.01) + 0.45(0.016) = 0.0147 Now Work Problems 31(a)–(b) There is a 1.47% probability that the integrated circuit is defective. To generalize the results of Formulas (1) and (2), we use the following definition: Definition A sample space S is partitioned into n subsets A1 ,A2 , p ,An provided that 1. The n subsets are pairwise disjoint. That is, Ai ¨ Aj = for all i j. 2. A1 ´ A2 ´ g ´ An = S 3. Ai for all i. That is, each subset is nonempty. Now let S be a sample space and A1 , A2 , p , An be a partition of the sample space. Let E be any event. Then E = (E ¨ A1) ´ (E ¨ A2) ´ g ´ (E ¨ An) Copyright © 2013 Pearson Education, Inc. 7804_Ch05CD_pp01-10.indd 4 9/8/11 12:57 PM Section 5.7 Bayes’s Rule 5 - 5 Because E ¨ A1 , E ¨ A2 , p , E ¨ An are disjoint, we have P(E) = P(E ¨ A1) + P(E ¨ A2) + g + P(E ¨ An) Finally, we replace P(E ¨ A1) with P(A1) # P(E A1) using the General Multiplication Rule. We do the same with the remaining probabilities. This gives us the Rule of Total Probability. In Other Words Rule of Total Probability The Rule of Total Probability can be thought of as a weighted average of the conditional probabilities, where the P(Ai) are the weights. Let E be an event that is a subset of a sample space S. Let A1 , A2 , p , An be a partition of the sample space S. Then P(E) = P(A1) # P(E A1) + P(A2) # P(E A2) + g + P(An) # P(E An) Example 4 Using the Rule of Total Probability Problem According to the U.S. Census Bureau, 27.3% of U.S. adult women have never married (single), 49.8% of U.S. adult women are married, and 22.8% of U.S. adult women are widowed, divorced, or separated (other). Of the single women, 6.7% are unemployed; of the married women, 2.6% are unemployed; of the “other” women, 4.9% are unemployed. What is the probability that a randomly selected U.S. adult woman is unemployed? Approach Define the following events: U: unemployed S: single M: married O: other We have the following probabilities: P(S) = 0.273 P(M) = 0.498 P(O) = 0.228 P(U S) = 0.067 P(U M) = 0.026 P(U O) = 0.049 Then, according to the Rule of Total Probability, P(U) = P(S) # P(U S) + P(M) # P(U M) + P(O) # P(U O) Solution P(U) = 0.273 # 0.067 + 0.498 # 0.026 + 0.228 # 0.049 = 0.042 Now Work Problem 33(a) The probability that a randomly selected U.S. adult woman is unemployed is 0.042 or 4.2%. 2 Use Bayes’s Rule to Compute Probabilities Let’s begin with an example. Example 5 Introduction to Bayes’s Rule Problem Refer to Examples 1 and 2. Suppose that a black ball was selected. What is the probability that the black ball came from urn I? Approach From Example 2, we know that 1 2 P(U2) = 3 3 4 4 36 P(B U1) = P(B U2) = P(B) = 7 13 91 We wish to know P(U1 B). Using the Conditional Probability Rule, we write P(U1 B) as P(U1) = P(U1 B) = P(U1 ¨ B) P(B) Copyright © 2013 Pearson Education, Inc. 7804_Ch05CD_pp01-10.indd 5 9/8/11 12:57 PM 5 - 6 Chapter 5 Probability But, P(U1 ¨ B) = P(U1) # P(B U1) and P(B) = P(U1) # P(B U1) + P(U2) # P(B U2) So we have P(U1 B) = P(U1 ¨ B) P(U1) # P(B U1) = # P(B) P(U1) P(B U1) + P(U2) # P(B U2) (4) Solution Substituting, we obtain P(U1 B) = Now Work Problem 29(c) P(U1) # P(B U1) P(B) 1#4 3 7 13 = = = 0.481 36 27 91 There is a 48.1% probability that a randomly selected black ball came from urn I. Equation (4) is a special case of Bayes’s Rule when the sample space is partitioned into two subsets. The general formula is given next. Bayes’s Rule Let A1 , A2 , p , An be a partition of a sample space S. Then, for any event E for which P(E) 7 0, the probability of event Ai for i = 1, 2, p , n, given the event E, is P(Ai E) = P(Ai ¨ E) P(E) = Example 6 P(Ai) # P(E Ai) P(A1) # P(E A1) + P(A2) # P(E A2) + g + P(An) # P(E An) Unemployed Women Problem According to the U.S. Census Bureau, 27.3% of U.S. adult women have never married (single), 49.8% of U.S. adult women are married, and 22.8% of U.S. adult women are widowed, divorced, or separated (other). Of the single women, 6.7% are unemployed; of the married women, 2.6% are unemployed; of the “other” women, 4.9% are unemployed. Suppose that a randomly selected U.S. adult woman is found to be unemployed. What is the probability that she is single? Approach Define the following events: U: S: M: O: unemployed single married other We have the following probabilities: P(S) = 0.273 P(M) = 0.498 P(O) = 0.228 P(U S) = 0.067 P(U M) = 0.026 P(U O) = 0.049 and, from Example 4, we know that P(U) = 0.042. We wish to determine the probability that a woman is single, given the knowledge that she is unemployed. That is, we wish to determine P(S U). We will use Bayes’s Rule as follows: P(S U) = P(S ¨ U) P(S) # P(U S) = P(U) P(U) Copyright © 2013 Pearson Education, Inc. 7804_Ch05CD_pp01-10.indd 6 9/8/11 12:57 PM Section 5.7 Bayes’s Rule 5 - 7 Solution 0.273(0.067) = 0.436 0.042 There is a 43.6% probability that a randomly selected unemployed woman is single. P(S U) = Now Work Problem 33(b) In Other Words A priori probabilities are probabilities computed prior to any knowledge. A posteriori probabilities are computed after gaining some knowledge. Example 7 Historical Note Thomas Bayes was born in 1702 in London, England. His father was a Nonconformist minister. Thomas followed in his father’s footsteps and was ordained as well. Bayes’s rule was presented in his paper “Essay towards solving a problem in the doctrine of chances” published posthumously in the Philosophical Transactions of the Royal Society of London in 1764. Today, Bayesian theory is used by many Internet search engines such as Google to help retrieve information. Bayesian theory is also heavily used by companies such as Microsoft and Intel. We say that all the probabilities P(Ai) are a priori probabilities. These are probabilities of events prior to any knowledge regarding the event. However, the probabilities P(Ai E) are a posteriori probabilities because they are probabilities computed after some knowledge regarding the event. In Example 6, the a priori probability of a randomly selected woman being single is 0.273. The a posteriori probability of a woman being single, knowing that she is unemployed, is 0.436. Notice the information that Bayes’s Rule gives us. Without any knowledge of the employment status of the woman, there is a 27.3% probability that she is single. But with the knowledge that the woman is unemployed, the likelihood of her being single increases to 43.6%. Let’s do one more example to illustrate Bayes’s Rule. Work Disability Problem Persons are classified as work disabled if they have a health problem that prevents them from working in the type of work they can do. Table 12 contains the proportion, by age, of Americans who are 16 years of age or older who are work disabled. Table 12 Age Event Proportion Work Disabled 16–24 A1 0.064 25–34 A2 0.097 35–44 A3 0.151 45–54 A4 0.229 55 and older A5 0.459 Source: U.S. Census Bureau If we let M represent the event that a randomly selected American who is 16 years of age or older is male, the Census Bureau also gives us the following probabilities: P(male 16924) = P(M A1) = 0.515 P(male 25934) = P(M A2) = 0.509 P(male 35944) = P(M A3) = 0.500 P(male 45954) = P(M A4) = 0.492 P(male 55 and older) = P(M A5) = 0.449 (a) If a work-disabled American aged 16 years of age or older is randomly selected, what is the probability that the American is male? (b) If a work-disabled American randomly selected is male, what is the probability that he is 25 to 34 years of age? Approach (a) We will use the Rule of Total Probability to compute P(M) as follows: P(M) = P(A1) # P(M A1) + P(A2) # P(M A2) + P(A3) # P(M A3) + P(A4) # P(M A4) + P(A5) # P(M A5) (b) We use Bayes’s Rule to compute P(25934 male) as follows: P(A2 M) = P(A2) # P(M A2) P(M) where P(M) is found from part (a). Copyright © 2013 Pearson Education, Inc. 7804_Ch05CD_pp01-10.indd 7 9/8/11 12:57 PM 5 - 8 Chapter 5 Probability Solution (a) P(M) = P(A1) # P(M A1) + P(A2) # P(M A2) + P(A3) # P(M A3) + P(A4) # P(M A4) + P(A5) # P(M A5) = (0.064)(0.515) + (0.097)(0.509) + (0.151)(0.500) + (0.229)(0.492) + (0.459)(0.449) = 0.477 There is a 47.7% probability that a randomly selected work-disabled American is male. (b) P(A2) # P(M A2) (0.097)(0.509) = = 0.104 P(M) 0.477 There is a 10.4% probability that a randomly selected work-disabled American who is male is 25 to 34 years of age. P(A2 M) = Notice that the a priori probability (0.097) and the a posteriori probability (0.104) do not differ much. This means that the knowledge that the individual is male does not yield much information regarding the age of the work-disabled individual. Now Work Problem 35 5.7 ASSESS YOUR UNDERSTANDING Vocabulary and skill building 1. If E and F are two events that are not disjoint, draw a Venn diagram that represents E ¨ F. 2. If E and F are two events that are not disjoint, draw a Venn diagram that represents E ´ F. 3. Suppose that S is a sample space. What would it mean to partition S into three disjoint subsets A1, A2, and A3? Draw a figure to help support your explanation. 4. What must be true regarding the sum of the probability of events that make up the partitions of the sample space? 5. Describe the Rule of Total Probability in your own words. 6. What is the difference between a priori and a posteriori probabilities? Which is Bayes’s Rule used for? In Problems 7–20, find the indicated probabilities by referring to the given tree diagram and by using Bayes’s Rule. P(E |A) 0.6 A P(A) 0.2 P (B) 0.55 P(C) 0.25 B C EC P(EC | A) 0.4 P(E |B) 0.7 E EC P(EC | B) 0.3 P(E |C) 0.4 E P(EC | C) 0.6 7. P(E A) 10. P(E c B) 13. P(E) 16. P(A E c) 19. P(B E) E 8. P(E B) 11. P(E C) 14. P(E c) 17. P(C E) 20. P(C E c) EC 9. P(E c A) 12. P(E c C) 15. P(A E) 18. P(B E c) 21. Suppose that events A1 and A2 form a partition of the sample space S with P(A1) = 0.55 and P(A2) = 0.45. If E is an event that is a subset of S and P(E A1) = 0.06 and P(E A2) = 0.08, find P(E). 22. Suppose that events A1 and A2 form a partition of the sample space S with P(A1) = 0.35 and P(A2) = 0.65. If E is an event that is a subset of S and P(E A1) = 0.12 and P(E A2) = 0.09, find P(E). 23. Suppose that events A1, A2, and A3 form a partition of the sample space S with P(A1) = 0.35, P(A2) = 0.45, and P(A3) = 0.2. If E is an event that is a subset of S and P(E A1) = 0.25, P(E A2) = 0.18, and P(E A3) = 0.14, find P(E). 24. Suppose that events A1, A2, and A3 form a partition of the sample space S with P(A1) = 0.3, P(A2) = 0.65, and P(A3) = 0.05. If E is an event that is a subset of S and P(E A1) = 0.05, P(E A2) = 0.25, and P(E A3) = 0.5, find P(E). 25. Use the information given in Problem 21 to find: (a) P(A1 E) (b) P(A2 E) 26. Use the information given in Problem 22 to find: (a) P(A1 E) (b) P(A2 E) 27. Use the information given in Problem 23 to find: (a) P(A1 E) (b) P(A2 E) (c) P(A3 E) 28. Use the information given in Problem 24 to find: (a) P(A1 E) (b) P(A2 E) (c) P(A3 E) 29. Urn I contains five black and seven white balls. Urn II contains six black, five white, and three red balls. We roll a fair die. If the die is a one, three or five, we randomly select a ball from urn I. However, if the die is a two, four, or six, we randomly select a ball from urn II. Copyright © 2013 Pearson Education, Inc. 7804_Ch05CD_pp01-10.indd 8 9/8/11 12:57 PM Section 5.7 Bayes’s Rule 5 - 9 (a)Determine the probability that the ball is black by drawing a tree diagram. (b)Determine the probability that the ball is black using the Rule of Total Probability. (c)Suppose that a randomly chosen ball is black. What is the probability that the ball came from urn I? (d)Suppose that a randomly chosen ball is red. What is the probability that the ball came from urn I? urn II? 30. Urn I contains eleven black and nine white balls and urn II contains three black, eight white, and six red balls. We roll a fair die. If the die is a one, three, or five, we randomly select a ball from urn I. However, if the die is a two, four, or six, we randomly select a ball from urn II. (a)Determine the probability that the ball is black by drawing a tree diagram. (b)Determine the probability that the ball is black using the Rule of Total Probability. (c)If a randomly chosen ball is black, what is the probability that the ball came from urn I? (d)If a randomly chosen ball is red, what is the probability that the ball came from urn I? urn II? 31. Urn I contains six black and nine white balls; urn II contains five black, twelve white, and nine red balls; and urn III contains eleven black, six white, and twelve red balls. We roll a fair die. If the die is a one, three, or five, we randomly select a ball from urn I. If the die is a two or four, we randomly select a ball from urn II. If the die is a six, we randomly select a ball from urn III. (a)Determine the probability that the ball is black by drawing a tree diagram. (b)Determine the probability that the ball is black using the Rule of Total Probability. (c)If a randomly chosen ball is black, what is the probability that the ball came from urn I? (d)If a randomly chosen ball is red, what is the probability that the ball came from urn I? urn II? urn III? 32. Urn I contains seven black and three white balls; urn II contains twelve black, five white, and six red balls; and urn III contains one black, eight white, and four red balls. We roll a fair die. If the die is a one, we randomly select a ball from urn I. If the die is a two, three, or four, we randomly select a ball from urn II. If the die is a five or six, we randomly select a ball from urn III. (a)Determine the probability that the ball is black by drawing a tree diagram. (b)Determine the probability that the ball is black using the Rule of Total Probability. (c)If a randomly chosen ball is black, what is the probability that the ball came from urn I? (d)If a randomly chosen ball is red, what is the probability that the ball came from urn I? urn II? urn III? 34. Elisa Test The standard test for the HIV virus is the Elisa test, which tests for the presence of HIV antibodies. If an individual does not have the HIV virus, the test will come back negative for the presence of HIV antibodies 99.8% of the time and will come back positive for the presence of HIV antibodies 0.2% of the time (a false positive). If an individual has the HIV virus, the test will come back positive 99.8% of the time and will come back negative 0.2% of the time (a false negative). Approximately 0.7% of the world population has the HIV virus. (a)What is the probability that a randomly selected individual has a test that comes back positive? (b)What is the probability that a randomly selected individual has the HIV virus if the test comes back positive? 35. Educational Attainment The data in the following table represent the proportion of Americans 25 years of age or older at various levels of educational attainment in 2008. Level Event Proportion Not a high school graduate A1 0.134 High school graduate A2 0.312 Some college, no degree A3 0.172 Associate’s degree A4 0.088 Bachelor’s degree A5 0.191 Advanced degree A6 0.103 Source: U.S. Census Bureau If we let M represent the event that a randomly selected American who is 25 years of age or older is male, we can also obtain the following probabilities from the Census Bureau data. P(M A1) = 0.505 P(M A2) = 0.482 P(M A3) = 0.467 P(M A4) = 0.433 P(M A5) = 0.480 P(M A6) = 0.513 (a)What is the probability that a randomly selected American 25 years of age or older is male? (b)What is the probability that an American male 25 years of age or older has an advanced degree? (c)What is the probability that an American male 25 years of age or older is not a high school graduate? 36. Educational Attainment Refer to Problem 35. If we let E represent the event that a randomly selected American who is 25 years of age or older is employed, we can also obtain the following probabilities from the Census Bureau data. Applying the Concepts P(E A1) = 0.412 P(E A2) = 0.586 P(E A3) = 0.659 33. Color Blindness The most common form of color blindness is red–green color blindness. People with this type of color blindness cannot distinguish between green and red. Approximately 8% of all males have red–green color blindness, while only about 0.64% of women have red–green color blindness. According to the U.S. Census Bureau, 49.1% of all Americans are male and 50.9% are female. (a)What is the probability that a randomly selected American is color blind? (b)What is the probability that a randomly selected American who is color blind is female? P(E A4) = 0.737 P(E A5) = 0.757 P(E A6) = 0.776 (a)What is the probability that a randomly selected American 25 years of age or older is employed? (b)What is the probability that an employed American 25 years of age or older has a bachelor’s degree? (c)What is the probability that an employed American 25 years of age or older is not a high school graduate? Copyright © 2013 Pearson Education, Inc. 7804_Ch05CD_pp01-10.indd 9 9/8/11 12:57 PM 5 - 10 Chapter 5 Probability 37. Voting Pattern The following data represent the proportion of Americans who voted in the 2008 presidential election at various levels of educational attainment. Level Event Proportion Grade school A1 0.069 High school graduate A2 0.590 College graduate A3 0.341 Source: Statistical Abstract of the United States, 2010 If we let D represent the event that a randomly selected American who voted in the 2008 presidential election voted Democratic, we can also obtain the following probabilities: P(D A1) = 0.720 P(D A2) = 0.570 P(D A3) = 0.510 (a)What is the probability that a randomly selected American who voted in the 2008 presidential election voted Democratic? (b)What is the probability that an American who voted Democratic has graduated from college? (c)What is the probability that an American who voted Democratic has a grade school education? 38. Murder Victims The following data represent the proportion of murder victims at various age levels in 2007. Level Event Proportion Less than 17 years A1 0.080 17–29 A2 0.431 30–44 A3 0.271 45–59 A4 0.142 At least 60 years A5 0.076 If we let M represent the event that a randomly selected murder victim was male, we can also obtain the following probabilities from the FBI data. P(M A1) = 0.638 P(M A4) = 0.734 P(M A2) = 0.860 P(M A5) = 0.630 P(M A3) = 0.771 (a)What is the probability that a randomly selected murder victim was male? (b)What is the probability that a randomly selected male murder victim was 17 to 29 years of age? (c)What is the probability that a randomly selected male murder victim was less than 17 years of age? 39. Espionage The CIA suspects that one of its operatives is a double agent. Past experience indicates that 95% of all operatives suspected of espionage are, in fact, guilty. The CIA decides to administer a polygraph to the suspected spy. It is known that if a person is guilty, the polygraph returns a result that indicates the person is guilty 90% of the time. If a person is innocent, the polygraph returns a result that indicates the person is innocent 99% of the time. What is the probability that this particular suspect is innocent, given that the polygraph indicates that he is guilty? Source: Federal Bureau of Investigation Copyright © 2013 Pearson Education, Inc. 7804_Ch05CD_pp01-10.indd 10 9/8/11 12:57 PM Answers 5.7 Assess Your Understanding 31. (a) 1. S U1 1 2 EF E 1 3 F U3 A2 S 5. Answers will vary. 7. P(E A) = 0.6 11. P(E C) = 0.4 15. P(A E) = 0.198 19. P(B E) = 0.636 23. P(E) = 0.1965 25. (a) P(A1 E) = 0.4783 27. (a) P(A1 E) = 0.4453 (c) P(A3 E) = 0.1425 29. (a) 1 2 1 2 U1 9. 13. 17. 21. P(Ec A) = 0.4 P(E) = 0.605 P(C E) = 0.165 P(E) = 0.069 W B 12 26 9 26 11 29 W R B 6 29 W R 617 = 0.3273 1885 (b) P(B) = P(U1) # P(B U1) + P(U2) # P(B U2) + P(U3) # P(B) = 617 = 0.3273 1885 (c) P(U1 B) = 0.6110 (d) P(U1 R) = 0;P(U2 R) = 0.626;P(U3 R) = 0.374 33. (a) P(colorblind) = 0.0425 (b) P(female colorblind) = 0.0766 35. (a) P(M) = 0.4810 (b) P(A6 M) = 0.1099 (d) P(A1 M) = 0.1407 37. (a) P(D) = 0.560 (b) P(A3 D) = 0.311 (c) P(A1 D) = 0.089 39. P(innocent testispositive) = 0.00058 A3 9 15 5 26 12 29 A1 B U2 1 6 3. To partition a sample space S into three subsets. A1, A2, and A3 means that S is separated into three nonempty subsets, where the subsets are all pairwise disjoint and A1 ´ A2 ´ A3 = S. 6 15 P(B U3) = (b) P(A2 E) = 0.5217 (b) P(A2 E) = 0.4122 5 12 Black 7 12 White 6 14 U2 Black 5 14 3 14 White Red 71 = 0.4226 168 (b) P(B) = P(U1) # P(B U1) + P(U2) # P(B U2) = 0.4226 (c) P(U1 B) = 0.4930 (d) P(U1 R) = 0;P(U2 R) = 1 P(B) = AN5-1 Copyright © 2013 Pearson Education, Inc. 7804_Ch05CDAns_ppA01-A01.indd 11 9/8/11 1:10 PM

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