39 a yes since pa 1 a 2 0 b pa 1 b pa 1 pb a 1 4020

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39. a. Yes, since P(A 1 A 2 ) = 0 b. P(A 1 B) = P(A 1 )P(B | A 1 ) = .40(.20) = .08 P(A 2 B) = P(A 2 )P(B | A 2 ) = .60(.05) = .03 c. P(B) = P(A 1 B) + P(A 2 B) = .08 + .03 = .11 d. 1 .08 P(A B) .7273 .11 = = 2 .03 P(A B) .2727 .11 = = 40. a. P(B A 1 ) = P(A 1 )P(B | A 1 ) = (.20) (.50) = .10 P(B A 2 ) = P(A 2 )P(B | A 2 ) = (.50) (.40) = .20 P(B A 3 ) = P(A 3 )P(B | A 3 ) = (.30) (.30) = .09 b. 2 .20 P(A B) .51 .10 .20 .09 = = + +
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Introduction to Probability 4 - 13 c. Events P(A i ) P(B | A i ) P(A i B) P(A i | B) A 1 .20 .50 .10 .26 A 2 .50 .40 .20 .51 A 3 .30 .30 .09 .23 1.00 .39 1.00 41. S 1 = successful, S 2 = not successful and B = request received for additional information. a. P(S 1 ) = .50 b. P(B | S 1 ) = .75 c. 1 (.50)(.75) .375 P(S B) .65 (.50)(.75) (.50)(.40) .575 = = = + 42. M = missed payment D 1 = customer defaults D 2 = customer does not default P(D 1 ) = .05 P(D 2 ) = .95 P(M | D 2 ) = .2 P(M | D 1 ) = 1 a. 1 1 1 1 1 2 2 P(D )P(M D ) (.05)(1) .05 P(D M) .21 P(D )P(M D ) P(D )P(M D ) (.05)(1) (.95)(.2) .24 = = = = + + b. Yes, the probability of default is greater than .20. 43. Let: S = small car S c = other type of vehicle F = accident leads to fatality for vehicle occupant We have P(S) = .18, so P(S c ) = .82. Also P(F | S) = .128 and P(F | S c ) = .05. Using the tabular form of Bayes Theorem provides: Events Prior Probabilities Conditional Probabilities Joint Probabilities Posterior Probabilities S .18 .128 .023 .36 S c .82 .050 .041 .64 1.00 .064 1.00 From the posterior probability column, we have P(S | F) = .36. So, if an accident leads to a fatality, the probability a small car was involved is .36. 44. Let A 1 = Story about Basketball Team A 2 = Story about Hockey Team W = "We Win" headline P(A 1 ) = .60 P(W | A 1 ) = .641 P(A 2 ) = .40 P(W | A 2 ) = .462
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Chapter 4 4 - 14 A i P(A i ) P(W | A 1 ) P(W A i ) P(A i | M ) A 1 .60 .641 .3846 .3846/.5694 = .6754 A 2 .40 .462 .1848 .1848/.5694 = .3246 .5694 1.0000 The probability the story is about the basketball team is .6754. 45. a. Let S = person is age 65 or older P(S) = 34,991,753 .12 281,421,906 = b. Let D = takes prescription drugs regularly P(D) = P(D S) + P(D S c ) = P(D | S)P(S) + P(D | S c )P(S c ) = .82(.12) + .49(.88) = .53 c. Let D 5 = takes 5 or more prescriptions P(D 5 S) = P(D 5 | S)P(S) = .40(.12) = .048 d. P(S | D 5 ) = 5 5 P(S D ) P(D ) P(D 5 ) = P(S D 5 ) + P(S c D 5 ) = P(D 5 | S)P(S) + P(D 5 | S c )P(S c ) = .40(.12) + (.28)(.88) = .048 + .246 = .294 P(S | D 5 ) = .048 .16 .294 = 46. a. P(Excellent) = .18 P(Pretty Good) = .50 P(Pretty Good Excellent) = .18 + .50 = .68 Note: Events are mutually exclusive since a person may only choose one rating. b. 1035 (.05) = 51.75 We estimate 52 respondents rated US companies poor. c. 1035 (.01) = 10.35 We estimate 10 respondents did not know or did not answer. 47. a. (2) (2) = 4 b. Let s = successful u = unsuccessful
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Introduction to Probability 4 - 15 Oil Bonds s u s u s u E 1 E 2 E 3 E 4 c. O = {E 1 , E 2 } M = {E 1 , E 3 } d. O M = {E 1 , E 2 , E 3 } e. O M = {E 1 } f. No; since O M has a sample point. 48. a. Number favoring elimination = .47(671) 315 b. Let F = in favor of proposal D = Democrat P(F | D) = .29 c. P(F) = .47 and P(F | D) = .29 Since P(F) P(F | D) they are not independent. d. I would expect Republicans to benefit most because they are the ones who had the most people in favor of the proposal. 49. Let I = treatment-caused injury D = death from injury N = injury caused by negligence M = malpractice claim filed $ = payment made in claim
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Chapter 4 4 - 16 We are given P(I) = 0.04, P(N | I) = 0.25, P(D | I) = 1/7, P(M | N) = 1/7.5 = 0.1333, and P($ | M) = 0.50 a. P(N) = P(N | I) P(I) + P(N | I c ) P(I c ) = (0.25)(0.04) + (0)(0.96) = 0.01 b. P(D) = P(D | I) P(I) + P(D | I c ) P(I c ) = (1/7)(0.04) + (0)(0.96) = 0.006 c. P(M) = P(M | N) P(N) + P(M | N c ) P(N c ) = (0.1333)(0.01) + (0)(0.99) = 0.001333 P($) = P($ | M) P(M) + P($ | M c ) P(M c ) = (0.5)(0.001333) + (0)(0.9987) = 0.00067 50. a. Probability of the event =
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