HW4.pdf - Probability and Statistics for Engineers...

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67 51. a. Let A = child has a food allergy, and R = child has a history of severe reaction. We are told that P ( A ) = .08 and P ( R | A ) = .39. By the multiplication rule, P ( A R ) = P ( A ) × P ( R | A ) = (.08)(.39) = .0312. b. Let M = the child is allergic to multiple foods. We are told that P ( M | A ) = .30, and the goal is to find P ( M ). But notice that M is actually a subset of A : you can’t have multiple food allergies without having at least one such allergy! So, apply the multiplication rule again: P ( M ) = P ( M A ) = P ( A ) × P ( M | A ) = (.08)(.30) = .024. 52. We know that P ( A 1 A 2 ) = .07 and P ( A 1 A 2 ) = .01, and that P ( A 1 ) = P ( A 2 ) because the pumps are identical. There are two solution methods. The first doesn’t require explicit reference to q or r : Let A 1 be the event that #1 fails and A 2 be the event that #2 fails. Apply the addition rule: P ( A 1 A 2 ) = P ( A 1 ) + P ( A 2 ) – P ( A 1 A 2 ) .07 = 2 P ( A 1 ) – .01 P ( A 1 ) = .04. Otherwise, we assume that P ( A 1 ) = P ( A 2 ) = q and that P ( A 1 | A 2 ) = P ( A 2 | A 1 ) = r (the goal is to find q ). Proceed as follows: .01 = P ( A 1 A 2 ) = P ( A 1 ) P ( A 2 | A 1 ) = qr and .07 = P ( A 1 A 2 ) = 1 2 1 2 1 2 ) ( ( ) ( ) P A P A A A P A A + + = .01 + q (1 – r ) + q (1 – r ) q (1 – r ) = .03. These two equations give 2 q – .01 = .07, from which q = .04 (and r = .25). 53. P ( B|A ) = ) ( ) ( ) ( ) ( A P B P A P B A P = (since B is contained in A , A B = B ) = 0833 .
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