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

This preview shows pages 1–2. Sign up to view the full content.

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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.