# HW8Answers(3.2) - ) P ( E | F ). We Frst calculate P ( E |...

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Math331, Spring 2008 Instructor: David Anderson Section 3.2 Homework Answers Homework: pgs. 87 - 88 #’s 1, 6, 10. 1. Let G be the event that Susan is guilty. Let L be the event that Robert lies. Then we are given that P ( G ) = . 65, P ( A | G ) = . 25, and P ( A | G c ) = 0. Therefore, P ( A ) = P ( A | G ) P ( G ) + P ( A | G c ) P ( G c ) = . 25 * . 65 + 0 = 0 . 1625 . 6. This will be an application of Theorem 3.2 once we set it up properly. Let A i be the event that we draw a red chip on the ith draw. Let B i be the event we draw a white chip on the ith draw. Note, that B i = A c i , but I am labelling them for clarity. We need the following probability: P ( A 1 B 2 A 3 B 4 + B 1 A 2 B 3 A 4 ) = P ( A 1 ) P ( B 2 | A 1 ) P ( A 3 | A 1 B 2 ) P ( B 4 | A 1 B 2 A 3 ) + P ( B 1 ) P ( A 2 | B 1 ) P ( B 3 | B 1 A 2 ) P ( A 4 | B 1 A 2 B 3 ) = 3 8 × 5 10 × 5 13 × 8 15 + 5 8 × 3 11 × 8 13 × 5 16 = 0 . 0712 10. Let F be the event that in the Frst 9 draws there are exactly three hearts. Let E be the event that the 10th draw is a heart. Then the event we want is exactly P ( EF ). We have P ( EF ) = P ( F
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Unformatted text preview: ) P ( E | F ). We Frst calculate P ( E | F ). We are therefore given that after 9 draws, three hearts have been found. Thus, 6 nonhearts have been found. There are 52-9 = 43 cards remaining, of which 13 - 3 = 10 are hearts. Thus, P ( E | F ) = 10 / 43. To calculate P ( F ) we need to count possibilities by considering draws as permutations of the 52 cards. Note that the number of permutations of 9 objects from 52 is exactly 52 P 9 = 52! / 43!. The number of ways to choose 3 hearts and 6 nonhearts is exactly ( 13 3 )( 39 6 ) However, for each of these combinations, there are 9! ways to permute them. Therefore, the probability of F is 9! p 13 3 Pp 39 6 P * 43! 52! = . 2536 . 1...
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## This note was uploaded on 03/31/2008 for the course MATH 331 taught by Professor Anderson during the Spring '08 term at University of Wisconsin.

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