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Unformatted text preview: Math 302.102 Fall 2010 Solutions to Assignment #3 1. Let A be the event that a randomly selected person is a man so that A c is the event that a randomly selected person is a woman. Let B be the event that a person is colour blind. We are told that P { B  A } = 0 . 05 and P { B  A c } = 0 . 0025. (a) If men and women each make up the same proportion of the population, then P { A } = 0 . 5 and P { A c } = 0 . 5, so that Bayes Rule tells us that P { A  B } = P { B  A } P { A } P { B } = P { B  A } P { A } P { B  A } P { A } + P { B  A c } P { A c } = (0 . 05)(0 . 5) (0 . 05)(0 . 5) + (0 . 0025)(0 . 5) = 20 21 . = 0 . 952381 . (b) If there are twice as many women as men in this population, then P { A } = 1 / 3, P { A c } = 2 / 3, and Bayes Rule tells us that P { A  B } = P { B  A } P { A } P { B } = P { B  A } P { A } P { B  A } P { A } + P { B  A c } P { A c } = (0 . 05)(1 / 3) (0 . 05)(1 / 3) + (0 . 0025)(2 / 3) = 10 11 . = 0 . 909091 . 2. Let A be the event that a randomly selected chip is good so that A c is the event that a ran domly selected chip is bad. Let B be the event that a chip passes the cheap chip test. We are told that P { A } = 0 . 8 and P { A c } = 0 . 2. Since all good chips pass the cheap chip test, P { B  A } = 1, but since 10% of bad chips also pass the cheap chip test, we have P { B  A c } = 0 . 1. (a) Using Bayes Rule we find P { A  B } = P { B  A } P { A } P { B } = P { B  A } P { A } P { B  A } P { A } + P { B  A c } P { A c } = (1)(0 . 8) (1)(0 . 8) + (0 . 1)(0 . 2) =...
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This note was uploaded on 04/07/2011 for the course MATH 302 taught by Professor Israel during the Spring '08 term at The University of British Columbia.
 Spring '08
 ISRAEL
 Math, Probability

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