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exam1practicesol - STAT4101 Practice Exam 1 solutions not...

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Unformatted text preview: STAT4101 Practice Exam 1 solutions not to be turned in 1a D , R , and I partition the set of Minnesotans, so by the law of total probability, P ( A ) = P ( A | D ) P ( D ) + P ( A | R ) P ( R ) + P ( A | I ) P ( I ) . We know that P ( D ) + P ( R ) + P ( I ) = 1. Let p = P ( D ). Then P ( R ) = 1- P ( I )- P ( D ) = 1- . 4- p = 0 . 6- p, so substituting this in, as well as all that we given, we have . 3 = 0 . 05 p + 0 . 7(0 . 6- p ) + 0 . 25(0 . 4) . Solving for p , we find that p = 0 . 338. So P ( D ) = p = 0 . 338 and P ( R ) = 0 . 6- p = 0 . 262. 1b By Bayes Rule, P ( D | A ) = P ( A | D ) P ( D ) P ( A ) = . 05(0 . 338) . 3 = 0 . 056 2a No. If they were mutually exclusive, P ( A B ) = P ( A )+ P ( B ) = 1 . 1, which is impossible. 2b By independence, P ( A B ) = P ( A ) P ( B ) = 0 . 6 . 5 = 0 . 3. 2c P ( A B ) = P ( A ) + P ( B )- P ( A B ) = 0 . 5 + 0 . 6- . 3 = 0 . 8....
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This note was uploaded on 02/23/2011 for the course MATH 444 taught by Professor Any during the Fall '10 term at Roosevelt.

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exam1practicesol - STAT4101 Practice Exam 1 solutions not...

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