324.midterm01.2005.solution

324.midterm01.2005.solution - 2005 STAT 324 midterm 1...

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2005 STAT 324 midterm 1 solution Moo K. Chung mchung@stat.wisc.edu Grade distribution 29 . 65 ± 6 . 18. 15 20 25 30 35 40 0 1 2 3 4 5 6 1. (a) Suppose that A and B are independent events such that the probability that neither occurs is 1 2 and the probability of B occurring is 1 3 . Determine the probability of A occurring. Solution. P (( A B ) c ) = 1 - P ( A B ) = 1 / 2. So P ( A B ) = 1 / 2. Since A and B are independent P ( A B ) = P ( A ) P ( B ) = P ( A ) / 3. By substituting terms into P ( A B ) = P ( A )+ P ( B ) - P ( A B ), we have 1 / 2 = P ( A )+1 / 3 - P ( A ) / 3. Hence P ( A ) = 1 / 4. (b) If A and B are independent events, prove that A c and B are also independent. Solution. Since A and B are independent, P ( A B ) = P ( A ) P ( B ). We need show P ( A c B ) = P ( A c ) P ( B ). P ( B ) = P ( A B )+ P ( A c B ) = P ( A ) P ( B )+ P ( A c B ). Hence
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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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324.midterm01.2005.solution - 2005 STAT 324 midterm 1...

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