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Unformatted text preview: ORIE 360/560 Engineering Probability and Statistics II Solution of Fall 2006 Prelim 1 Problem 1 Let A = { the chosen die is crooked } , B = { the chosen die shows a 4 two times in a row } . We are looking for the conditional probability P ( A  B ). By the Bayes rule P ( A  B ) = P ( B  A ) P ( A ) P ( B  A ) P ( A ) + P ( B  A c ) P ( A c ) . We know that P ( A ) = 3 / 10, P ( A c ) = 7 / 10, P ( B  A ) = 1, P ( B  A c ) = (1 / 6) 2 = 1 36 . Substituting, we obtain P ( A  B ) = 3 10 1 3 10 1 + 7 10 1 36 = . 939 . Problem 2 ( a ) The density has to integrate to 1. Therefore, 1 = integraldisplay  f X ( x ) dx = integraldisplay 1 ( x + 1) dx + integraldisplay 2 c dx = 1 2 + 2 c . Solving for c we obtain c = 1 / 4. ( b ) The general formula for the cdf is F X ( x ) = integraltext x f X ( y ) dy . For x < 1 this results in F X ( x ) = 0. For 1 x < 0 this gives us F X ( x ) = integraldisplay x 1 ( y + 1) dy = 1 2 ( x + 1) 2 ....
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This note was uploaded on 04/13/2008 for the course ORIE 360 taught by Professor Ehrlichman during the Fall '07 term at Cornell University (Engineering School).
 Fall '07
 EHRLICHMAN

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