hw3soln - ORIE 3500/5500 Fall'10 HW 3 Solutions Homework 3...

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ORIE 3500/5500, Fall ’10 HW 3 Solutions Homework 3 Solutions Problem 1 Since P ( X = a i , Y = b j ) is either 0 or 1 14 , we know that all the entries of the first column and the second row of the table are equal to 1 14 . Since P ( X = 1) = 1 14 , we know that the rest of the entries in the first row have to be 0. Similarly, P ( Y = 5) = 1 14 , the other entries in column 5 have to be 0. Continuing in this way, we get the following table: a i /b j 1 2 3 4 5 P ( X = a i ) 1 1 14 0 0 0 0 1 14 2 1 14 1 14 1 14 1 14 1 14 5 14 3 1 14 1 14 1 14 1 14 0 4 14 4 1 14 1 14 0 0 0 2 14 5 1 14 1 14 0 0 0 2 14 P ( Y = b j ) 5 14 4 14 2 14 2 14 1 14 1 Problem 2 There are 5 2 = 10 ways of testing 5 transitors, each of which is equally likely. Let g be a letter meaning a good transitor. Also d is a letter for a defective one. P ( N G = 2 , N D = 4) = P ( g, g, d, d, d ) = 1 10 P ( N G = 3 , N D = 4) = P ( g, d, g, d, d ) + P ( d, g, g, d, d ) = 1 5 P ( N G = 4 , N D = 2) = P ( d, d, g, g, g ) = 1 10 P ( N G = 4 , N D = 3) = P ( g, d, d, g, g ) + P ( d, g, d, g, g ) = 1 5 P ( N G = 5 , N D = 2) = P ( d, d, g, d, g ) + P ( d, d, d, g, g ) = 1 5 P ( N G = 5 , N D = 3) = P ( g, d, d, d, g ) + P ( d, g, d, d, g ) = 1 5 Problem 3 (a) Z 2 0 Z 1 0 Cx (3 - 2 Cy ) dxdy = Z 2 0 C (3 - 2 Cy ) 1 2 x 2 1 0 dy = Z
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