Unformatted text preview: 5.(b) Here, since the renewal property does not hold and there are not too many doors, it is simplest to write down all the paths that lead out. (In fact, using this method you can actually determine the probability mass function of X and directly compute E ( X ) and E ( X 2 ).) 8. (a) Since U and V are independent, U uniform over (0, 1), V uniform over (0, 2), we have P ( V > U ) = i 1 P ( V > U  U = u ) f ( u ) du = i 1 P ( V > u  U = u ) du by substitution and f ( u ) = 1 = i 1 P ( V > u ) du since U and V independent = i 1 (1u 2 ) du = 3 / 4 (2) (b) note P ( V > U  I V = 1) = 1 while P ( V > U  I V = 0) = P ( V > U  < V < 1) = 1 / 2 by symmetry. Therefore P ( V > U ) = (1)(1 / 2) + (1 / 2)(1 / 2) = 3 / 4 9. P ( X = x ) = I 1 P ( X = x  U = u ) f ( u ) du = I 1 (1u ) x1 udu = 1 / [ x ( x + 1)]...
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 Spring '08
 Chisholm
 Probability theory, Bayesian probability, Probability mass function

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