practice1_sol

# practice1_sol - ORIE 3500/5500 Fall ’10 Solutions for...

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Unformatted text preview: ORIE 3500/5500, Fall ’10 Solutions for Practice Problems Solutions for Selected Practice Problems Problem 1 Let X A be the number of times coin A comes up heads, and let X B be the number of times coin B comes up heads. Let S A and S B denote the events that coin A or coin B is biased, respectively. We want to compute P ( S A | X A > X B ). By Bayes’ formula, P ( S A | X A > X B ) = P ( X A > X B | S A ) P ( S A ) P ( X A > X B | S A ) P ( S A ) + P ( X A > X B | S B ) P ( S B ) . We have P ( X A > X B | S A ) = P ( X A = 1 ,X B = 0 | S A ) + P ( X A = 2 ,X B = 0 | S A ) + P ( X A = 2 ,X B = 1 | S A ) = (2 × . 7 × . 3 × . 5 2 ) + (0 . 7 2 × . 5 2 ) + (0 . 7 2 × 2 × . 5 2 ) = 0 . 4725 . Similarly, P ( X A > X B | S B ) = P ( X A = 1 ,X B = 0 | S B ) + P ( X A = 2 ,X B = 0 | S B ) + P ( X A = 2 ,X B = 1 | S B ) = (2 × . 5 2 × . 3 2 ) + (0 . 5 2 × . 3 2 ) + (0 . 5 2 × 2 × . 7 × . 3) = 0 . 1725 . Also, since we have no information which coin is biased, P ( S A ) = P ( S B ) = 1 / 2 . It follows that P ( S A | X A > X B ) = (0 . 4725 × . 5) (0 . 4725 × . 5) + (0 . 1725 × . 5) ≈ . 732 ....
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practice1_sol - ORIE 3500/5500 Fall ’10 Solutions for...

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