# prelim1soln - ORIE 3500/5500, Fall 10 Prelim 1 Solution...

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Unformatted text preview: ORIE 3500/5500, Fall 10 Prelim 1 Solution Prelim 1 Solution Problem 1 Introduce events B 11 : the two coins were taken from box 1; B 22 : the two coins were taken from box 2; B 12 : the two coins came from different boxes; HH : both coins show tail; TT : both coins show head. We need to compute P B 11 B 22 HH TT = P B 11 HH TT + P B 22 HH TT . Note : this is NOT equal to P B 11 B 22 HH + P B 11 B 22 TT . Clearly: P ( B 11 ) = P ( B 22 ) = 1 / 4 , P ( B 12 ) = 1 / 2 , P ( HH | B 11 ) = (1 / 4) 2 = 1 / 16 , P ( HH | B 12 ) = (1 / 4)(1 / 3) = 1 / 12 , P ( HH | B 22 ) = (1 / 3) 2 = 1 / 9 , P ( TT | B 11 ) = (3 / 4) 2 = 9 / 16 , P ( TT | B 12 ) = (3 / 4)(2 / 3) = 1 / 2 , P ( TT | B 22 ) = (2 / 3) 2 = 4 / 9 . Therefore, by the law of total probability, P ( HH ) = P ( B 11 ) P ( HH | B 11 ) + P ( B 12 ) P ( HH | B 12 ) + P ( B 22 ) P ( HH | B 22 ) = 49 / 576 , P ( TT ) = P ( B 11 ) P ( TT | B 11 ) + P ( B 12 ) P ( TT | B 12 ) + P ( B 22 ) P ( TT |...
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## prelim1soln - ORIE 3500/5500, Fall 10 Prelim 1 Solution...

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