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Fall 2010 HW05 - University of Illinois Fall 2010 ECE 313...

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University of Illinois Fall 2010 ECE 313: Problem Set 5: Solutions Bayes’ Formula and binary hypothesis testing 1. [Dissecting a vote] (a) The number of judges voting yes has the binomial distribution with parameters n = 3 and p, and M is the event that either two or all three judges vote yes. So P ( M ) = ( 3 2 ) p 2 (1 - p )+ p 3 = 3 p 2 - 2 p 3 . (b) Given A is true, M is equivalent to the event that at least one of the two other judges votes yes, so by the same reasoning as in part (a), P ( M | A ) = P (at least one of judge two or judge three votes yes | A ) = P (at least one of judge two or judge three votes yes) = 2 1 p (1 - p ) + p 2 = 2 p - p 2 . Another way to do this problem is to refer to the sample space: Ω = { x 1 x 2 x 3 : x i ∈ { 0 , 1 }} , where x i = 1 means judge i votes yes. Then MA = { 110 , 101 , 111 } and P ( MA ) = 2 p 2 (1 - p ) + p 3 . So, by the definition of conditional probability, P ( M | A ) = P ( MA ) P ( A ) = 2 p 2 (1 - p )+ p 3 p = 2 p - p 2 . (c) By the definition of conditional probability, P ( A | M ) = P ( AM ) P ( M ) = P ( A ) P ( M | A ) P ( M ) = p (2 p - p 2 ) 3 p 2 - 2 p 3 = 2 - p 3 - 2 p . This answer is shown in the following plot: The limit as p 0 is 2/3. This makes sense, because if p is small, by far the most likely way for the committee to reach a yes decision is to have two judges vote yes and one judge vote no, so by symmetry the conditional probability the first judge votes yes is close to 2/3. The limit as p 1 is one. This makes sense because if p is close to one, with probability close to one, all three judges vote yes. Conditioning on M
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