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Unformatted text preview: } and 2H = event { There are two Heads observed in the tossed coins (second part of the experiment) }. Run the experiment 100 times first with flat‐die‐probabilities (p=1/6) and a fair‐coin (p=0.5). By counting outcomes of interest validate computationally that: P( D3 | 2H ) = [ P( 2H | D3 ) P( D3 ) ] / P( 2H ) Answer varies. In my trial, I obtained P( D3 | 2H ) = 0.2308, P( 2H | D3 ) = 0.3158, P( D3 ) = 0.19, P( 2H ) = 0.26, which validates the Bayes rule if you plug in the numbers. b. Now...
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This note was uploaded on 01/22/2014 for the course STATISTICS 35 taught by Professor Dinov during the Spring '09 term at UCLA.

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