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Unformatted text preview: EE464 Homework #2 Solutions Problem #1 Let us define the following events: A â‰¡ { N heads } , B 1 â‰¡ { Fair Coin } and B 2 â‰¡ { Loaded Coin } . Then we want to plot P ( B 1  A ) vs p l . Using Bayesâ€™ Theorem we can write P ( B 1  A ) = P ( A  B 1 ) P ( B 1 ) P ( A  B 1 ) P ( B 1 )+ P ( A  B 2 ) P ( B 2 ) . Now note that P ( B 1 ) = P ( B 2 ) = 1 / 2. Furthermore, since the flips are independent we see that P ( A  B 1 ) = (1 / 2) n and that P ( A  B 2 ) = p n l . Therefore, we can write P ( B 1  A ) = (1 / 2) n +1 (1 / 2) n +1 +(1 / 2) p n l = (1 / 2) n (1 / 2) n + p n l . A plot for the case N = 10 is shown below. Problem #2 We first use independence to prove the equality. We can write P ( A  B ) = P ( A âˆ© B ) P ( B ) , P ( B ) > 0 by definition, and similarly for P ( A  B C ), so that we will only consider the cases for which P ( B ) 6 = 0 and P ( B ) 6 = 1. Now by inde pendence we can simplify and write P ( A  B ) = P ( A ) P ( B ) P ( B ) = P ( A ). So now it suffices to show that P ( A  B C ) = P ( A ). To do this we use total probability to write P ( A ) = P ( A âˆ© B ) + P ( A âˆ© B C ) = P ( A âˆ© B ) + P ( A  B C ) P ( B C ). By independence we rewrite as P ( A ) = P ( A ) P ( B ) + P ( A  B C ) P ( B C ). Now we note that P ( B ) = 1 P ( B C ) so that P ( A ) = P ( A )(1 P ( B C )) + P ( A  B C ) P ( B C ). Rearranging gives P ( A ) = P ( A )+ P ( B C )( P ( A  B C ) P ( A ) so that P ( B C )( P ( A  B C ) P ( A )) = 0. Now since we have assumed that P ( B C ) 6 = 0, it must be true that P ( A  B C ) = P ( A ), thus proving the forward 1 implication. QED For the reverse implication we assume that P ( A  B ) = P ( A  B C ) = p . Then we can write as before P ( A ) = P ( A  B ) P ( B ) + P ( A  B C ) P ( B C ) = pP ( B ) + pP ( B C ) = p ( P ( B ) + P ( B C )) = p . Therefore P ( A ) = P ( A  B ) = P ( A âˆ© B ) P ( B ) which implies that P ( A âˆ© B ) = P ( A ) P ( B ). QED Problem #3 (a) We first note that there are 81 possible outcomes ( N 1 ,N 2 ) each hav ing probability = 1 / 81. Let B â‰¡ { Î£ is odd } . Now we are looking for p â‰¡ P (Î£ = 7  B ). Using conditional probability we can write p = P (Î£=7 âˆ© B ) P ( B ) . Note that since { Î£ = 7 } âŠ† B , we can write p = P (Î£=7) P ( B ) ....
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 Spring '06
 Caire
 Conditional Probability, Probability, Probability theory, Empirical Probability vs, Probability vs Number

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