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solution03 - Solutions to Homework 03[1 The number of ways...

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Solutions to Homework 03 [1] The number of ways of picking 20 raccoons out of N is N 20 ! The number of ways of picking 5 tagged raccoons out of 10 and 15 untagged raccoons out of N-10 is 10 5 ! N - 10 15 ! The probability of picking 5 tagged out of 20 is p ( N ) = 10 5 ! N - 10 15 ! N 20 ! To find the N that maximizes p ( N ), we need to find the range of N where p ( N ) increases with N. This is equivalent to the range of N where p ( N ) /p ( N - 1) 1. p ( N ) p ( N - 1) = N - 10 15 ! N - 1 20 ! N 20 ! N - 11 15 ! = ( N - 10)( N - 20) ( N - 25) N 1 ( N - 10)( N - 20) ( N - 25) N 40 N Therefore, when N = 40, p ( N ) has the maximum of p (40) = . 284. [2] (i) P [ A | B ] = P [ A B ] P [ B ] for P [ B ] > 0. Now 0 P [ A B ] , 0 < P [ B ] 0 P [ A | B ] Also, ( A B ) B P [ A B ] P [ B ] P [ A | B ] 1 Therefore, 0 P [ A | B ] 1 ii. First note that B S = B . Then P [ S | B ] = P [ B S ] P [ B ] = P [ B ] P [ B ] = 1 1
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iii. We’ll use the following result A C = φ ( A B ) ( C B ) = φ Then, P [ A C | B ] = P [( A C ) B ] P [ B ] = P [( A B ) ( C B )] P [ B ] = P [( A B )] + P [( C B )] P [ B ] = P [ A | B ] + P [ C | B ] [3] Let X denote the input and Y denote the output.
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