20082ee131A_1_HW3SOL

20082ee131A_1_HW3SOL - EE 131A Homework #3 Spring 2008...

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Unformatted text preview: EE 131A Homework #3 Spring 2008 Solution K. Yao 1. a. 4/499 = 0.008 b. (5/500)(4/499)=0.00008 c. (495/500)(494/499)=0.98 2. a. To check for independence between A and B, verify whether P ( B | A ) = P ( B )? P ( B | A ) = 4 / 499 = 0 . 008; P ( B ) = (495 / 500)(5 / 499) + (5 / 500)(4 / 499) = 0 . 01 . Thus, A and B are not independent. b. P ( B ) = 5 / 500 = 0 . 01 = P ( B | A ) . 3. P ( A B C ) = P ( A ( B C )) = P ( A )+ P ( B C )- P ( A ( B C )) = P ( A )+ P ( B )+ P ( C )- P ( B C )- P ( A ( B C )) . By distributive law, P ( A ( B C )) = P (( A B ) ( A C )) = P ( A B )+ P ( A C )- P ( A B C ) . Upon its substitutiong back to the origina equation, we have P ( A B C ) = P ( A )+ P ( B )+ P ( C )- P ( A B )- P ( A C )- P ( B C )+ P ( A B C ) . 4. I [- 1 , 1]. P [ I ] = k length( I ). 1 = P [[- 1 , 1]] = k length([- 1 , 1]) = k 2. k = 1 2 ....
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20082ee131A_1_HW3SOL - EE 131A Homework #3 Spring 2008...

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