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20082ee131A_1_HW3SOL

# 20082ee131A_1_HW3SOL - EE 131A 1 a 4/499 = 0.008...

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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 . P [ I ] = 1 2 × length( I ). a. A = {- 1 x < 0 } , P [ A ] = 1 2 × 1 = 1 2 ; B = {| x - 0 . 5 | < 1 } = {- 0 . 5 < x 1 } , P [ B ] = 1 2 × 3 2 = 3 4 . C = { x > 0 . 75 } = { 0 . 75 < x 1 } , P [ C ] = 1 2 × 1 4 = 1 8 . A B = ( - 1 2 , 0), P [ A B ] = 1 2 × 1 2 = 1 4 ; A C = , P [ A C ] = 0. b. A B = {- 1 x 1 } = S , P [ A B ] = 1; A C = {- 1 x < 0 } ∪ { 0 . 75 < x 1 } , P [ A C ] = 1 2 { 1 + 1 4 } = 5 8 ; A B C = {- 1 x 1 } = S , P [ A B C ] = 1.

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