20071ee131A_1_131a3sol

# 20071ee131A_1_131a3sol - EE 131A Homework#3 Solution Winter...

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EE 131A Homework #3 Winter 2007 Solution K. Yao 1. There are 2 4 = 16 possible events taken from S. They are: ; { a } ; { b } ; { c } ; { d } ; { a,b } ; { a,c } ; { a,d } ; { b,c } ; { b,d } ; { c,d } ; { a,b,c } ; { a,b,d } ; { b,c,d } ; { a,c,d } ; { a,b,c,d } . 2. a. i =1 A i = { x : 0 x < 1 } . b. Since all the A i ’s are mutually exclusive, then i =1 A i = . 3. Solve for C 2 0 C n - 2 4 C n 4 = 1 2 × C 2 2 C n - 2 2 C n 4 C n - 2 4 = C n - 2 2 / 2 ( n - 2)! 4!( n - 6)! = ( n - 2)! 2 × 2!( n - 4)! , n 6 , 1 4 × 3 = 1 2 × ( n - 4)( n - 5) ( n - 4)( n - 5) = 6 n 2 - 9 n + 14 = 0 = ( n - 2)( n - 7) . Thus, n = 7 . 4. a. 4/499 = 0.008 b. (5/500)(4/499)=0.00008 c. (495/500)(494/499)=0.98 5. 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.

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## This note was uploaded on 06/09/2010 for the course EE 131A taught by Professor Lorenzelli during the Spring '08 term at UCLA.

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20071ee131A_1_131a3sol - EE 131A Homework#3 Solution Winter...

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