20065ee131A_1_EE131AF06HW2Solution

# 20065ee131A_1_EE131AF06HW2Solution - Solution of Assignment...

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Unformatted text preview: Solution of Assignment 2, EE 131A Due: Wednesday October 18, 2006 Problem 1. Prove that parenleftbigg n k parenrightbigg = parenleftbigg n- 1 k- 1 parenrightbigg + parenleftbigg n- 1 k parenrightbigg . Solution: (a) parenleftbigg n- 1 k- 1 parenrightbigg + parenleftbigg n- 1 k parenrightbigg = ( n- 1)! ( k- 1)! ( n- k )! + ( n- 1)! k ! ( n- k- 1)! = ( n- 1)! k + ( n- 1)! ( n- k ) k ! ( n- k )! = ( n- 1)!( k + n- k ) k ! ( n- k )! = n ! k ! ( n- k )! = parenleftbigg n k parenrightbigg . (b) Let X be a set of size n and x ∈ X an element of X . Then the subsets of size k of X can be divided to two classes; (i) the subsets which do not contain x , there are ( n- 1 k ) such subsets; (ii) the subsets which contain x , there are ( n- 1 k- 1 ) such subsets. Problem 2. A box contains 8 white and 10 black balls. Suppose 4 balls are drawn (without replacement). Find the probability of drawing at least one white ball. Solution: The sample space S consists of all subsets of size 4 of the set of the 18 balls. Thus | S | = ( 18 4 ) = 3060 . Then P [ at least one white ball ] = 1- P [ no white ball ] = 1- ( 10 4 ) ( 18 4 ) = 95 102 ....
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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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20065ee131A_1_EE131AF06HW2Solution - Solution of Assignment...

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