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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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 Spring '08
 LORENZELLI
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