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Unformatted text preview: EE 465 : Homework 2 Solution 1. Each flip after the first will independently result in a changeover with probability 1 2 . Thus, we have a changeover possibility in n 1 locations, implying: P ( k changeovers ) = n 1 k ¶ ‡ 1 2 · k ‡ 1 2 · n 1 k = n 1 k ¶ ‡ 1 2 · n 1 2. a) X= number of red balls removed before the first black ball is chosen X i = n 1 , if redball iistakenbeforeany black ball , otherwise ⇒ X = number of timesthatX i = 1 ⇒ X = n X i =1 X i b) E [ X ] = E [ n X i =1 X i ] = n X i =1 E [ X i ] E [ X i ] = 1 X j =0 jP ( X i = j ) = 0 · P ( X i = 0) + 1 · P ( X i = 1) = P ( X i = 1) P ( X i = 1) = P ( Redball iischosenbeforeall mblack balls ) = 1 m + 1 since the i th red ball and the m black balls are equally likely to be the one chosen earliest. Therefore, E [ X ] = ∑ n i =1 E [ X i ] = ∑ n i =1 1 m +1 = n m +1 3. a) Φ X i ( t i ) = Φ(0 , ,..., , 1 , ,..., 0) with the 1 at the i th place b) Φ( t 1 ,t 2 ,...,t n ) = E ( e t 1 X 1 + t 2 X 2 + ... +...
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This note was uploaded on 08/05/2009 for the course EE 465 taught by Professor Chow during the Fall '04 term at USC.
 Fall '04
 Chow

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