131A_1_131A_1_prac-sol-part-2-2010

# 131A_1_131A_1_prac-sol-part-2-2010 - EE 131A Probabilities...

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EE 131A Practice Problems Solutions Probabilities Fall 2005 Instructor: Vwani Roychowdhury 1. P [ nth is new ] = m X i =1 P [ nth is new | nth = i ] p i = m X i =1 P [ i is not in the first ( n - 1)] p i = m X i =1 [(1 - p i )] n - 1 p i (a) Let deﬁne: A 1 = A is hit and A 0 = A is not hit. P [ A 0 ] = X k =1 P [ A 0 ∩ { B is hit at the kth time } ] = X k =1 P A (1 - P B )[(1 - P A )(1 - P B )] k - 1 = P A (1 - P B ) 1 - (1 - P B )(1 - P A ) (b) Probability that both are hit is equal to: X k =1 [(1 - P A )(1 - P B )] k - 1 P A P B = P A P B 1 - (1 - P A )(1 - P B ) (c) the probability that the duel ends after the n th round of shots is equal to the probability that both hit each other exactly at the n th round which is equal to: [(1 - P A )(1 - P B )] n - 1 [ P A (1 - P B ) + P B (1 - P A ) + P A P B ] 1

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(d) The conditional probability that the duel ends after the n th round of shots given that A is not hit is Prob [ game ends at n | A isn 0 t hit ] which is equal to: P [ game ends at n A isn 0 t hit
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## This note was uploaded on 01/08/2011 for the course EE 131 taught by Professor R during the Spring '10 term at UCLA.

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131A_1_131A_1_prac-sol-part-2-2010 - EE 131A Probabilities...

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