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20105ee131A_1_ps3_2010fall_CW_revised

# 20105ee131A_1_ps3_2010fall_CW_revised - EE 131A Probability...

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EE 131A Problem Set 3 Probability Wednesday, October 13, 2010 Instructor: Lara Dolecek Due: Wednesday, October 20, 2010 Reading: Chapters 3 and 4 of Probability, Statistics, and Random Processes by A. Leon-Garcia 100 points total 1. (In)dependence and Sets. Problem 2.82, page 90 of ALG P ( A ) = 1 / 2 P ( B ) = 1 / 2 P ( C ) = 1 / 2 P ( A B ) = P ( A ) × P ( B ) = 1 / 4 P ( B C ) = P ( B ) × P ( C ) = 1 / 4 P ( A C ) = P ( A ) × P ( C ) = 1 / 4 P ( A B C ) = 1 / 4 However, P ( A | B C ) = P ( A B C ) P ( B C ) = 1 6 = 1 2 = P ( A ) So we can see that A , B , and C are not independent events. 2. Sequential dependent experiments. Problem 2.107, page 92 of ALG (a) Let the pair ( b, w ) to be the state of b black balls and w white balls. Then we have the trellis diagram: (2,2) (2,1) (1,2) (2,0) (1,1) (0,2) (2,0) (1,0) (0,1) (0,2) (2,0) (0,0) (0,2) 1/2 1/2 2/3 2/3 1/2 1 1/2 1 1 1 1 1 1/3 1/3 1

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(b) The following are all the probabilities for outcome sequences with length 2 and 3. P ( bb ) = 1 / 6 , P ( bw ) = 1 / 3 P ( ww ) = 1 / 6 , P ( wb ) = 1 / 3 P ( bbw ) = 1 / 6 , P ( wwb ) = 1 / 6 P ( bww ) = 1 / 6 , P ( wbb ) = 1 / 6 P ( bwb ) = 1 / 6 , P ( wbw ) = 1 / 6 (c) P ( no blacks after 3 draws ) = P ( bbw ) + P ( bwb ) + P ( wbb ) = 1 / 2 . P ( no whites after 3 draws ) = P ( wwb ) + P ( wbw ) + P ( bww ) = 1 / 2 .
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20105ee131A_1_ps3_2010fall_CW_revised - EE 131A Probability...

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