hw2-sol - AMS 311(Fall 2011 Joe Mitchell PROBABILITY THEORY...

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AMS 311 (Fall, 2011) Joe Mitchell PROBABILITY THEORY Homework Set # 2 – Solution Notes (1). (12 points) Suppose that two fair dice have been tossed and the total of their top faces is found to be divisible by 3. What is the probability that both of them have landed 6? The sample space consists of 36 equally likely outcomes, S = { (1 , 1) ,..., (6 , 6) } . We want to compute P ( E | F ), where E = “both are 6”= { (6 , 6) } and F = “sum is divisible by 3” = { (1,2), (1,5), (2,1), (2,4), (3,3), (3,6), (4,2), (4,5), (5,1), (5,4), (6,3), (6,6) } . P ( E | F ) = P ( E F ) P ( F ) = P ( { (6 , 6) } ) P ( { (1 , 2) , (1 , 5) , (2 , 1) , (2 , 4) , (3 , 3) , (3 , 6) , (4 , 2) , (4 , 5) , (5 , 1) , (5 , 4) , (6 , 3) , (6 , 6) } ) = 1 12 (2). (12 points) Suppose for simplicity that the number of children in a family is 1, 2, or 4, with probability 1/3 each. Little Alice (a girl) has no brothers. What is the probability that she is an only child? (Set the problem up carefully. Remember to deFne the sample space, and any events that you use!) We can de±ne the sample space as S = { B, G, BB, BG, GB, GG, BBBB, BBBG, BBGB, BBGG, BGBB, BGBG, BGGB, BGGG, GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG } . Let A = { G,GG,GGGG } be the event that there are no boys in the family (ie, Alice has no brothers). Let B i be the event that the family has i children. (e.g., B 1 = { B,G } and B 2 = { BB,BG,GB,GG } ) The events B 1 , B 2 , B 4 partition the sample space S . Note that the outcomes of the sample space are not equally likely. However, the 2 outcomes within B 1 are equally likely, the 4 outcomes within B 2 are equally likely, and the 16 outcomes within B 4 are equally likely. We want to compute the probability that Alice is an only child (event B 1 happens), given that Alice has no brothers (event A happens). That is, we want to compute P ( B 1 | A ) = P ( B 1 A ) P ( A ) = P ( A | B 1 ) P ( B 1 ) P ( A | B 1 ) P ( B 1 ) + P ( A | B 2 ) P ( B 2 ) + P ( A | B 4 ) P ( B 4 ) = 1 2 · 1 3 1 2 · 1 3 + 1 4 · 1 3 + 1 16 · 1 3 = 8 13 , where we have used the fact that P ( A | B 1 ) = 1 / 2, P ( A | B 2 ) = 1 / 4, and P ( A | B 4 ) = 1 / 16. (3).

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This note was uploaded on 10/20/2011 for the course AMS 311 taught by Professor Tucker,a during the Fall '08 term at SUNY Stony Brook.

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hw2-sol - AMS 311(Fall 2011 Joe Mitchell PROBABILITY THEORY...

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