hw2-sol

hw2-sol - AMS 311(Fall 2009 Joe Mitchell PROBABILITY THEORY...

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AMS 311 (Fall, 2009) 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 4. 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 =“bothare6”= { (6 , 6) } and F = “sum is divisible by 4” = { (1,3), (2,2), (2,6), (3,1), (3,5), (4,4), (5,3), (6,2), (6,6) } . P ( E | F )= P ( E F ) P ( F ) = P ( { (6 , 6) } ) P ( { (1 , 3) , (2 , 2) , (2 , 6) , (3 , 1) , (3 , 5) , (4 , 4) , (5 , 3) , (6 , 2) , (6 , 6) } ) = 1 9 (2). (12 points) Suppose for simplicity that the number of children in a familyis1 ,2 ,or3 ,w i thprobab i l i ty 1/3 each. Little Bobby has no sisters. What is the probabilitytha theisanon lych i ld?(Se ttheprob lemup carefully. Remember to deFne the sample space, and any eventstha tyouuse !) We can de±ne the sample space as S = { B, G, BB, BG, GB, GG, BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG } . Let A = { B,BB,BBB } be the event that there are no girls in the family (ie, Bobby has no sisters). Let B i be the event that the family has i children. (e.g., B 1 = { B } and B 2 = { BB,BG,GB,GG } ) P ( B 1 A )= P ( B 1 ) · P ( A | B 1 )=(1 / 3)(1 / 2) = 4 / 24 P ( B 2 A )= P ( B 2 ) · P ( A | B 2 )=(1 / 3)(1 / 4) = 2 / 24 P ( B 3 A )= P ( B 3 ) · P ( A | B 3 )=(1 / 3)(1 / 8) = 1 / 24 We want to compute: P ( B 1 | A )= P ( B 1 A ) P ( A ) = 4 / 24 4 / 24 + 2 /
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hw2-sol - AMS 311(Fall 2009 Joe Mitchell PROBABILITY THEORY...

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