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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 = “both are 6”= { (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 family is 1, 2, or 3, with probability 1/3 each. Little Bobby has no sisters. What is the probability that he is an only child? (Set the problem up carefully. Remember to define the sample space, and any events that you use!) We can define 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 / 24 + 1 / 24 = 4 / 7 (3). (12 points) English and American spellings are colour and color , respectively. A man staying at a Parisian hotel writes this word, and a letter taken at random from his spelling is found to be a vowel. If 40 percent of the English-speaking men at the hotel are English and 60 percent are Americans, what is the probability that the writer is an Englishman?
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