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Unformatted text preview: MTH 4120: Introduction to Probability, PS13A Problem Set 2 SOLUTIONS 1. Suppose that two fair dice have been tossed and the total of the faces is divisible by 4. What is the probability that both of the dice are showing a 6? Solution The sample space consists of 36 equally likely outcomes, S = (1 , 1) ,..., (6 , 6). Since we know that the sum of the face must be divisible by 4, our reduced sample space is F = (1 , 3) , (2 , 2) , (2 , 6) , (3 , 1) , (3 , 5) , (4 , 4) , (5 , 3) , (6 , 2) , Let E be the event that both dice are showing 6. Then E = (6 , 6). Let F be the event that the sum of the faces is 4. Then F = (1 , 3) , (2 , 2) , (2 , 6) , (3 , 1) , (3 , 5) , (4 , 4) , (5 , 3) , (6 , 2) , (6 , 6). P ( E  F ) = P ( EF ) P ( F ) = 1 / 36 9 / 36 = 1 9 2. There is a 5050 chance that Barbara carries the gene for hemophilia. If she is a carrier, then each of her children has a 5050 chance of having hemophilia. Suppose that Barbara has had three children, and none of them have the disease. What is the probability that Barbara is a carrier? If she has a fourth child, what is the probability that the child will have hemophilia? Solution Let E be the event that Barbara is a carrier of the gene for hemophilia. Let F be the event that Barbaras first three children do not have hemophilia. Since Barbara has a 5050 chance of carrying the gene, we know that P ( E ) = 0 . 5. (a) We want to find the probability that Barbara is a carrier of the gene given that none of her three children have hemophilia, P ( E  F ). P ( E  F ) = P ( EF ) P ( F ) = P ( F  E ) P ( E ) P ( F  E ) P ( E ) + P ( F  E c ) P ( E c ) = = (0 . 5) (0 . 5) 3 (0 . 5) 3 (0 . 5) + 1 (0 . 5) = 1 9 Notice we needed to assume independence among the three children to write that P ( F  E ) = (0 . 5) 3 . (b) Let G be the event that the fourth child has hemophilia. We want to find P ( G  F ), since we are still assuming that that the first three children do not have hemophilia. To compute P ( G  F ), we condition on whether or not Barbara is a carrier: P ( G  F ) = P ( G  EF ) P ( E  F ) + P ( G  E c F ) P ( E c  F ) = (0 . 5) 1 9 + 0 8 9 = 1 18 3. A woman has agreed to participate in an ESP experiment. She is asked to pick, randomly, two distinctA woman has agreed to participate in an ESP experiment....
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This note was uploaded on 12/06/2009 for the course MATH 4120 taught by Professor Carlosj.moreno during the Spring '08 term at CUNY Baruch.
 Spring '08
 CARLOSJ.MORENO
 Probability

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