Math135_hw3-soln

Math135_hw3-soln - Math 135, Spring 2011: HW 3 Solutions...

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Unformatted text preview: Math 135, Spring 2011: HW 3 Solutions September 21, 2011 This homework is due at the beginning of class on Thursday, February 3 . You are free to talk with each other and get help. However, you should write up your own answers and understand everything you write. Problem 1. A woman has 2 children, one of whom is a boy born on a Tuesday. What is the probability that both children are boys? (You may assume that boys and girls are equally likely and independent, and that children are equally likely to born on any given day of the week.) SOLUTION. In this problem, one of the first questions that comes to mind is, Why does Tuesday matter? To determine whether this information matters, let us work from the definitions. We are asked to find the probability that the woman has two boys, given that one of her two children is a boy born on a Tuesday. The problem does not tell us which of the two children is a boy. Let the days of the week be numbered 1 , 2 , 3 , , 7, where 1=Mon, 2=Tues, 7=Sunday. Since the woman has two children, the sample space consists of ordered pairs ( Bi,Bj ) , ( Bi,Gj ) , ( Gi,Bj ) , ( Gi,Gj ) where B or G represents the gender, and i and j each take on any integer value between 1 and 7 and represent the day of the week that the child is born. Furthermore, ( Bi,Gj ) represents the event that the firstborn child is a boy born on day i, and the secondborn is a girl born on day j, etc. Note that all of these outcomes are equally likely. Since we know that one of the children was a boy born on a Tuesday, however, we can restrict ourselves to considering the following outcomes in our sample space: the 7 outcomes ( B 2 ,Gj ), where j = 1 , 2 , , 7; the 7 outcomes ( Gj,B 2) where j = 1 , 2 , , 7, the 7 outcomes ( B 2 ,Bj ), where j = 1 , 2 , 7, and the six outcomes ( Bj,B 2) where j = 1 , 3 , , 7, because we already counted the pair ( B 2 ,B 2). Hence the event C = { The woman has two children, one of whom is a boy born on a Tuesday } has a total of 27 outcomes. The event D = { both children are boys } has 49 outcomes, but the event D C = { both children are boys, one of whom was born on a Tuesday } has only 13 outcomes. Hence P (woman has two boys | woman has two children, one of whom is Tuesday-born boy) = P ( D | C ) = P ( D C ) P ( C ) = 13 27 which is slightly less than 1/2. One way to think about this is that if you only know that the woman has two children, then the probability that both are boys is 1/4; if you know that one of them is a boy, the probability that both are boys rises to 1/3; and if you know that one of them is a boy born on a Tuesday, the probability that both are boys rises to nearly 1/2....
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This note was uploaded on 10/24/2011 for the course MATH 135 taught by Professor Staff during the Fall '08 term at Duke.

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Math135_hw3-soln - Math 135, Spring 2011: HW 3 Solutions...

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