Date: 02/06/99 at 13:31:41 From: Brett Rogers Subject: Bayes Theorem This problem has caused a lot of debate at my office. Could you please explain the mathematical difference between the following two situations: 1) A woman has two children. At least one of them is a girl. What is the probability that her other child is a boy? 2) A woman has two children. She is standing in front of you with her 10-year-old daughter. What is the probability that her other child is a boy? Thank you for your time. Respectfully, Brett Rogers
Date: 03/04/99 at 18:53:23 From: Doctor Stacey Subject: Re: Bayes Theorem A woman has two children. Let us assume that the probability of a girl is one-half: P(g) = .5 Now, what are the possibilities? P(g,g)=.25 P(g,b)=.25 P(b,g)=.25 P(b,b)=.25 Then, in the first case, the situation becomes narrowed; we only look at those results where there is at least one girl. Now, given that, we can have (g,g), (g,b), or (b,g). The probability of a boy, given that there is at least one girl, is 2/3. Now let us look at the next situation. The woman's 10-yr-old daughter is right there. So we know that we cannot have both (g,b) and (b,g) be possibilities. We do not know whether this girl is the first or second child, but she has to be one or the other. So, we have possibilities of either (g,g), (g,b), or (g,g), (b,g). In either case, the probability that the other child is a boy is 1/2. I hope this helps you out. Feel free to write back if you need more assistance. - Doctor Stacey, The Math Forum http://mathforum.org/dr.math/
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