Probability of Two Male Children
Date: 7/5/96 at 17:56:22 From: Bret McClory Subject: Changing Probability: Two Boys Hi! My name is Bret McClory, I'm 16, and I have a math problem that has been bugging me. It's more of a logic question. Several weeks ago, in one of the weekly periodicals I read, a person presented this logical statement: If a family has two children, and the older child is a boy, there is a 50 percent chance the family will have two boys. However, in a family with two children, if all we know is that one child is a boy (no age specified) there is a 1/3 chance of that family having two male children. This boggles me, as I can plainly see the first scenario - it's quite simple: the younger child may be a boy or girl (1/2). It is a mystery to me how by not knowing whether the known boy is older or younger could change the probability of the family as a whole. However, I have discussed this problem with my grandfather and he explains it like so: any two-child family has four possible outcomes: b:b, g:g, b:g, g:b. He states that, in the first case where the known boy is also known to be the older child, we thereby eliminate the g:g and g:b combinations, leaving two left. And obviously one of those two is b:b so our odds of having two male children is 1/2. By his same logic, in the 2nd instance all that is known is that one child is a boy, therefore of course the g:g combination cannot exist, leaving b:b, b:g, g:b. Hence, there is a 1/3 chance of two boys. Although his explanation is very clear, I fail to understand how the odds of the entire family can be swayed by whether or not a boy is older or not. Example: I see a common ordinary boy. I am told that he has a sibling. Am I to believe now that his sibling has a 2/3 chance of being a girl?? Now I am told he is the older child. Magically the odds change and his younger sibling has a 1/2 chance of being a girl now? It baffles me. Bret McClory
Date: 7/5/96 at 20:8:17 From: Doctor Anthony Subject: Re: Changing Probability: Two Boys This is an example of conditional (Baysian) probability. You are told additional information (a condition e.g oldest child is a boy), and this restricts the sample space, changing the probabilities. A simple example will illustrate the point. Suppose I have 5 coins in a bag, one of which is a double-headed penny. I choose a coin at random and toss it four times. Each time I get a head. What is the probability that I have the double-headed penny? The sample space is now reduced to getting 4 heads, and we have the two possibilities to consider: I have the double-headed coin and throw 4 heads = 1/5 * 1 = 1/5 I have a normal coin and throw 4 heads = 4/5 * (1/2)^4 = 1/20 Since one or other of these MUST happen we say:- The sample space is now 1/5 + 1/20 = 1/4 and this is the denominator of the probability calculation Probability that I have double-headed coin = (1/5)/(1/4) = 4/5 The more information you provide, the more you change the sample space (the denominator) of the probability calculation. In the case of sons or daughters in a family of four, if you exclude gg or gb, then as you said you can only have bb or bg, so 1/2 probability that second child is a girl, whereas if one child (unspecified) is a boy, the probability space is now bb, bg, gb and the chance of two boys is now only 1/3. The thing to remember is that CONDITIONAL probability can dramatically change the commonsense idea of what a particular probability should be. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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