I don't think this could be right. Since each family has precisely one boy, then a family of N kids has N-1 girls and 1 boy. The probability of a family having N kids is 1/2^N, so the fraction of boys is sum( 1/N * 1/(2^N). After the 1st 3 terms, there are at least 1/2 + 1/4*1/2 + 1/8*1/3 fraction of boys, which is 16/24 or 2/3 .66666. So it is greater than 2/3 boys. ----- Message from ms@alum.mit.edu --------- Date: Wed, 7 Jul 2010 19:30:52 -0400 From: Mike Speciner <ms@alum.mit.edu> Reply-To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Probability puzzle To: Dan Asimov <dasimov@earthlink.net>, math-fun <math-fun@mailman.xmission.com>
Making the usual simplifying assumptions, and handwaving furiously:
Since boys and girls are equally likely, it takes on average two tries to get a boy, and it's given that for each family, exactly one child is a boy. So half the kids are girls.
--ms
On Wednesday 07 July 2010 18:41:04 Dan Asimov wrote:
(This problem is said to be one that Google asked prospective employees in job interviews. My phrasing is almost identical to how it was presented to me.)
PUZZLE: Suppose that in a very large population, each family has children until the first boy is born, and then has no more. What is the percentage of girls in the population?
(Of course, make all the usual simplifying assumptions in this kind of problem.)
--Dan
I sleep as fast as possible so I can get more rest in the same amount of time.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
----- End message from ms@alum.mit.edu -----