Probability Involving Gender of Children
Date: 04/18/2005 at 13:02:10 From: Mansoor Subject: child gender probability If a family plans to have six children, and the probability that a particular child is a girl is 1/2, find the probability that the six child family contains exactly two girls. I find these gender problems very confusing. Can you please explain this to me using whatever formula I need? Thank you.
Date: 04/21/2005 at 11:19:25 From: Doctor Wilko Subject: Re: clhild gender probability Hi Mansoor, Thanks for writing to Dr. Math! I'll provide two methods that you can use to approach problems like these. One way that is worth doing, to convince yourself, is to list all the ways you can have a six-child family. The first child could be a boy or girl, the second child could be a boy or girl, and so on for the six children. Because there are two choices for each child (boy/girl), the total number of different possible arrangements of a six-child family is: 2 * 2 * 2 * 2 * 2 * 2 = 2^6 = 64 different possibilities 1. b, b, b, b, b, b ---(six boys) 2. b, b, b, b, b, g ---(five boys, one girl) 3. b, b, b, b, g, b ---(five boys, one girl) 4. --- . | . |---(everything in between) . | . --- 64. g, g, g, g, g, g ---(six girls) The hard part is keeping these all straight and not double counting or missing any of the possibilities. If you do this correctly, you'll see that out of the 64 possible six-child families, 15 of them will contain exactly two girls. Therefore, P(exactly two girls out of six children) = 15/64 = 0.234375 = 23.4375% A second and much faster method to solve this sort of problem is to use something known as the Binomial Distribution Formula. You can read more about how it works here: Binomial Probability Formula http://mathforum.org/library/drmath/view/66627.html In this case, we can apply the formula like this: P(exactly two girls out of six children) = 6C2 * (1/2)^2 * (1/2)^4 = 15 * (1/4) * (1/16) = 0.234375 = 23.4375% Here are some other related links from our archives about coin flipping, because these boy/girl probability problems can be modeled by coin flips (heads/tails): Getting Two Heads in Four Tosses of a Coin http://mathforum.org/library/drmath/view/56664.html Coin Flipping http://mathforum.org/library/drmath/sets/select/dm_coin_tossing.html Does this help? Please write back if you need anything else. - Doctor Wilko, The Math Forum http://mathforum.org/dr.math/
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