Probability

**kvm113** · May 5th 2009, 01:07 PM

A family has eleven children. The probability of having a girl is 1/2. What is the probability of having no more than 1 boy?

Please show your work. Thanks in advance.

**putnam120** · May 5th 2009, 01:42 PM

First off I think a mod should move this since I don't think this is the right place for this topic.

So let P(x)= probability of having x boys.
Then you want $P(0)+P(1)$
$P(0) = \left(\frac{1}{2}\right)^{11}$ since there are no boys.
$P(1)={11\choose 1}\left(\frac{1}{2}\right)^{11}$ there are ${11\choose 1}$ ways to have one boy out of a group of 11 people.

**kvm113** · May 5th 2009, 02:15 PM

So, if the question was like "A family has five children. The probability of having a girl is 1/2. What is the probability of having no more than 3 boys?"

Is this how you would do that?

$ P(0) = \left(\frac{1}{2}\right)^{5} $
$ P(1) = {5\choose 1}\left(\frac{1}{2}\right)^{5} $
$ P(2) = {5\choose 2}\left(\frac{1}{2}\right)^{5} $
$ P(3) = {5\choose 3}\left(\frac{1}{2}\right)^{5} $

**putnam120** · May 5th 2009, 04:49 PM

Yes, that's correct.

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