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.
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 $\displaystyle P(0)+P(1)$
$\displaystyle P(0) = \left(\frac{1}{2}\right)^{11}$ since there are no boys.
$\displaystyle P(1)={11\choose 1}\left(\frac{1}{2}\right)^{11}$ there are $\displaystyle {11\choose 1}$ ways to have one boy out of a group of 11 people.
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?
$\displaystyle
P(0) = \left(\frac{1}{2}\right)^{5}
$
$\displaystyle
P(1) = {5\choose 1}\left(\frac{1}{2}\right)^{5}
$
$\displaystyle
P(2) = {5\choose 2}\left(\frac{1}{2}\right)^{5}
$
$\displaystyle
P(3) = {5\choose 3}\left(\frac{1}{2}\right)^{5}
$