Probability of having exactly $2$ boys in a family of $6$ children The question is: What is the probability of having exactly two boys in a family of $6$ children.
I set up the equation with $\binom{6}{2}(0.5)^2$, but I don't know whether this is correct nor do I know how to achieve the answer. Can someone help?
 A: Just for fun, you can use the linearity of expectation. For $1<k<6$, let $X_k$ be the random variable that is $1$ if the $k$-th oldest child is the second oldest boy in a two-boy family and $0$ otherwise. Then the desired probability is $\sum_{k=1}^6E(X_k)$, which is the expected number of children in a family that are the second-oldest boy in a family of two boys. Since there can be only one such child, this total expectation is the desired probability, that one exists.
The expectation $E(X_k)$ is the probability that the $k$-th oldest child is the second oldest boy in a two boy family, and it is
\begin{align} &p({\mbox {there is exactly one boy among the $k-1$ oldest children}})\\
\cdot\ &p({\mbox {child $k$ is a boy}})\\
\cdot\ &p({\mbox {children $k+1\dots6$ are girls}})\\
=\ &2^{-(k-1)}(k-1)
\cdot\ 2^{-1}\cdot\ 2^{-(6-k)}\\
=\ &(k-1)2^{-6}.
\end{align}
Then the desired probability is $\sum_{k=1}^6(k-1)2^{-6}=\frac {k(k-1)}{2}2^{-6}.$
A: It is : $\binom{6}{2}\cdot (0.5)^6$. Can you work out the decimals...
A: The binomial probability that an event with probability $p$ occurs exactly $k$ times in $n$ trials is 
$$\binom{n}{k}p^k(1 - p)^{n - k}$$
where $p^k$ is the probability that the event occurs exactly $k$ times, $(1 - p)^{n - k}$ is the probability that the event does not occur during the other $n - k$ trials, and $\binom{n}{k}$ counts the number of ways in which exactly $k$ of the $n$ trials result in the event occurring.
If we assume that the probability that a boy is born is $1/2$, then the probability that a girl is born is also $1/2$ since $1 - 1/2 = 1/2$.  Hence, the probability that a family with six children will have exactly two boys is 
$$\binom{6}{2}\left(\frac{1}{2}\right)^2\left(1 - \frac{1}{2}\right)^4 = \binom{6}{2}\left(\frac{1}{2}\right)^6$$
as you found.
