[Math] Probability of having exactly $2$ boys in a family of $6$ children

Question

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?

Answer 1

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 $\sum_{k=1}^6(k-1)2^{-6}=\frac {k(k-1)}{2}2^{-6}.$