4.2 Conditional Probability

Define the following events:

$$A:=\text{Both children are boys}$$ $$B:=\text{One child is a boy}.$$ We want $\Pr[A|B]$. By Bayes theorem, we get that $\Pr[A|B]=\dfrac{\Pr[B|A]\cdot\Pr[A]}{Pr[B]}$.

Clearly, $\Pr[B|A]$=1. $\Pr[A]=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$ and $\Pr[\bar{B}]=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4} \Rightarrow \Pr[B]=1-\frac{1}{4}=\frac{3}{4}$. Hence, $\Pr[A|B]=\frac{1}{3}$

A simpler argument works by enumerating all outcomes where at least one child is a boy, representing each outcome as an ordered pair of first the younger child's age and then the older child's age. The possibilities are BG, BB, and GB. Just 1 out of these 3 outcomes has two boys, so the conditional probability is 1/3.