How do I find the probability of a three children family having exactly two boys given that at least one of their children is a boy?
Do I use the dependent formula $$P(A \text{ and } B) = P(A) \times P(B \text{ given that }A \text{ has occurred})$$ or do I use the conditional probability form of $P(B|A)$?
Best Answer
$\Pr[B\mid A]$ is the same as $\Pr[\text{$B$ given that $A$ has occurred}]$. Therefore, if you divide both sides of $\Pr[\text{$A$ and $B$}]=\Pr[A]\cdot\Pr[\text{$B$ given that $A$ has occurred}]$ by $\Pr[A]$, you get $$ \Pr[B\mid A]=\frac{\Pr[\text{$A$ and $B$}]}{\Pr[A]}, $$ which is the conditional probability formula.
This can be used to solve your problem. Write $$ \begin{aligned} \Pr[\text{exactly 2 boys}\mid\text{at least 1 boy}]&=\frac{\Pr[\text{exactly 2 boys and at least 1 boy}]}{\Pr[\text{at least 1 boy}]}\\ &=\frac{\Pr[\text{exactly 2 boys}]}{\Pr[\text{at least 1 boy}]}. \end{aligned} $$ The second line follows from the first because there being exactly two boys implies that there is at least one boy.
If you treat the situation as a Bernoulli process, you can compute both of the needed probabilities.