How to find the probability of a family having two boys out of three? 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 $P(B|A)$?
 A: $\Pr[B\mid A]$ 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.
A: given that  ,at least  one children  is boy,consider different  situation .first  consider that we  have information   about one boy,then probability that in  remained two  children  one of them  is boy  is $1/2=0.5$ if  it has two boys then probability is $1$,but if three boy,then probability that  they have  exactly two  boy,is zero,i think it would help you
