Probability of $2$ boys in a family. In a family there are 3 children with minimum $1$ boy.What is the probability there are exactly $2$ boys in the family?
I think I have to use combinatorics to solve this problem.
I have solved some problem in probability but this one seems to me totally different.Answer of this problem is 3/7.But I cant't figure it out.Please help me.
 A: There are $2^3=8$ combinations, with no restrictions. There are $7$ combinations with at least a boy, because there is only $1$ combination with no boys (namely, three girls).
The number of combinations with exactly two boys is $\binom 32=3$. (These are $BBG$, $BGB$ and $GBB$). Then the probability is $3/7$.
A: This is a conditional probability. You are given there is at least 1 boy and asked to find the probability that there are two boys.
Let even $A=$ three children with 2 boys, $B=$three children with at least 1 boy, then
$$P(A|B)=\frac{P(A\cap B)}{P(B)}$$
where, $$P(A\cap B)=\binom{3}{2}\left(\frac{1}{2}\right)^3$$
$$P(B)=1-\left(\frac{1}{2}\right)^3$$
Hence $P(A|B) =\frac{3}{7}$.
A: What can happen if a family is blessed with $3$ children? Possibilities:


*

*BBB

*BBG

*BGB

*BGG

*GBB

*GBG

*GGB

*GGG


Each of these events have the same probability. Note that $7$ of these possibilities remain if it is demanded that at least one of the children is a boy. The last (GGG) falls off. In how many of these  $7$ possibilities are there exactly $2$ boys? Just counting we find: in $3$. So the answer to your question is $\frac{3}{7}$. 
Counting was enough here and on purpose I did not use combinatorics. That can be used to find out the mentioned values $7$ and $3$. For that see the answer of @ajotatxe.
