If in a family of 3 children, there is at least 1 boy, what is the probability that there are exactly 2 boys among the children?
My attempt:
I calculated all the possibilities of the family by having at least one boy,
$1$B and $2$G – $2*2$$*$$3 \choose 1$
$2$B and $1$G – $2$$*$$3 \choose 1$
$3$B – $1$
That sums up to 18.
So, the probability of having exactly 2 boys,
$$\frac{2*3}{18}$$
But the answer given is $\frac{3}{7}$
Any help would be appreciated.
Best Answer
Let $B$ be the number of boys.
What you are looking for is
$$P(B=2 | B\geq 1) = \frac{P([B=2]\cap [B\geq 1])}{P(B\geq 1)}=\frac{P(B=2)}{P(B\geq 1)}$$
Assuming $B$ follows a binomial distribution of parameters $p=1/2$ and $n=3$, you have $$P(B=2) = \frac{\binom{3}{2}}{2^3} = \frac{3}{8}$$ and $$P(B\geq 1) = 1 - P(B=0)= 1 - \frac{\binom{3}{0}}{2^3} = \frac{7}{8}$$
which leads to $3/7$ as expected.