A father has 3 children with at least one boy. The probability that he has 2 boys and 1 girl is

Consider the following events:
A: Father has at least one boy.
B: Father has 2 boys and one girl
Then,
A = one boy and2 girls, 2 boys and one girl, 3 boys and no girl
$\mathrm{A}\cap \mathrm{B}$ = 2 boys and one girl
Now, the required probability is

$=\mathrm{P}(\mathrm{A}/\mathrm{B})=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{A}\right)}=\frac{1}{3}$