Probability that last child is a boy Johnny has 4 children. It is known that he has more daughters
than sons. Find the probability that the last child is a boy.
I let A be the event that the last child is a boy, P(A) = $\frac{1}{2}$.
and B be the event that he as more daughters than sons. But im not sure how to calculate P(B) and what are the subsequent steps to take after.
Appreciate any help.
Thanks

The number of girls in the family would have a binomial distribution, so the prior probability that there are 3 or 4 girls in the family would be:
$$\begin{align}
\mathsf P(B) & = {4\choose 3}(\tfrac 1 2)^3(\tfrac 1 2)+{4\choose 4}(\tfrac 1 2)^4
\\ & = \frac 5{16}
\end{align}$$
Now for the probability that the last child in the family is a boy and that there are more girls than boys in the family is equal to: the prior probability that the first three children are girls and the last is a boy:
$$\begin{align}
\mathsf P(A\cap B) & = \frac{1}{16}
\end{align}$$
Thus the posterior probability, that the last child is a boy given that their are more girls in the family than boys is:
$$\begin{align}
\mathsf P(A\mid B) & = \frac{\mathsf P(A\cap B)}{\mathsf P(B)}
\\ & = {\frac 1 {16}}\bigg/\frac 5 {16}
\\ & = \dfrac 1 5
\end{align}$$
