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

Picking up Robert Israel's suggestion: If a couple keeps having children until they have at least a boy and a girl, and ignoring twins, ... it's difficult. 
If they had children until they reached their goal, the probability is 1/16 each that they stop at 3 boys and 1 girl, at 1 boy and 3 girls, 3/4 that they stopped with 2 or 3 children, 1/8 that they had five or more. 
But they may not be finished yet. They may have four sons or four daughters and be waiting for another one. After being married for some time, there is a probability p that they have enough time for four children, and a probability q < p that they have enough time for five children. Both p and q grow with time, but don't reach 1. 
The probabilities if there are four children: p/16 for 1B + 3G, p/16 for 3B + 1G, (p - q)/16 each for 4B or 4G. Since there are more girls, p/16 for 1B + 3G and the last is a boy, (p-q)/16 for four girls. With conditional probabilities, the probability that the last is a boy is p / (2p - q). 
It would depend on the length of the relationship. If they are together for four years, it would be quite unlikely that they could have five children, (q would be small compared to p) and the probability of GGGB would be only slightly larger than GGGG, so the probability would be only a bit higher than 0.5. If they are together for many years, having GGGG and no fifth child is unlikely, so the chance that the last one is a boy is closer to 1. 
That also means that statistics for currently living parents and historical data would show different numbers. 
Obviously for different parental strategies the result would be different. 
