Question 1079029
We can assume that every pregnancy has the same probability of turning a boy or a girl. And that every pregnancy is independent from the rest.

We have two scenarios of the number of pregnancy.

1) The couple has 3 children. So let X be the number of girls in 3 pregnancies. X is a binomial with n=3 and p=0.5

P(X=2)= 3C2*(0.5)^2*0.5^1 = 3*(0.5)^3 = 3/8.

2)the couple has 4 children. So let Y be the number of girls in 4 pregnancies. Y is a binomial with n=4 and p=0.5

P(Y=2)=4C2*(0.5)^2*(0.5)^2 = 6*(0,5)^4 = 6/16 = 3/8.

So as every scenario is valid, the Probability is th sum of every probability.

P(Having two girls in 3 or 4 pregnancies) = 3/8 + 3/8 = 3/4 

@natolino_