In Culture A, parents have children until they have a boy and a girl (not necessarily in that order). In Culture B parents have children until they have 2 boys, regardless of the number of girls. Assuming children are born one at a time, that boys and girls are equally probable, and that families can have as many children as they like until they decide to stop, calculate (a) the expected number of children a Culture A family will have, and (b) the expected number of children a Culture B family will have. (Hint: condition on the sex of the first child born.) (c) What is the expected number of boys and girls in each case?
For part a , I get the expected number is 3. Is it correct? In my thinking that Let X be the number of children. And if the first child is boy or girl, then （X-1）~geom(1/2). Thus, E(X-1）=2. And E(X)=1+E(X-1)=1+2=3.
In part b I got the expected number is 4. So, part a and b is done.
For Part C the question ask boys and girls in each case?Does is mean I need to find 4 expected numbers , 1 for boys and 1 for girls in each case?
Second Question
Using the hint, if the first child is female, then the expected number of further children is equal to what it was at the outset, because we are no nearer to the requirement of 2 males being met. So if we expect E children, then in the case of the first child being female, we then expect a total of E + 1 children
If the first child is male, then the expected future children is the same as the expectation for the first male, which can be worked out as $M = 1 + \frac{1}{2}M $
gives M = 2 (you've calculated similar with geometric series in question 1)
so for 2 males
$E = 1 + P(first female)E + P(first(male)) \times 2$
$E = 1 + \frac{1}{2}E + \frac{1}{2} \times 2$
$ E = 4 $