Question: In a society each couple can have at most two children. Each couple stops having a children once they have a son. What is the expected proportion of females in the total population of the society?
My Attempt: There are three favourable outcomes: M, FM, FF. Thus the proportions of females are respectively $1/3, 1/2$ and $3/4$ with probabilities $1/2, 1/4$ and $1/4$ respectively. Therefore I got the probability to be $(1/3 \times 1/2 + 1/2 \times 1/4+ 3/4\times 1/4)$. I am not sure whether this is correct. Also do we divide the expected number of females by the expected number of total to find the expected proportion or do we find the expected value of the proportions as I have done?
You can't weight proportions the way you did, because both numerators and denominators vary.
To solve the problem along the lines you suggested:
Assuming we are counting all the people in the family (including both the mother and the father) we see that there is a $\frac 12$ chance that the family has $3$ members and $1$ $F$, a $\frac 14$ chance that it has $4$ members and $2\,F's$ and a $\frac 14$ chance that it has $4$ members and $3\,F's$. Thus, if there are exactly $N$ families the total number of $F's$ is $$N\times \left( \frac 12+\frac 24+\frac 34\right)=\frac 74N$$ while the total number of people is $$N\times \left( \frac 32+\frac 44+\frac 44\right)=\frac {14}4N$$ So the desired ratio is $\boxed {\frac 12}$.
It's worth confirming that this is also the gender ratio amongst the kids alone, which implies that it is stable across generations.