I was given the following problem:
In a country in which people only want boys, every family continues to have children until they have a boy. If they have a girl, they have another child. If they have a boy, they stop. What is the proportion of boys to girls in the country?
Could you please check my solution?
Let's start by saying that the probability of having a boy or a girl is 50%. Then we have:
- probability of having a family with 1 boy: $\frac12$;
- probability of having a family with 1 girl and 1 boy: $\frac12 \cdot \frac12$;
- probability of having a family with 2 girls and 1 boy: $\left(\frac12\right)^3 \cdot \frac12$;
- ...
- probability of having a family with $n$ girls and 1 boy: $\left(\frac12\right)^n \cdot \frac12$.
As we could expect, we have $\sum_{n = 1}^\inf \left(\frac12\right)^n = 1$.
Suppose there are $f$ families in the whole country. Then:
- $f \cdot \frac12$ families have 1 boy;
- $f \cdot \left(\frac12\right)^2$ families have 1 girl, 1 boy;
- ...
- $f \cdot \left(\frac12\right)^n$ families have $n - 1$ girls, 1 boy.
Obviously, $\sum_{n = 1}^\inf f \cdot \left(\frac12\right)^n = f$.
Finally:
- total boys: $f$ (because all families have exactly a boy);
- total girls: $\sum_{n = 1}^\inf f \cdot \left(\frac12\right)^n \cdot (n - 1) = f$.
In conclusion, the ratio is still 50% : 50%.
This result makes perfect sense to me, but I'm wondering if I got everything right?
