Question

Suppose in a society where there are equal numbers of men and women. There is a 50% chance for each child that a couple gives birth to is a girl and the genders of their children are mutually independent. Suppose in this strange and primitive society every couple prefers a girl and they will continue to have more children until they get a girl and once they have a girl they will stop having more children, what will eventually happen to the gender ratio of population in this society?

Answer 1

Answer:

eventually the gender ratio of population in this society will be 50% male and 50% female.

Step-by-step explanation:

For practical purposes we will think that every couple is healthy enough to give birth as much children needed until giving birth a girl.

As the problem states, "each couple continue to have more children until they get a girl and once they have a girl they will stop having more children". Then, every couple will have one and only one girl.

1. This girl would be the n-th child with a probability $(0.5)^n$.

We will denote for P(Bₙ) the probability of a couple to have exactly n boys.

Observe that statement 1 implies that:

$P(B_{n-1})=(0.5)^{n}$.

Then, the average number of boys per couple is given by

$\sum^{\infty}_{n=1}(n-1)P(B_{n-1})=\sum^{\infty}_{n=1}(n-1)(\frac{1}{2} )^n=\sum^{\infty}_{n=2}n(\frac{1}{2} )^n=\\\\=\sum^{\infty}_{m=2}\sum^{\infty}_{n=m}(\frac{1}{2} )^n=\sum^{\infty}_{m=2}(\frac{1}{2} )^{m-1}=\sum^{\infty}_{m=1}(\frac{1}{2} )^{m}=1.$

This means that in average every couple has a boy and a girl. Then eventually the gender ratio of population in this society will be 50% male and 50% female.