Get the probability, with Pascal's triangle.

Question

I have an statement that says:

If a marriage has 5 children,

That they be more women than men??

Making the pascal triangle, and adding the variables

$m = men, w = women$

$1m^5$ $5m^4w$ $10m^3w^2$ $10m^2w^3$ $5mw^4$ $1w^5$

The probability of men is $\frac{31}{32}$ and same with women.

But how does that help me get, the probability that they are more women, than men?

The correct answer should be $\frac{1}{2}$, but i cant understand how to solve it.

Answer 1

If we neglect the possibility that a child is intersex, there are two possible outcomes for the sex of each child. Hence, there are $2^5$ possible sequences of the children's sexes in our sample space.

The favorable cases consist of those sequences in which the majority of children are female. Thus, we need to count sequences of length $5$ in which there are at least $3$ female children. There are $\binom{5}{k}$ sequences with exactly $k$ female children since we must select which $k$ of the $5$ positions (for first, second, third, fourth, and fifth child) are occupied by female children. Hence, the number of favorable cases is $$\binom{5}{3} + \binom{5}{4} + \binom{5}{5}$$ Dividing this by the $2^5$ possible sequences of the children's sexes gives the desired probability.

Answer 2

The interpretation is as follows.

There is one combination of 5 men. 5 men here.

There are 5 combinations of 4 men and 1 woman. +20 men. Thats 25 men and 5 women.

If you keep going like this, you will end up with 80 men and 80 women. $\frac{1}{2}$ probability of each.