What is the probability that in a family of 5 children, girls outnumber boys. (Assuming probability of a boy is ${\frac{1}{2}}$) How I look at this is that the total number of outcomes is $5!$ and in the given event, either there are $3,4$ or $5$ girls in the family. Now I should calculate the total number of possible outcomes where there are 3 girls and 4 girls and 5 girls and hence find the probability. But is this approach correct? Also is there any shorter approach to this question?

The total number of families is $2^5 =32$: from $GGGGG$ to $BBBBB$, we have an sequence of $5$ children and for each child we can have a boy or a girl. All of these sequences are equally likely.

How many sequences have no boys, 1 boy or 2 boys?

* no boys: just one ($GGGGG$)
* one boy: just $5$: the one boy can be the first, second, etc. till the fifth.
* two boys: just $10$: the two boys an be at $\binom{5}{2}$ places: we choose the two positions out of the $5$ without order and no replacement.



So in $16 = 1+5+10$ of the outcomes, girls outnumber boys. So the probability is $\frac{1}{2}$. This makes sense in hindsight: there can be no "draw": either there are more boys or more girls. And by symmetry, these must have the same probability.