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?
If for a random chosen family with $5$ children $G$ denotes the number of girls and $B$ denotes the number of boys, then - if the probability on a boy equals the probability of a girl - $(G,B)$ and $(B,G)$ have the same distribution.

This implies that: $$P(G&gt;B)=P(B&gt;G)$$

Next to that we have: $$P(G&gt;B)+P(B&gt;G)=1$$

Now draw conclusions.
You have to create a tree diagram to get all possible sample space. then see how many space have girls outnumbered. I can't draw it here.
There are 32 possible outcomes. And 16 outcomes have no of girls 3 or more. 
So probability of outnumbered girls is 16/32 = 1/2

The possible outcomes are like these
BBBBB, BBBBG, BBBGB, BBBGG, BBGBB,BBGBG,BBGGB,BBGGG, ---------------GGGGG