Suppose that 15 percent of the families in a certain community have no children

Suppose that 15 percent of the families in a certain community have no children, 20 percent have 1, 35 percent have 2, and 30 percent have 3 children; suppose further that each child is equally likely (and independently) to be a boy or a girl. If a family is chosen at random from this community, then B, the number of boys, and G, the number of girls, determine the conditional probability mass function of the size of a randomly chosen family containing 2 girls.

There are exactly THREE ways this can happen:

1) family has EXACTLY 2 girls
2) family has 2 girls and 1 boy
3) family has all 3 girls

The first one is pretty simple. Given that you are going to "select" exactly two children, find the probability that they are BOTH girls (it's a coin flip, so p = 50% = 0.5):

0.5² = 0.25

So the probability that the family has EXACTLY 2 girls is the probability that the family has exactly two children times the probability that those two children will be girls:

35% / 4 = 8.75%

Now find the probability that, given the family has exactly 3 children, that EXACTLY two are girls. Now you flip 3 times but only need to "win" twice--this is a binomial experiment.

There are 3 choose 2 = 3 ways to have exactly two girls: 1st, 2nd, or 3rd is a boy...interestingly the probability of having ANY PARTICULAR permutation is just 0.5³ = 1/8 (because it's still 0.5 * 0.5 for two girls, then 0.5 for 1 boy).

So the chance of EXACTLY 2 girls is: 3/8

Now find the probability for having EXACTLY 3 girls...that's easy, there's only one way, you just have all 3 girls, probability is just 1/8
--> add these up

3/8 + 1/8 = 4/8 = ½
-->

So now use the percent of families with exactly 3 children to find this portion of the probability:

30% / 2 = 15%
--> add the two probabilities...here is it in full detail

p(contains 2 girls) = p(2 children) * p(2 girls, 2 children) + p(3 children) * p(2 or 3 girls, 3 children)
-->

35% /4 + 30% * (3/8 + 1/8)
-->

8.75% + 15% = 23.75%

Edit:

You can reason that the probability that a family of three has at least 2 girls SHOULD be 50% since, it SHOULD be JUST as likely as having at least two boys...since these cannot BOTH happen (you cannot BOTH have 2 girls AND 2 boys if you only have 3 children)...then their respective probabilities should add to give 1 (i.e. one or the other ALWAYS happens).

Since they are equally likely it's just 100%/2 = 50% = 1/2