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.
My attempt
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):
$0.5^2 = 0.25$
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:
$\frac{1}{4} \cdot 35\% = 8.75\%$
Now find the probability that, given the family has exactly 3 children, that exacly 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 $0.5^3 = 1/8$ (because it's still $0.5 \times 0.5$ for two girls, then $0.5$ for one boy).
So the chance of exactly 2 girls is: $\frac{3}{8}$
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 $\frac{1}{8}$.
Now, add these up
$\frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$
So now use the percent of families with exactly 3 children to find this portion of the probability:
$\frac{1}{2} \cdot 30\% = 15\%$
Hence, add the two probabilities… here is it in full detail
$$\begin{eqnarray}\mathbb{P}(\text{contains 2 girls}) &=& \mathbb{P}(\text{2 children}) \times \mathbb{P}(\text{2 girls, 2 children}) + \\ &\phantom{+=}& \mathbb{P}(\text{3 children}) \times \mathbb{P}(\text{2 or 3 girls, 3 children}) \end{eqnarray}$$
$\frac{1}{4} 35\% + 30\% \times \left(\frac{3}{8} +\frac{ 1}{8}\right)$
$8.75\% + 15\% = 23.75\%$
Is my attempt correct?
Best Answer
It’s correct as far as it goes, but it’s incomplete. You’ve shown that $23.75$% of the families have at least two girls, but that doesn’t answer the question. What you’re to find is probability mass function of the family size given that the family has two girls. In other words, you want to calculate $$\Bbb P(B+G=x\mid G\ge 2)$$ for the various possible values of $x$.
This is very easy and obvious for $x=0$ and $x=1$, so I’ll skip to $x=2$.
You calculated that $8.75$% of all the families have exactly two girls and no boys. What fraction of the families with at least two girls is this? It’s $$\frac{8.75}{23.75}=\frac7{19}\;,$$ so the conditional probability that a randomly chosen family has exactly two children given that it has at least two girls is $7/19$: $\Bbb P(B+G=2\mid G\ge 2)=7/19$.
From here you should be able to finish it, I think.