In a city, the distribution of X = # of children in each family is $$\boxed{\begin{array}{c|r:r:r} x & 0& 1& 2& 3\\ \hline P(x)& 0.1 & 0.4 & 0.3 & 0.2 \end{array}}$$
Let G be the # of girls in each family. We will suppose that all the children are equally likely to be girls or boys.
1)What is the probability that G = 2 conditional on X = 3?
So I'm trying to figure this out and from my understanding the conditional probability should be denoted as $P(G = 2 | X = 3)$ which should equal $\frac{P(G =2, X = 3)}{P(X=3)}$. I'm just having trouble trying to find the top probability, P(G =2, X = 3). If having a girl is equally as likely having a boy shouldn't this probably just be 1/2? But when I use $\frac{1/2}{0.2}$ I get a whole number, which can't be correct.
2)Calculate the marginal probability that G = 2.
Why isn't this just 1/2 if it's equally as likely as having a boy?
3)Find the marginal probability distribution function for G.
4)Calculate E(G).
I know I just have to find $P_G(X) * X$ for each X.
"Conditional on $X=3$" means if the family has three children, what is the probability there are two girls?
Clearly this is a Binomial distribution so the answer to the first part is $$\binom32(\frac 12)^3=\frac 38$$
For the second part, you can calculate the probability that a two-child family have two girls, which is, by similar argument, $\frac 14$
Therefore the total probability $G=2$ is given by $$\frac 38\times0.2+\frac 14\times 0.3$$
Now you can work out, in a similar way, the respective probabilities $G=0, G=1$ and $G=3$ and obtain the probability distribution of $G$ and thus derive the expected value.
I hope this helps.