Discrete random variable, joint density question Discrete random variables $X$ and $Y$ have the joint density shown in the table below.
$ \ \ \ \ X \ 1 \ \ \ 2$
$Y$
$1 \ \ \ \ \ \ \frac{1}{10} \ \ \ \frac{2}{10}$
$2 \ \ \ \ \ \ \frac{3}{10} \ \ \ \frac{4}{10}$
What is the conditional density of $Y$ given $X = 1$?
What is the procedure for doing this problem? I have an exam tomorrow and we never covered this in the lectures. I have just found it on an old exam paper which means it could potentially be on tomorrows exam.

Given $X=1$ we are looking at the first column. The "ratio"in that column is $1$ to $3$. 
More formally, we want $\Pr(Y=1|X=1)$ and $\Pr(Y=2|X=1)$. Note that $\Pr (X=1)=\frac{1}{10}+\frac{3}{10}=\frac{4}{10}$. 
So 
$$\Pr(Y=1|X=1)=\frac{\Pr(X=1 \cap Y=1)}{\Pr(X=1)}=\frac{1/10}{4/10}=\frac{1}{4}.$$
By a similar argument, $\Pr(Y=2|X=1)=\dfrac{3}{4}$.
Basically, we confine attention to the column $Y=1$, and scale up the probabilities in that column so they add to $1$.
