Conditional probability problem with urn An urn contains 3 white, 6 red, and 5 black balls. Six of these balls are randomly selected from the urn. Let $X$= number of white balls selected and $Y$= number of black balls selected. Compute the conditional probability mass function of $X$ given that $Y=3$. 
For the conditional probability  mass function, I need to find $\frac{P(X=x|Y=3)}{P(Y=3)}$. 
I tried breaking it up and thinking of each piece individually. 
For $P(Y=3)$: I have 14 balls total in the urn, with 5 of the 14 black. Now if I choose 6 balls at random, what's the probability that 3 are black? 
But I don't know where to go from here. Is there a simple way to think about these types of problems?
 A: For the conditional mass function, you need to calculate $\Pr(X=x|Y=3)$. This is not quite what you wrote. To do the calculation, note that
$$\Pr(X=x|Y=3)=\frac{\Pr((X=x)\cap (Y=3))}{\Pr(Y=3)}.$$ 
First we calculate $\Pr(Y=3)$. There are $14$ balls, of which $5$ are black. There are $\dbinom{14}{6}$ ways to choose $6$ balls. The number of ways to choose $3$ black and $3$ non-black is $\dbinom{5}{3}\dbinom{9}{3}$. For $\Pr(Y=3)$, divide. 
Now we need to calculate $\Pr((X=x)\cap (Y=3))$ for the possible values of $x$, which range from $0$ to $3$. So there are four different calculations to do. We will deal with $x=2$, and you can do the others. 
So we want $\Pr((X=2)\cap (Y=3))$. Thus we want the probability of $3$ black, $2$ white, and therefore $1$ red. The number of ways to choose $6$ balls is, as before, $\dbinom{14}{6}$. The number of ways to choose $3$ black, $2$ white, and $1$ red is $\dbinom{5}{3}\dbinom{3}{2}\dbinom{6}{1}$. Divide.
When we put things together, the denominators $\dbinom{14}{6}$ cancel, and therefore we do not need to compute them.  So the conditional probability is 
$$\frac{\binom{5}{3}\binom{3}{2}\binom{6}{1}}{\binom{5}{3}\binom{9}{3}}.$$
