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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?

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    $\begingroup$ I assume you can find $P(Y=3)$, and that the balls are selected without replacement. For $P(X = t | Y = 3)$, you need to find the probability that $t$ white balls, $3-t$ red balls and $3$ black balls are selected. Can you do this? $\endgroup$ – Macavity Mar 27 '13 at 4:44
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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 $$\frac{\binom{5}{3}\binom{3}{2}\binom{6}{1}}{\binom{5}{3}\binom{9}{3}}.$$

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  • $\begingroup$ Thank you, that helped. Does this follow a binomial distribution? $\endgroup$ – Alti Mar 27 '13 at 5:16
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    $\begingroup$ It is not binomial, since it is essentially sampling without replacement. More closely related to the hypergeometric. $\endgroup$ – André Nicolas Mar 27 '13 at 5:41

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