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?

  • 1
    $\begingroup$ 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

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 $$\frac{\binom{5}{3}\binom{3}{2}\binom{6}{1}}{\binom{5}{3}\binom{9}{3}}.$$

  • $\begingroup$ Thank you, that helped. Does this follow a binomial distribution? $\endgroup$ – Alti Mar 27 '13 at 5:16
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.