Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A box contains 3 red balls, 4 blue balls, and 6 green balls. Balls are drawn one-by-one without replacement until all the red balls are drawn. Let $D$ be the number of draws made.


a)$P(D\le 9)$



For part a) I know the answer should be $\binom{10}{6}/\binom{13}{9}$ but not sure how to get this.

And I am wondering the relationship between (a) and (b). And how to calculate hypergeometric distribution in general.

For part c) I could use the basic definition to get the answer but I am sure there is an easier way to do this.

share|cite|improve this question
up vote 2 down vote accepted

You can look at red and not red.

For (a), you want the three red balls to be in the first nine positions. I would have thought there were $\binom{9}{3}$ ways of choosing these positions which you divide by the total number of ways of positioning the red balls $\binom{13}{3}$. This gives the same numerical answer as you have, suggesting there is more than one approach.

For (b), you can either look at $\Pr(D \le 9) - \Pr(D \le 8)$, or look at two red balls in the first eight positions and one in the ninth.

For (c), following the definition might be easiest to understand and calculate electronically. An alternative approach for (c) would be to say that the expected number of balls before the first red, strictly between the first and second red, strictly between the second and third red, and after the third red are each the same and so are each $\frac{6+4}{4}=2.5$, and so taking three of these plus the three red balls gives $E[D]=3\times 2.5+3$.

share|cite|improve this answer
(c) is correct, but is there an easy way to see that those four expectations are the same? – Michael Lugo Nov 2 '11 at 14:36
@Michael: not only are the four expectations the same, but the distributions are too. The classic explanation is that cutting a continuous or equally-spaced-discrete circle in $k$ random places gives $k$ pieces with identical (but not independent) distributions; the same thing happens with a line and $k-1$ random cuts into $k$ pieces (just go back to the circle and make the first of $k$ cuts to create a line with end points). – Henry Nov 2 '11 at 17:08
For a), you can also think of picking the six remaining non-red balls out of the ten balls that aren't red. – user5137 Nov 2 '11 at 17:33

For (c), number the non-red balls $1, 2, \ldots, 10$; let $I_k$ be the indicator of the event that the non-red ball numbered $k$ is drawn before the last red ball to be drawn. Then $D = (I_1 + I_2 + \cdots + I_{10}) + 3$.

Then $E(I_k)$ is the probability that the non-red ball $k$ is not last among itself and the three red balls; it's last with probability $1/4$ and therefore not last with probability $3/4$.

share|cite|improve this answer
why these indicator r.v.s have the same probability? – geraldgreen Nov 3 '11 at 18:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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