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I have this problem which I have been struggling with for a while now

An urn contains n red balls, n white balls and n black balls. You draw k balls at random without replacement ($k\leqq n$). Find an expression for the probability that you do not get all colors.

I tried to solve this in the following way: I note that it should be logical to think P(not getting all colours in k draws) = P(getting exactly one color only in k draws OR get exactly two different colors only in k draws).

Therefore, I choose the events $A_1=\{$get one color k times$\}$ and $A_2=\{$get two colors k times$\}$

Therefore, we seek $P(A_1 \cup A_2)$.

Clearly, $A_1 = \frac{{n\choose k}}{3n\choose k}$ and $A_2 = \frac{{2n\choose k}}{3n\choose k}$. And because the both events are disjoint, we simply get

$P(A_1 \cup A_2) = P(A_1) + P(A_2) = \frac{1}{3n\choose k}\big({n\choose k}+{2n\choose k}\big)$.

According to my textbook, however, the answer should be

$\frac{3}{3n\choose k}\big({2n\choose k}-{n\choose k}\big)$.

What am I doing wrong? Are the answers equivalent because of some mystical binomial-identity?

jvdhooft
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tarkovsky123
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3 Answers3

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You have to multiply by 3, because there are three ways to select the colors. Furthermore, when considering all ways to pick two colors, you also include the scenario where you select only one color. These must be subtracted, hence you get:

$$\frac{{3 \choose 2}\left({2n \choose k}-2{n \choose k}\right)+{3 \choose 1}{n \choose k}}{3n \choose k} = \frac{3({2n \choose k} - {n \choose k})}{3n \choose k}$$

jvdhooft
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  • Okey, thanks. The above comment states however that my $A_1$ and $A_2$ are not disjoint. How can they not be disjoint? – tarkovsky123 Apr 05 '18 at 17:26
  • @tarkovsky123 When considering ${2n \choose k}$ for $A_2$, you also consider picking $k$ balls of the same color. Therefore, $A_1$ and $A_2$ are not disjoint. – jvdhooft Apr 05 '18 at 17:30
  • $2n \choose k$ means you pick $k$ among the set containing $n$ balls of 1 color and n balls of another color. In this configuration it is possible that you actually pick $k$ balls of the same color which falls into your $n \choose k$ possibilities you already counted. – Frostic Apr 05 '18 at 17:46
  • Okey that makes sense. But if I was to try and compute $P(A_1 \cup A_2) = P(A_1) + P(A_2) + P(A_1 \cap A_2)$, how would I, with my events above, compute $P(A_1 \cap A_2)$ ? I see now that its obvious they are not disjoint, but how to explicity calculate the intersection of the events? – tarkovsky123 Apr 06 '18 at 09:00
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Let us take the colors blue, red and green and let $B,R,G$ stand for the events that no blue, red, green balls are selected respectively.

Then you are looking for $P(B\cup R\cup G)$ and with inclusion/exclusion and symmetry we find:$$P(B\cup R\cup G)=3P(B)-3P(B\cap R)=3\left[\frac{\binom{2n}{k}-\binom{n}{k}}{\binom{3n}{k}}\right]$$

drhab
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  • Thanks!! Amazing how easy this solution is. I have a very hard time in how to effectively choose the most appropriate events for this kind of problems. If I had just set up the same configuration of events that you just did, I would have solved it in an instant using inclusion/exclusion. – tarkovsky123 Apr 06 '18 at 08:55
  • Glad to help. Indeed an instant is enough. Choosing the right configuration is quite often a matter of intuiton and maturity. Both grow by practicizing maths and that is exactly what you are doing. – drhab Apr 06 '18 at 09:23
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The reader may want to consult this MSE link by way of enrichment where a much more general question was answered. Substituting the parameters from the present question value by value into the formula that was proved there yields for the complementary probability of seeing all three colors the form (in the following we have $n\ge 1$)

$${3n\choose k}^{-1} {3\choose 3} \sum_{p=0}^3 {3\choose p} (-1)^{3-p} {pn\choose k} \\ = {3n\choose k}^{-1} \times \left({3n\choose k} - 3{2n\choose k} + 3{n\choose k} - {0\choose k}\right).$$

This gives for $k=0$ (no balls drawn) the probability of seeing all three colors

$$1\times ( 1-3+3-1 ) = 0.$$

For $1\le k\le 3n$ we obtain

$${3n\choose k}^{-1} \times \left({3n\choose k} - 3{2n\choose k} + 3{n\choose k}\right).$$

The complementary probability then becomes

$$1 - {3n\choose k}^{-1} \times \left({3n\choose k} - 3{2n\choose k} + 3{n\choose k}\right) \\ = {3n\choose k}^{-1} \left( 3{2n\choose k} - 3{n\choose k} \right).$$

This confirms the two answers that were first to appear.

Marko Riedel
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