An urn contains n red balls, n white, n black. What is the probability of not getting all colors?

Question

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?

Answer 1

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:$$P(B\cup R\cup G)=3P(B)-3P(B\cap R)=3\left[\frac{\binom{2n}{k}-\binom{n}{k}}{\binom{3n}{k}}\right]$$

Answer 2

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:

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

Answer 3

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.