# probability balls in urn

In a jar, there are 7 red balls, 5 yellow balls, and 3 green balls. Four balls are drawn randomly without replacement.

A. What is the probability that the balls drawn are of three different colors?

B. What is the expected number of different colors among the balls drawn?

C. What is the variance of the number of different colors among the balls drawn?

here is what I tried for A.

Probability of a red ball on the first draw: 7/15

Probability of a yellow ball on the second draw: 5/14

Probability of a green ball on the third draw: 3/13

Total Probability: 7/15 * 5/14 * 3/13 = 1/26

I'm not sure if I'm missing something, also I have no idea how to answer B and C, any suggestions?

• On (A) you might see "yellow, green, red" or some other order Commented May 9, 2023 at 10:13
• For (B) and (C) it would also be worth working out the probabilities for just 1 colour appearing on all 3 balls, and for 2 colours appearing. Then you have a distribution on $\{1,2,3\}$ (the probabilities should add up to $1$) and so you can find its mean and variance Commented May 9, 2023 at 10:20
• You only accounted for three of the four draws. It is easier to solve the problem if you do not consider the order of selection, as in Vadim Chernetsov's answer. Commented May 9, 2023 at 10:28

The number in (A) is the sum of the coefficients of $$x^2yz, xy^2z, xyz^2$$ in
$$(1+x)^7(1+y)^5(1+z)^3$$
i.e. $$\binom{7}{2}\binom{5}{1}\binom{3}{1}+\binom{7}{1}\binom{5}{2}\binom{3}{1}+\binom{7}{1}\binom{5}{1}\binom{3}{2}$$.
For A: $$\text{Probability}=\frac{\binom{7}{2}\binom{5}{1}\binom{3}{1}+\binom{7}{1}\binom{5}{2}\binom{3}{1}+\binom{7}{1}\binom{5}{1}\binom{3}{2}}{\binom{15}{4}}=\frac{6}{13}.$$