probability balls in urn

Question

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?

Answer 1

For A: $$ \text{Probability}=\frac{\binom{7}{2}\binom{5}{1}\binom{3}{1}+\binom{7}{1}\binom{5}{2}\binom{3}{1}+\binom{7}{1}\binom{5}{1}\binom{3}{2}}{\binom{15}{4}}=\frac{6}{13}. $$

Answer 2

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. $\binom{7}{2}\binom{5}{1}\binom{3}{1}+\binom{7}{1}\binom{5}{2}\binom{3}{1}+\binom{7}{1}\binom{5}{1}\binom{3}{2}$.