Find expected value of unique types given different numbers of items in each type

Question

I'm working on a practice problem for an exam and I haven't been able to find out a way to solve this. There are 4 bins that contain different amounts of different colored balls. I.e. 1 red, 1 green, 2 yellow, 6 blue. Draw 3 balls randomly without replacement. Let the r.v. Y represent the number of different types of balls in the selection. Find the expectation of Y. I've tried using indicators variables to represent the number of unique colors drawn, but I'm not able to get any farther. How would this be solved?

Answer 1

This is not an answer.

This is a long-winded comment.

First see Linearity of Expectation, which contains a proof that the formula applies even when the events are dependent.

Assume that $(a,b,c,d)$ represent the number of Color-1, ..., Color-4, respectively.

The probability of drawing $3$ balls that do not contain (for example) Color-1 is

$$Q_1 = \frac{\binom{b+c+d}{3}}{\binom{a+b+c+d}{3}}.$$

I could be mistaken, because my knowledge of Linearity of Expectation is somewhat superficial. To the best of my knowledge, the desired computation is

$$\sum_{i=1}^4 P_i ~: ~P_i = 1 - Q_i.$$

Alternatively:

$$4 - \sum_{i=1}^4 Q_i.$$