# Probability: Inclusive Exclusive Problem or combinatorics?

A box contains 30 red balls, 30 white balls, and 30 blue balls. If 10 balls are selected at random, without replacement, what is the probability that at least one color will be missing from the selection?

My approach: We are asked to calculate the probability in which at least one color will be missing. Therefore, we can instead calculate the probabilities in which all of the colors are the same and subtract from 1.

$\Pr(\text{all white balls})=\Pr(\text{all white balls})=\Pr(\text{all blue balls})$

since they all have the same number of balls.

I thought

$\Pr(\text{all white balls})=\cfrac{\binom{30}{10}}{\binom{90}{10}}$

So I thought solution is: $1-3 \cdot\cfrac{\binom{30}{10}}{\binom{90}{10}}$ which is wrong. Can someone please explain why my thought process is incorrect? Thank you in advance.

Your approach is almost correct; but if one color is missing then that means that all balls are the other TWO colors, so there will be $60 \choose 10$ ways for this to happen for each choice of missing color. Then you also have to handle inclusion-exclusion because the cases of one missing color overlap if all balls are of the same color. In the end, after you work it out you should get
$$\frac{3{60 \choose 10} - 3{30 \choose 10}}{90 \choose 10}$$