There is a box containing $$20$$ green marbles, $$20$$ blue marbles, and $$20$$ purple marbles. You draw $$10$$ marbles at random without replacement. What is the probability that you do not get all the colors?

The solution in the book:

$$\Large 3\frac{\binom{20}{10} \binom{40}{0}}{\binom{60}{10}} + 3\frac{\binom{40}{10} \binom{20}{0}}{\binom{60}{10}}$$

I believe the book seperated into cases.

Case 1: All the marbles are exactly $$1$$ color.

Case 2: All the marbles are exactly $$2$$ colors.

I feel like the solution is wrong because there is overcounting in the second case. If we lump together $$2$$ colors , such as green and blue marbles, that gives us $$40$$ marbles and choose $$10$$. However this also includes cases such as all green, since we could draw all $$10$$ green.

• For what it's worth, I completely agree with you. Commented Sep 30, 2021 at 13:36
• I should mention, the solution was actually given by an instructor, not a book.
– john
Commented Sep 30, 2021 at 14:59

$$\displaystyle \small 3 \cdot \binom{40}{10} \binom{20}{0} ~$$ counts all outcomes where we pick marbles of two colors and also counts outcomes with marbles of single color twice. Take $$40$$ green and blue marbles for example. $$\binom{40}{10}$$ counts all outcomes where we have $$10$$ marbles of both colors or $$10$$ marbles of just green or blue color. We then also count Blue and Purple and Purple and Green as part of $$3 \cdot {40 \choose 10}$$. So we need to subtract $$3 \cdot {20 \choose 10}$$ so outcomes with marbles of single color get counted only once.
$$\displaystyle 3 \cdot \frac{\binom{40}{10} \binom{20}{0}}{\binom{60}{10}} - 3 \cdot \frac{\binom{20}{10} \binom{40}{0}}{\binom{60}{10}}$$