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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.

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

1 Answer 1

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Yes you are correct. The second term,

$ \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.

So instead of addition of the first term, it should have been subtracted. The answer should be,

$ \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}} $

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