**
10
**
distinct balls. A person draws a set of
**
4
**
balls from the bag and puts them back. The person draws another set of
**
6
**
balls from the bag and puts them back again. What is the probability of drawing exactly
**
3
**
balls that was common to both the sets?

2/735
8/21
1/8
3/10

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For the sake of argument, suppose all the balls are initially colored white, and when you draw the 4 balls initially, you color them black. Now we can rephrase the question as follows:

Well, for this to happen, you must have drawn 3 white balls ( $\binom{6}{3}$ ways) and 3 black balls ( $\binom{4}{3}$ ways) for a total of $\binom{6}{3} \cdot \binom{4}{3} = 80$ ways to draw exactly 3 black balls. Since there are 10 balls total, there are $\binom{10}{6} = 210$ total ways to draw 6 balls.

Then the probability that you draw exactly 3 black balls is $\frac{80}{210} = \boxed{\frac{8}{21}}$