3 balls

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:

A bag has 6 white balls and 4 black balls. You draw 6 balls from the bag. What is the probability that exactly 3 of them are black?

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 $\frac{80}{210} = \boxed{\frac{8}{21}}$

Note that the bag contains $\large 10$ distinct balls.

We start out by selecting $\large 3$ balls from the bag of $\large 10$ balls that would be common to the set of $\large 4$ balls and the set of $\large 6$ balls that were drawn. The number of ways of selecting $\large 3$ balls from a bag of $\large 10$ balls is $\Large {10 \choose 3}$ . These $\large 3$ balls were common to both our sets, thus, we need to select $\large 3$ more balls from the balls that remain to make our set of $\large 6$ balls and after that $\large 1$ more ball from the remaining to make our set of $\large 4$ balls.

Selecting $\large 3$ balls from the remaining $\large 7$ balls that would constitute our set of $\large 6$ balls. The number of ways of doing this is $\Large {7 \choose 3}$ . And finally, selecting $\large 1$ more ball from the remaining $\large 4$ balls which can be done in $\Large {4 \choose 1}$ ways.

The total ways to select $\large 6$ balls and $\large 4$ balls from the bag of $\large 10$ balls is $\Large {10 \choose 4} \times {10 \choose 6}$ .

Thus, the probability is : $\huge \frac { {10 \choose 3} \times {7 \choose 3} \times {4 \choose 1} } { {10 \choose 4} \times {10 \choose 6} }$

The correct answer to the problem is $\Large \frac{8} {21}$