Probability that exactly 2 balls are white 
What I did.
I let $X$ be number of withdraws before $x$ white balls. We can call our succes in this case to be gettin a white ball and probability is of course $p = \frac{3}{6} = \frac{1}{2}$. So we see $X$ is negative binomial r.v with $n=4$ trials. So, 
$$ P(X=2) = { 4 - 1 \choose 2 - 1} \left( \frac{1}{2} \right)^2 \left( \frac{1}{2} \right)^2 $$
Which gives 
$$ P(X=2) = \boxed{\dfrac{3}{16} }$$
Am I interpreting the problem correctly?

$P(2W|4) = \binom{4}{2}\cdot (\frac{1}{2})^4$
$ = 6\cdot \frac{1}{16} = \frac{3}{8}$
