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?
 A: First, note that we have a finite number of trials, $n = 4 ($although the game goes on forever, we are only concerned with the first $4$ balls.$)$ Each trial is a Bernoulli trial - that is, each trial has only one of two outcomes: white and not white. Define a success as the event that a white ball is drawn. Then the probability of success $p$ is $p =\dfrac{1}{2}$. Since each ball is replaced after it is drawn, we have sampling with replacement, and thus independence.
Since we are dealing with a finite number of independent Bernoulli trials with a constant probability of success $p$, we use the binomial distribution
Let $X$ be the number of white balls (successes) that appear in $n = 4$ trials. Then we want to find $P(X=2)$
$$P(X=k)=\dbinom{n}{k}p^k(1-p)^{n-k}$$
Then,
$$P(X=2)=\dbinom42\left(\dfrac{1}{2}\right)^2\left(1-\dfrac{1}{2}\right)^{4-2}$$
$$=\dbinom42\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}\right)^2$$
$$=\dfrac38=0.375$$
A: As far as I understand your solution, you are computing the probability that it takes $4$ draws to get $2$ white.  This is not what the question asked.
You should simply be using the binomial distribution, and the answer is $\frac{6}{16}=\frac38$.
A: $P(2W|4) = \binom{4}{2}\cdot (\frac{1}{2})^4$
$ = 6\cdot \frac{1}{16} = \frac{3}{8}$
