mjhjmtu

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 $p = frac36 = frac12$. So we see $X$ is negative binomial r.v with $n=4$ trials. So,

$$ P(X=2) = 4 - 1 choose 2 - 1 left( frac12 right)^2 left( frac12 right)^2 $$

Which gives

$$ P(X=2) = boxeddfrac316 $$

Am I interpreting the problem correctly?

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 $p =dfrac12$. 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 $P(X=2)$

$$P(X=k)=dbinomnkp^k(1-p)^n-k$$

Then,
$$P(X=2)=dbinom42left(dfrac12right)^2left(1-dfrac12right)^4-2$$
$$=dbinom42left(dfrac12right)^2left(dfrac12right)^2$$
$$=dfrac38=0.375$$

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 $frac616=frac38$.

$P(2W|4) = binom42cdot (frac12)^4$

$ = 6cdot frac116 = frac38$