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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?

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  • $\begingroup$ $\LaTeX$ Tip: use \Bigr( \frac 12 \Bigr) to get $\Bigr( \frac 12 \Bigr)$ $\endgroup$ – Mohammad Zuhair Khan Oct 11 '18 at 3:23
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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, $$P(X=2)=\dbinom42\left(\dfrac{1}{2}\right)^2\left(1-\dfrac{1}{2}\right)^{4-2}$$ $$=\dbinom42\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}\right)^2$$ $$=\dfrac38=0.375$$

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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 $\frac{6}{16}=\frac38$.

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  • $\begingroup$ Oh! So, this is bernoulli trials and we have 4 trials and we call $X$ be the number of white balls and so $P(X=2) = {4 \choose 2} (1/2)^2 (1/2)^2 = 6/16 = 3/8 $. Is this correct now? $\endgroup$ – Mikey Spivak Oct 11 '18 at 3:36
  • $\begingroup$ $\checkmark\ \!$ $\endgroup$ – David Oct 11 '18 at 3:41
  • $\begingroup$ I dont know why I was thinking on negtive binomial. I always get confused with the wording of this problems. $\endgroup$ – Mikey Spivak Oct 11 '18 at 3:41
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$P(2W|4) = \binom{4}{2}\cdot (\frac{1}{2})^4$

$ = 6\cdot \frac{1}{16} = \frac{3}{8}$

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