# 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?

• $\LaTeX$ Tip: use \Bigr( \frac 12 \Bigr) to get $\Bigr( \frac 12 \Bigr)$ – Mohammad Zuhair Khan Oct 11 '18 at 3:23

## 3 Answers

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, $$P(X=2)=\dbinom42\left(\dfrac{1}{2}\right)^2\left(1-\dfrac{1}{2}\right)^{4-2}$$ $$=\dbinom42\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}\right)^2$$ $$=\dfrac38=0.375$$

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

• 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? – Mikey Spivak Oct 11 '18 at 3:36
• $\checkmark\ \!$ – David Oct 11 '18 at 3:41
• I dont know why I was thinking on negtive binomial. I always get confused with the wording of this problems. – Mikey Spivak Oct 11 '18 at 3:41

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

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