# Black and white balls-probability

Problem : There are "k" white and "m" black balls in the box. We draw the balls without replacement. What is the probability, that durning the n-th draw ball will be white. I don't have idea how to start.

• Whenever I have a problem that I"m not sure how to solve or start, I use actual numbers for my variables and see what happens when I do some basic calculations Usually a pattern of sorts emerges. Like, I would start with 5 white balls and 4 black balls. Then look at the probability that the 6th ball drawn w/o replacement is white. But then consider the 2nd ball, 3rd ball, etc.... You should always be playing around with numbers... – Eleven-Eleven Oct 15 '18 at 17:17
• @Eleven-Eleven Thanks for the advice. – PabloZ392 Oct 15 '18 at 17:20

You can start to calculate that at the n-th draw the ball is white , where $$n=1,2,3$$. Suppose you have 3 white (w) balls and $$2$$ black ($$b$$) balls. Let denote $$P(X_n=w)$$ the probability that at the n-th draw the ball is white. Then we have

$$P(X_1=w)=\frac{3}{3+2}=\frac35$$

For $$P(X_2=w)$$ there are two ways: a) $$ww$$ and b) $$bw$$. The probabilities are

a) $$\frac35\cdot \frac{3-1}{5-1}=\frac{3\cdot 2}{5\cdot 4}=\frac{6}{20}$$

b) $$\frac25\cdot \frac{3}{5-1}=\frac{2\cdot 3}{5\cdot 4}=\frac{6}{20}$$

Therefore $$P(X_2=w)=\frac{6}{20}+\frac{6}{20}=\frac{3}{5}$$

Equivalent calculations can be done for $$n=3$$

This could be a start to solve the problem.

You can look at the probability distribution function for the geometric distribution. This looks at the distribution to get the first "success" (which in your case is pulling a white ball. Let $$X$$ be a random variable such that: $$X \sim Geo\big(k ,\, p(k)\big)$$ $$\Pr(X=k)=(1-p)^{k-1}p$$

Edit: As @callculus pointed out, this is how you would solve it with replacement. Refer to their answer for a solution without replacement.

• The drawing is without replacement. – callculus Oct 17 '18 at 13:54