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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.

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  • $\begingroup$ 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... $\endgroup$ – Eleven-Eleven Oct 15 '18 at 17:17
  • $\begingroup$ @Eleven-Eleven Thanks for the advice. $\endgroup$ – PabloZ392 Oct 15 '18 at 17:20
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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.

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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.

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

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