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12 white and 8 black balls are placed in an urn. What is the expected time of waiting on taking out the black ball from the urn, if we are taking balls without replacement.

It should be solved by geometric distribution, which looks like this:

$P(X=k)=(1-p)^{k-1}\,p \; $ for number of trials up to and including the first success

But I am not really sure how to define the probability, because as we know, probability is still changing, because number of balls is decreasing as we are taking the balls from the urn.

With the probability we can then compute:$\,EX=\sum_{k=1}^{\infty}k\,P$

$EX=\sum_{k=1}^{\infty}k\,(\frac{12-(k+1)}{20-(k+1)})^{k-1}(\frac{8}{20-(k+1)})$

Can this equation be solution to the example?

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They are asking for the expectation without replacement, so you can expect a finite sum. The probability that the $i$th ball taken out is the first black ball is $$P_i=\frac{\binom {12}{i-1}\binom{8}{1}}{\binom {20}{i}}$$ Now just sum this up $$\sum _{i=1}^{12}iP_i$$ To obtain your expected value.

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    I have one question. Isn't the hypergeometric distribution used when you are taking more than one ball in one round. If yes, then I am not really sure about your solution. Maybe it is not obvious from my description, but we are taking only one ball each round so shouldn't be there geometric distribution? – 2017-01-19
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    If you would give me explanation I would be happy. – 2017-01-19
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    You can think of the first $i$ rounds as one whole round in which you take out $i$ balls, in which the white balls are the special set and the black balls the ordinary one. If you like, you can also think of it combinatorially: there are $\binom {20}{i}$ ways to choose the first $i$ balls to be drawn, and there are $\binom {12}{i-1}$ ways to choose the first $i-1$ balls from the white set and $8$ ways to choose the $i$th ball from the black set. – 2017-01-19