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?
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.