Expected value for the number of tries to draw the black ball from the bag We have a bag with $4$ white balls and $1$ black ball. We are drawing balls without replacement. Find expected value for the number of tries to  draw the black ball from the bag. 

Progress. The probability to draw a black ball from first trial is $1/5$. The problem is how to find the probability to draw black ball from $2$nd, $3$rd, $ \ldots, 5$th trial. When I know all this probabilities I can find expected value as $1\cdot(1/5) + 2 p_2 + \dots + 5 p_5$.

It is as if you will create a word with $4$ W's and $1$ B. For example $BWWWW$ or $WWWBW$ etc. How many such words can you create? Answer: $5$ and any such word is equally likely. 
In other words: the probability that the black ball will be drawn at any place - not only the first - is equal to $1/5$. Not conditional probability, but probability. Do not get confused, that if you have drawn $4$ White balls then the probability of drawing the black ball in the fifth draw is $1$. This is the conditional probability. "A priori" it is equally likely that the black ball will be drawn at any given point from $1$ to $5$. So, $$E[X]=\frac{1}{5}\cdot 1+ \frac{1}{5}\cdot2+\ldots+\frac15\cdot 5=\frac15(1+2+3+4+5)=3 $$ (where $X$ denotes the number of trials).
