A sack contains $12$ white balls and $8$ black balls. The balls are taken out one by one randomly and without replacement. Let $X$ be the number of times a black ball was taken immediately after a white ball. Calculate $E(X)$.
Answer by book: $4.8$
I calculated the probability for taking a black ball after a white in two consecutive steps: $p=6:25$. It made sense to me that there are $19$ "tests" to the indicator (because the probability can't happen in the first ball) so by linearity $E(X)=19 \cdot 6:25=4.56$ which is wrong, but $20 \cdot 6:25$ gives the right answer.
But why are there $20$ steps? Thanks in advance.
Best Answer
There are $19$ steps, but your first calculation is wrong
Taking a particular pair, the probability it is white-black is $\frac{12 \times 8}{20 \times 19}=\frac{24}{95}$ not your $\frac{12 \times 8}{20 \times 20}=\frac{6}{25}$. Draws are without replacement
The expected number of such pairs is then $19\times \frac{24}{95}=\frac{24}{5}=4.8$