I am trying problem no. 2 from Purdue HW here .
Balls are drawn from an urn containing $w$ white balls and $b$ black balls until a white ball appears. Find the mean value $m$ and variance $σ^2$ of the number of black balls drawn, assuming each ball is placed back into the urn after being drawn. (In other words, for each attempt we have $w$ white balls and $b$ black balls in the urn.)
I did not understand the part in the solution that says "the mean and variance of $Y$ are $\frac{1}{p}$ and $\frac{1}{p^2} − \frac{1}{p}$, respectively". $Y$ is a Bernoulli variable. Shouldn't the mean be $p$ and variance $p(1-p)$??
And if I had to solve the same problem without replacement of balls how would I solve it?
I tried solving it by taking an example like $3$ white and $3$ black balls. And I got a mean of $$\frac{3}{6} + \frac{3}{6}\cdot\frac{2}{5} + \frac{3}{6}\cdot\frac{2}{5}\cdot\frac{1}{4}= \frac{3}{4}$$ which seems correct to me. But how do I express the same in terms of variable of $w$ and $b$?