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 $b$?

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    $\begingroup$ No, $Y$ is geometric. It is stated in the text. $\endgroup$ – Jimmy R. Oct 3 '16 at 8:07
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    $\begingroup$ Right. I missed it. $\endgroup$ – roang Oct 3 '16 at 8:11
  • $\begingroup$ (Concerning without replacement) Actually you used the legitimate equation $\mu=\sum_{n=1}^{\infty}\Pr(X\geq n)$ to find the mean. It is correct. $\endgroup$ – drhab Oct 3 '16 at 9:33
  • $\begingroup$ I deleted my answer, because it is not providing a closed formula for mean and variance. $\endgroup$ – drhab Oct 3 '16 at 9:39

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