Non symmetric urn and balls problem.

Question

Consider an urn containing $N$ balls, of which m are white and $N - m$ are black. Balls are randomly selected from the urn according to the following rule:

• If a black balls is selected, it is "observed" and the it is returned to the urn.

• If a white balls is selected, it is "observed" and then it is replaced with a black ball.

Let $X$ be a random variable that counts the number of white balls in a sample of size $n$.

Determine the expected value of $X$.

To clariy the problem consider the case $N=8$, $m=3$ and $n=2$, then:

$P\{X=0\} = \frac{5}{8}\times\frac{5}{8}=\frac{25}{64}$

$P\{X=1\} = \frac{5}{8}\times\frac{3}{8} + \frac{3}{8}\times\frac{6}{8}=\frac{33}{64}$

$P\{X=2\} = \frac{3}{8}\times\frac{2}{8} = \frac{6}{64}$

It is easy to see that the total number of balls in the urn remains fixed, and that "equivalent" permutations of balls do not occur with the same probability, so the usual procedure of counting permutations with the desired property does not work.

Any help would be great.

Thanks in advance.

Edit: Fixed a typo.

Answer 1

Let the white balls be numbered $1$ to $m$. Don't bother to number the black balls. For $i=1$ to $m$, let $X_i=1$ if ball $i$ is selected at least once, and let $X_i=0$ otherwise. Then the number of white balls recorded is $X_1+\cdots+X_m$.

The probability that ball $i$ is not selected $n$ times in a row is $\left(\frac{N-1}{N}\right)^n$. Thus $$\Pr(X_i=1)=1-\left(\frac{N-1}{N}\right)^n.$$ This is $E(X_i)$. By the linearity of expectation, the expected number of white balls recorded is $$m\left(1-\left(\frac{N-1}{N}\right)^n\right).$$

Remark: Trying to get at the distribution of the number of white balls recorded is likely to be a painful process. However, as in many other problems, the linearity of expectation makes the expectation quite accessible.