Question

An urn initially contains $b$black and $w$white balls. At each stage, we add $r$black balls and then withdraw, at random,$r$balls from the $b+w+r$balls in the urn. Show that

$F$[number of white halls after stage $t$]

$={\left(\frac{b+w}{b+w+r}\right)}^{t}w$

Short answer

We prove that,

$E[$number of white balls after stage $t]$

$={\left(\frac{b+w}{b+w+r}\right)}^{t}w$

Step-by-step solution

Given information

Given in the question that, An urn initially contains $b$black and $w$white balls. At each stage, we add $r$black balls and then withdraw, at random, $r$balls from the $b+w+r$ balls in the urn.

Explanation

Characterize random variable $N_t$as the counter of white balls after stage $t$. We should work out the distribution of $N_t$given $N_{t-1}$. We realize that there exist $N_{t-1}$white balls in the urn and we draw (without substitution) $r$balls. Consequently, the quantity of white balls drawn at stage $t$given $N_{t-1}$has Hypergeometric distribution with the complete number of components in urn$b+w+r$, we have $N_{t-1}$fruitful components in urn and we draw $r$components. Hence, we end up with $N_{t-1}-$drawn white balls after stage $t$. Involving the equation for the mean of Hypergeometric distribution, we have that

$E\left(N_t\mid N_{t-1}\right)=\frac{b+w}{b+w+r}\cdot N_{t-1}$

Use the relation to obtain that

$E\left(N_t\right)=\frac{b+w}{b+w+r}E\left(N_{t-1}\right)$

Using the mathematical induction, we have that

$E\left(N_t\right)={\left(\frac{b+w}{b+w+r}\right)}^{t}E\left(N_0\right)$

Now, observe that $E\left(N_0\right)$is the number of white balls at the beginning, which is equal to $w$. Hence

$E\left(N_t\right)={\left(\frac{b+w}{b+w+r}\right)}^{t}w$

Final answer

We prove that,

$E[$number of white balls after stage $t]$

$={\left(\frac{b+w}{b+w+r}\right)}^{t}w$