Q.31 - A First Course in Probability

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 $t$ ]

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

Step-by-Step Solution

Answer

We prove that,

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

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

Step 1: 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. 

Step 2: 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

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

Step 3: Final answer

We prove that, 

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

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