Question

An urn has n white and m black balls that are removed one at a time in a randomly chosen order. Find the expected number of instances in which a white ball is immediately followed by a black one.

Short answer

The expected number of instances in which a white ball is immediately followed by a black one.

$\text{The expected number of instances is}\frac{nm}{n+m}$

Step-by-step solution

Given Information

An urn has n white and m black balls that are removed one at a time in a randomly chosen order.

Explanation

Assume that an urn has n white and m black balls that are removed one at a time in a randomly chosen order. Let X represents the number of instances in which a white ball is immediately followed by a black one, and let $E_{j,}1\le j\le n+m-1$, denote the event:

$E_j="j\text{th ball is white,}(j+1)\text{th ball is black}"$

Then

$P\left\{E_j\right\}=$

$P\left\{j\text{th ball is white}\right\}P\left\{\right(j+1)\text{th ball is black}\mid j\text{th ball is white}\}=$

$\left(\frac{n}{n+m}\right)\left(\frac{m}{n+m-1}\right)$

Explanation

$\text{If we define variables}{I}_j\text{in the following way:}$

$I_j=\{\begin{array}{ll}1,& \text{if}{E}_j\text{occurs}\\ 0,& \text{if}{E}_j\text{does not occur}\end{array}$

then

$X=\sum _{j=1}^{n+m-1}{I}_j$

and therefore the expected number of instances in which a white ball is immediately followed by a black one is

$E\left[X\right]=E\left[\sum _{j=1}^{n+m-1}{I}_j\right]=\sum _{j=1}^{n+m-1}E\left[I_j\right]=\sum _{j=1}^{n+m-1}P\left\{E_j\right\}$

$=\sum _{j=1}^{n+m-1}\left(\frac{n}{n+m}\right)\left(\frac{m}{n+m-1}\right)$

$=(n+m-1)\left[\left(\frac{n}{n+m}\right)\left(\frac{m}{n+m-1}\right)\right]$

$=\frac{nm}{n+m}$

Step 4: Final Answer

The expected number of instances in which a white ball is immediately followed by a black one.

$=\frac{nm}{n+m}$