Question

An urn contains $2$ white and $2$ black balls. A ball is drawn at random. If it is white, it is not replaced into the urn. Otherwise, it is replaced with another ball of the same colour. The process is repeated. Find the probability that the third ball drawn is black.

Answer 1

Here, there are four possibilities,

First Ball	Second Ball	Event
White	White	$E_1$
White	Black	$E_2$
Black	White	$E_3$
Black	Black	$E_4$

Let $E$ denote the event of drawing a black ball in the third attempt
$P\left(E|E_1\right)=1$; because there is no white ball to be selected.
$P\left(E|E_2\right)=\frac{3}{4}$, because there are $3$ black balls and $1$ white ball
$P\left(E|E_3\right)=\frac{3}{4}$ because again there are $3$ black and $1$ white ball
$P\left(E|E_4\right)=\frac{4}{6}=2/3$, because there are $4$ black and $2$ white balls

The four events $E_1,E_2,E_3,$ and $E_4$are mutually exclusive and exhaustive. Using the the theorem of total probability,

$\therefore P\left(E\right)=P\left(E_1\right)P\left(E|E_1\right)+P\left(E_2\right)P\left(E|E_2\right)+P\left(E_3\right)P\left(E|E_3\right)+P\left(E_4\right)P\left(E|E_4\right)$

$=\frac{1}{6}\times 1+\frac{1}{3}\times \frac{3}{4}+\frac{1}{5}\times \frac{3}{4}+\frac{3}{10}\times \frac{2}{3}$

$=\frac{1}{6}+\frac{1}{4}+\frac{3}{20}+\frac{1}{5}$

Ans. $=\frac{23}{30}$.