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 along with another ball of the same color. The process is repeated so that the probability that the third ball drawn is black is $1-k$. Find the value of $k$

Answer 1

There are four possibilities for this event

First Ball	Second Ball	Event
White	White	E1
White	Black	E2
Black	White	E3
Black	Black	E4

$P\left(E_1\right)=\frac{2}{4}\times \frac{1}{3}=\frac{2}{12}=\frac{1}{6}$

$P\left(E_2\right)=\frac{2}{4}\times \frac{2}{3}=\frac{4}{12}=\frac{1}{3}$

$P\left(E_3\right)=\frac{2}{4}\times \frac{2}{5}=\frac{4}{20}=\frac{1}{5}$

$P\left(E_4\right)=\frac{2}{4}\times \frac{3}{5}=\frac{6}{20}=\frac{3}{10}$

Let $E$ denotes the event of drawing a black ball in the third attempt

$P\left(\frac{E}{E_1}\right)=1$

Because there is no white ball left to be selected

$P\left(\frac{E}{E_2}\right)=\frac{3}{4}$

Because there are $3$ black balls and $1$ white ball left

$P\left(\frac{E}{E_3}\right)=\frac{3}{4}$

Because again there are $3$ black balls and $1$ white ball left

$P\left(\frac{E}{E_4}\right)=\frac{4}{6}=\frac{2}{3}$

Because there are $4$ black balls and $2$ white balls left

$P\left(E\right)=P\left(E_1\right)P\left(\frac{E}{E_1}\right)+P\left(E_2\right)P\left(\frac{E}{E_2}\right)+P\left(E_3\right)P\left(\frac{E}{E_3}\right)+P\left(E_4\right)P\left(\frac{E}{E_4}\right)$

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

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

$=\frac{10+15+9+12}{60}$

$=\frac{46}{60}=\frac{23}{30}$

So,

$$\begin{gathered} 1-k=\frac{23}{30} \\ \Rightarrow k=1-\frac{23}{30} \\ \Rightarrow k=\frac{7}{30} \end{gathered}$$

Hence, the value of $k$ is $\frac{7}{30}$.