Question

Three urns have the following composition of balls:
Urn I : $1$ white, $2$ black
Urn II : $2$ white, $1$ black
Urn III : $2$ white, $2$ black
One of the urns is selected at random and a ball is drawn. It turns out to be white. Find the probability that it came from urn III.

Answer 1

Let $P\left(E\right)$ be the probability of choosing $1$ white ball from Urn.

$P\left(E_i\right)=$ Probability of selecting one of three urns .

Therefore,

$P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}$

$P\left(E/E_1\right)=\frac{{}^1{C}_1}{{}^3{C}_1}=\frac{1}{3}$

$P\left(E/E_1\right)=\frac{{}^2{C}_1}{{}^3{C}_1}=\frac{2}{3}$

$P\left(E/E_1\right)=\frac{{}^2{C}_1}{{}^4{C}_1}=\frac{1}{2}$

Therefore,

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

$\Rightarrow P\left(E_3/E\right)=\frac{\frac{1}{3}\cdot \frac{1}{2}}{\frac{1}{3}\cdot \frac{1}{3}+\frac{1}{3}\cdot \frac{2}{3}+\frac{1}{3}\cdot \frac{1}{2}}=\frac{3}{2+4+3}=\frac{1}{3}$

Hence the probability that the ball came from urn III is $\frac{1}{3}$.