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 Urn is selected at random and a ball is drawn . It turns out of to be white. Find the probability that it come from Urn-III.

Probability of selecting $Urn-I, Urn-II, Urn-III$ are

$P\left(E_1\right)=P\left(E2\right)=P\left(E_3\right)=\frac{1}{3}$

Having choosen that urn, the Probability of drawn a white ball  $P\left(\frac{W}{E_1}\right)=\frac{1}{3}$, similarly

 $P\left(\frac{w}{E_2}\right)=\frac{2}{3},P\left(\frac{w}{E_3}\right)=\frac{2}{4}$.

We have to find the probability, $P\left(\frac{E_3}{w}\right)$

by baye’s theorem 

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

$=\frac{\frac{1}{3}\cdot \frac{2}{4}}{\frac{1}{3}\cdot \frac{1}{3}+\frac{1}{3}\cdot \frac{2}{3}+\frac{1}{3}\cdot \frac{2}{4}}=\frac{1}{3}$