Three boxes numbered, $I, II, III$  contain the balls as follows :

 

 

 

 

 

One box is randomly selected and a ball is
deawn from it. If the ball is red, then find the
probability that it is from Box-$II$.

Let$B_1,B_2,B_3$be the events of selecting $I, II, III$ respectively

$\text{ i.e. }P\left(B_1\right)=P\left(B_2\right)=P\left(B_3\right)=\frac{1}{3}$

Probability of selecting Red ball from frist box

$P\left(R/B_1\right)=\frac{3}{6}=\frac{1}{2}$

Probability of selecting Red ball from 2nd box

$P\left(R/B_2\right)=\frac{1}{4}$

Probability of selecting Red ball from 3rd box

$P\left(R/B_3\right)=\frac{3}{12}=\frac{1}{4}$

Assuming that ball is Red, Probability it is from box $II$

$P\left(B_2/R\right)=\frac{P\left(B_2\right)P\left(R/B_2\right)}{P\left(B_1\right)P\left(R/B_1\right)+P\left(B_2\right)P\left(R/B_2\right)+P\left(B_3\right)P\left(R/B_3\right)}$

$=\frac{\frac{1}{3}\times \frac{1}{4}}{\frac{1}{3}\cdot \frac{1}{2}+\frac{1}{3}\cdot \frac{1}{4}+\frac{1}{3}\cdot \frac{1}{4}}=\frac{\frac{1}{3}\times \frac{1}{4}}{\frac{1}{3}\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\right]}$

$=\frac{\frac{1}{4}}{\frac{1}{2}+\frac{1}{2}}=\frac{\frac{1}{4}}{1}=\frac{1}{4}$  $\therefore P\left(B_2/R\right)=\frac{1}{4}$