Bag A contains 2 red and 3 white balls, Bag B contains 3 red and 2 white balls. If a ball is drawn at random and is found to be red, then the probability that it was drawn from bag B, is :

$\frac{3}{5}$

Let E₁ be the event that ball is drawn from bag A, E₂ be the event that it is drawn from bag B and E that the ball is red. We have to find P(E₂/E).

Since both bags are equally likely to be selected, we have

$P(E_1)=P(E_2)=\frac{1}{2}$

Also $P(E/E_1)=\frac{2}{5}$ and $P(E/E_2)=\frac{3}{5}$

By Bayes theorem, we have

$P(E_2/E)=\frac{P(E_2).P(E/E_2)}{P(E_1)P(E/E_1)+P(E_2).P(E/E_2)}=\frac{\frac{1}{2}×\frac{3}{5}}{\frac{1}{2}×\frac{2}{5}+\frac{1}{2}×\frac{3}{5}}=\frac{\frac{3}{10}}{\frac{1}{5}+\frac{3}{10}}$

$⇒\frac{\frac{3}{10}}{\frac{2+3}{10}}=\frac{\frac{3}{10}}{\frac{5}{10}}⇒\frac{3}{10}×\frac{10}{5}=\frac{3}{5}$