A bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. One ball is drawn at random from a randomly chosen bag and is found to be red. The probability that it was drawn from B is

$\frac{25}{52}$

Let E₁ be the event that the ball is drawn from bag A, E₂ the event that it is drawn from bag B and E that the ball is red.

We have to find $P(\frac{E_2}{E}).$

Since both the bags are equally likely to be selected,

we have $P(E_1)= P(E_2)=\frac{1}{2}.$ Also $P(\frac{E}{E_1})=\frac{3}{5}$ and $P(\frac{E}{E_2})=\frac{5}{9}$.

Hence by Baye's theorem, we have $P(\frac{E_2}{E})=\frac{P(E_2)P(\frac{E}{E_2})}{P(E_1)P(\frac{E}{E_1})+P(E_2)P(\frac{E}{E_2})}=\frac{\frac{1}{2}.\frac{5}{9}}{\frac{1}{2}.\frac{3}{5}+\frac{1}{2}.\frac{5}{9}}=\frac{25}{52}.$