$A$ 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 

Let $E_1$be the event that the ball is drawn from bag $A,$$E_2$ the event that it is drawn from bag $B$ and $E$ that the ball is red. 

We have to find $P\left(E_2/E\right).$

Since both the bags are equally likely to be selected, 

We have $P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}$

Also $P\left(E/E_1\right)=3/5$ and $P\left(E/E_2\right)=5/9$.

Hence by Baye's theorem, we have

$\begin{array}{l}P\left(E_2/E\right)=\frac{P\left(E_2\right)P\left(E/E_2\right)}{P\left(E_1\right)P\left(E/E_1\right)+P\left(E_2\right)P\left(E/E_2\right)}\\ =\frac{\frac{1}{2}\cdot \frac{5}{9}}{\frac{1}{2}\cdot \frac{3}{5}+\frac{1}{2}\cdot \frac{5}{9}}=\frac{25}{52}.\end{array}$