Q.Bag $A$ contains 3 red and 2 white balls and bag $B$ contains 2 red and 5 white balls. A bag is selected at random, a ball is drawn and put into the other bag, and then a ball is drawn from that bag. Find the probability that both the balls drawn are of same colour.

Bag A: 3 red, 2 white (5 total). Bag B: 2 red, 5 white (7 total). Each bag is equally likely to be picked first (prob $1/2$).

If A is picked first: red transferred (prob $3/5$) makes B into 3R+5W=8, so second draw from B is red with prob $3/8$, giving joint $\frac{3}{5}\times\frac{3}{8}=\frac{9}{40}$. White transferred (prob $2/5$) makes B into 2R+6W=8, second draw white with prob $3/4$, giving $\frac{2}{5}\times\frac{3}{4}=\frac{3}{10}$. So $P(\text{same}\mid\text{start A})=\frac{9}{40}+\frac{3}{10}=\frac{21}{40}$. …

Law of total probability across four branches: which bag is chosen first, then which colour is transferred - each branch's second draw …

Forgetting that the receiving bag's total ball count becomes 8 (not staying at 5 or 7) after the transfer, …