Hello Gautam,

Let us consider A1 is the events that a red ball is transferred from a bag I to Bag II and  A2 Event is that green ball is transferred from the bag I to Bag II.

P(A1) = 4/6 & P(A2) = 2/6

Assume B be the event that the ball drawn from Bag is Green.

(1) when a Green ball is transferred.

P(B/A1) = 6/9

(2) when a Red ball is transferred.

P(B/A2) = 5/9

Now, P(A2/B) = [P(B/A2) × P(A2)] ÷ [(P(A1) × P(B/A2)) + (P(A2) × P(B/A1)]

Similarly, P(A2/B) = [((2/6) × (6/9)) ÷ ((4/6) × (5/9)) + ((2/6) × (6/9))]

= 3/8

Probability that the ball drawn fromBag I to Bag II is 3/8.

Hope this information was useful to you.

Good Luck!!