There are three bags B1, B2, and B3. Bag B1 contains 5 red and 5 green balls. B2 contains 3 red and 5 green balls, and B3 contains 5 red and 3 green balls. Bags B1, B2, and B3 have probabilities 3/10, 3/10, and 4/10 respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?

Correct option is C.
Probability that the chosen ball is green equals 39/80
P(B1) = 3/10
P(B2) = 3/10
P(B3) = 4/10
Bag 1
Red Balls: 5
Green Balls: 5
Total: 10
Bag 2
Red Balls: 3
Green Balls: 5
Total: 8
Bag 3
Red Balls: 5
Green Balls: 3
Total: 8
(1) P(Ball is Green) = P(B1)P(G/B1) + P(B2)P(G/B2) + P(B3)P(G/B3) = (3/10) × (5/10) + (3/10) × (5/8) + (4/10) × (3/8) = 39/80
(2) P(Ball chosen is Green | Ball is from 3rd Bag) = 3/8
P(Ball is from 3rd Bag | Ball chosen is Green) = P(B3)P(G/B3) / [P(B1)P(G/B1) + P(B2)P(G/B2) + P(B3)P(G/B3)]
P(B3) = 4/10 = (4/10) × (3/8) / (39/80) = 4/13.