Bag I contains 3 red and 2 green balls and Bag II contains 5 red and 3 green balls. A ball is drawn from one of the bag at random and it is found to be green. Then the probability that it is drawn from Bag I is

Let $B_1$ and $B_2$ be the events that the ball is drawn from Bag I and Bag II respectively, and $G$ the event that the ball is green. Since a bag is chosen at random, $P(B_1) = \frac{1}{2}$ and $P(B_2) = \frac{1}{2}$.

Bag I contains 5 balls (3 red, 2 green), so $P(G|B_1) = \frac{2}{5}$. Bag II contains 8 balls (5 red, 3 green), so $P(G|B_2) = \frac{3}{8}$.

By Bayes' theorem, $P(B_1|G) = \frac{P(G|B_1) P(B_1)}{P(G)}$. The total probability of drawing a green ball is $P(G) = P(G|B_1)P(B_1) + P(G|B_2)P(B_2) = \frac{2}{5} \cdot \frac{1}{2} + \frac{3}{8} \cdot \frac{1}{2} = \frac{1}{5} + \frac{3}{16} = \frac{16+15}{80} = \frac{31}{80}$.

Substituting, $P(B_1|G) = \frac{\frac{2}{5} \cdot \frac{1}{2}}{\frac{31}{80}} = \frac{\frac{1}{5}}{\frac{31}{80}} = \frac{1}{5} \cdot \frac{80}{31} = \frac{16}{31}$.

The probability the green ball came from Bag I is $\frac{16}{31}$, so the answer is D.