MHTCET 23 Apr 2025 Shift 1 Paper

Concept:This question is based on conditional probability and Bayes' theorem.Explanation:Let $B_1$ be the event of selecting Bag I and $B_2$ be the event of selecting Bag II.Let $G$ be the event that the drawn ball is green.Since one bag is chosen at random, $P(B_1) = P(B_2) = \frac{1}{2}$.Bag I has 2 green balls out of 5 total balls, so $P(G|B_1) = \frac{2}{5}$.Bag II has 3 green balls out of 8 total balls, so $P(G|B_2) = \frac{3}{8}$.Now find the total probability of drawing a green ball:$$P(G) = P(G|B_1)P(B_1) + P(G|B_2)P(B_2)$$$$P(G) = \frac{2}{5}\times\frac{1}{2} + \frac{3}{8}\times\frac{1}{2} = \frac{1}{5} + \frac{3}{16} = \frac{16+15}{80} = \frac{31}{80}$$By Bayes' theorem, the required probability is:$$P(B_1|G) = \frac{P(G|B_1)P(B_1)}{P(G)}$$$$P(B_1|G) = \frac{\frac{2}{5}\times\frac{1}{2}}{\frac{31}{80}} = \frac{\frac{1}{5}}{\frac{31}{80}} = \frac{1}{5}\times\frac{80}{31} = \frac{16}{31}$$Therefore, the probability that the green ball came from Bag I is $\frac{16}{31}$.Answer:$\frac{16}{31}$, which matches option D.