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MHT CET202523 Apr 2025Morning ShiftMathematicsProbabilityActual

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

Options

  1. A8 31
  2. B12 31
  3. C14 31
  4. D16 31

Correct answer

D. 16 31

Step-by-step solution

Let B₁ and B₂ 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(B₁) = 1 2 and P(B₂) = 1 2 . Bag I contains 5 balls (3 red, 2 green), so P(G|B₁) = 2 5 . Bag II contains 8 balls (5 red, 3 green), so P(G|B₂) = 3 8 . By Bayes' theorem, P(B₁|G) = P(G|B₁) P(B₁) P(G) . The total probability of drawing a green ball is P(G) = P(G|B₁)P(B₁) + P(G|B₂)P(B₂) = 2 5 1 2 + 3 8 1 2 = 1 5 + 3 16 = 16+15 80 = 31 80 . Substituting, P(B₁|G)

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