MHT CET · Maths · Probability
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
- A \(\frac{8}{31}\)
- B \(\frac{12}{31}\)
- C \(\frac{14}{31}\)
- D \(\frac{16}{31}\)
Answer & Solution
Correct Answer
(D) \(\frac{16}{31}\)
Step-by-step Solution
Detailed explanation
\(P(B_1|G) = \frac{P(G|B_1)P(B_1)}{P(G|B_1)P(B_1) + P(G|B_2)P(B_2)}\) \(P(B_1|G) = \frac{(\frac{2}{5})(\frac{1}{2})}{(\frac{2}{5})(\frac{1}{2}) + (\frac{3}{8})(\frac{1}{2})}\)
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