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Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ballsis transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred is red, is:





Solution

Conditional Probability – Bayes' Theorem

Given: One ball is transferred from Bag I to Bag II, and then a ball is drawn from Bag II and is black.

Goal: Find the probability that the transferred ball was red, given that a black ball was drawn.

Using Bayes' theorem: \[ P(R|A) = \frac{P(R \cap A)}{P(A)} = \frac{\frac{3}{10} \cdot \frac{5}{10}}{\frac{3}{10} \cdot \frac{5}{10} + \frac{4}{10} \cdot \frac{6}{10} + \frac{3}{10} \cdot \frac{5}{10}} = \frac{15}{54} = \boxed{\frac{5}{18}} \]

✅ Final Answer: \( \boxed{\frac{5}{18}} \)



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