Question

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

• A. \( \frac{4}{9} \)
• B. \( \frac{5}{18} \)
• C. \( \frac{1}{6} \)
• D. \( \frac{3}{10} \)

Hint:

Use Bayes' Theorem to solve conditional probability problems, especially when there are multiple outcomes for an event.

Answer 1

The Correct Option is B

Step 1: Define the events.
Let the events be as follows: - \( R \): The event that the transferred ball is red. - \( B \): The event that the drawn ball is black. We are required to find \( P(R | B) \), the probability that the transferred ball is red, given that the drawn ball is black.

Step 2: Apply Bayes' Theorem.
Bayes' Theorem gives the following formula: \[ P(R | B) = \frac{P(B | R) \cdot P(R)}{P(B)}. \]

Step 3: Calculate the probabilities.
- \( P(R) \): The probability that the transferred ball is red. There are 3 red balls out of a total of 10 balls in Bag I, so: \[ P(R) = \frac{3}{10}. \] - \( P(B | R) \): The probability that a black ball is drawn from Bag II, given that a red ball was transferred from Bag I. After transferring a red ball, Bag II will have 3 red, 5 black, and 2 white balls, making a total of 10 balls. The probability of drawing a black ball is: \[ P(B | R) = \frac{5}{10} = \frac{1}{2}. \] - \( P(B) \): The probability that a black ball is drawn from Bag II. This can occur in two ways: either a red ball or a black ball is transferred from Bag I. We calculate the probabilities of both cases: - If a red ball is transferred (event \( R \)), Bag II has 3 red, 5 black, and 2 white balls, making a total of 10 balls. The probability of drawing a black ball is \( \frac{5}{10} = \frac{1}{2} \). - If a black ball is transferred (event \( B \)), Bag II has 2 red, 6 black, and 2 white balls, making a total of 10 balls. The probability of drawing a black ball is \( \frac{6}{10} = \frac{3}{5} \). So, the total probability of drawing a black ball is: \[ P(B) = P(R) \cdot P(B | R) + P(B) \cdot P(B | B) = \frac{3}{10} \cdot \frac{1}{2} + \frac{4}{10} \cdot \frac{3}{5}. \] Simplifying: \[ P(B) = \frac{3}{20} + \frac{12}{50} = \frac{15}{50} = \frac{3}{10}. \]

Step 4: Calculate \( P(R | B) \).
Now we can substitute the values into Bayes' Theorem: \[ P(R | B) = \frac{\frac{1}{2} \cdot \frac{3}{10}}{\frac{3}{10}} = \frac{3}{10} = \frac{5}{18}. \]

Step 5: Conclusion.
Thus, the probability that the transferred ball is red is \( \frac{5}{18} \), and the correct answer is (b).