Question

Bag P contains 6 red and 4 blue balls, and Bag Q contains 5 red and 6 blue balls. A ball is transferred from Bag P to Bag Q, and then a ball is drawn from Bag Q. What is the probability that the ball drawn is blue?

• A. \( \frac{7}{15} \)
• B. \( \frac{8}{15} \)
• C. \( \frac{4}{19} \)
• D. \( \frac{8}{19} \)

Hint:

Use the Law of Total Probability: \[ P(B) = P(R_T) P(B | R_T) + P(B_T) P(B | B_T) \] where \( P(A | B) \) is the conditional probability.

Answer 1

The Correct Option is B

Step 1: Understanding the possible events.
- A ball is first transferred from Bag P to Bag Q. - Then, a ball is drawn from Bag Q. - The goal is to find the probability that the drawn ball is blue. Step 2: Defining the probability of transferring each type of ball.
- Probability of transferring a red ball from Bag P to Bag Q: \[ P(R_T) = \frac{6}{10} = \frac{3}{5} \] - Probability of transferring a blue ball from Bag P to Bag Q: \[ P(B_T) = \frac{4}{10} = \frac{2}{5} \] Step 3: Probability of drawing a blue ball from Bag Q.
Case 1: If a red ball is transferred to Bag Q: - Bag Q now has 6 red and 6 blue balls. - Probability of drawing a blue ball: \[ P(B | R_T) = \frac{6}{12} = \frac{1}{2} \] Case 2: If a blue ball is transferred to Bag Q: - Bag Q now has 5 red and 7 blue balls. - Probability of drawing a blue ball: \[ P(B | B_T) = \frac{7}{12} \] Step 4: Total probability using the Law of Total Probability.
\[ P(B) = P(R_T) \cdot P(B | R_T) + P(B_T) \cdot P(B | B_T) \] \[ = \left(\frac{3}{5} \times \frac{1}{2} \right) + \left(\frac{2}{5} \times \frac{7}{12} \right) \] \[ = \frac{3}{10} + \frac{14}{60} \] \[ = \frac{18}{60} + \frac{14}{60} = \frac{32}{60} = \frac{8}{15} \] Thus, the probability of drawing a blue ball is \( \frac{8}{15} \), and the correct answer is (B).