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).