Question

For \( k = 1, 2, 3 \), the box \( B_k \) contains red balls and \( (k+1) \) white balls. Let \( P(B_1) = \frac{1}{2}, P(B_2) = \frac{1}{3}, P(B_3) = \frac{1}{6} \). A box is selected at random and a ball is drawn from it. If a red ball is drawn, then the probability that it came from box \( B_2 \) is

• A. \( \frac{35}{78} \)
• B. \( \frac{14}{39} \)
• C. \( \frac{10}{63} \)
• D. \( \frac{13}{78} \)

Hint:

When dealing with conditional probability, use Bayes' Theorem to reverse the probability and find the desired result.

Answer 1

The Correct Option is B

Step 1: Use Bayes' Theorem.
Use Bayes' Theorem to calculate the probability. Bayes' Theorem states that: \[ P(B_2 | \text{Red}) = \frac{P(\text{Red} | B_2) P(B_2)}{P(\text{Red})} \]
Step 2: Conclusion.
The probability that the red ball came from box \( B_2 \) is \( \frac{14}{39} \). Final Answer: \[ \boxed{\frac{14}{39}} \]