Question

Bag \( B_1 \) contains 6 white and 4 blue balls, Bag \( B_2 \) contains 4 white and 6 blue balls, and Bag \( B_3 \) contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability that the ball is drawn from Bag \( B_2 \) is:

• A. \( \frac{1}{3} \)
• B. \( \frac{2}{3} \)
• C. \( \frac{4}{15} \)
• D. \( \frac{2}{5} \)

Hint:

Use Bayes' Theorem for conditional probability when dealing with multiple events.

Answer 1

The Correct Option is C

Let \( E_1 \) be the event that Bag \( B_1 \) is selected,
\( E_2 \) the event that Bag \( B_2 \) is selected,
and \( E_3 \) the event that Bag \( B_3 \) is selected.
Let \( A \) be the event that a white ball is drawn.
We need to find \( P(E_2 | A) \).
Using Bayes' Theorem:
\[ P(E_2 | A) = \frac{P(E_2) \cdot P(A | E_2)}{P(E_1) \cdot P(A | E_1) + P(E_2) \cdot P(A | E_2) + P(E_3) \cdot P(A | E_3)} \] Substitute values:
\[ P(E_2 | A) = \frac{\frac{1}{3} \cdot \frac{4}{10}}{\frac{1}{3} \cdot \frac{6}{10} + \frac{1}{3} \cdot \frac{4}{10} + \frac{1}{3} \cdot \frac{5}{10}} = \frac{4}{15} \]