Question

There are two bags. Bag-1 has 4 white and 6 black balls, and Bag-2 has 5 white and 5 black balls. A die is rolled, and if it shows a multiple of 3, a ball is drawn from Bag-1; otherwise, from Bag-2. If the ball drawn is not black, the probability it was not drawn from Bag-2 is:

• A. \(\frac{4}{9}\)
• B. \(\frac{3}{8}\)
• C. \(\frac{2}{7}\)
• D. \(\frac{4}{19}\)

Hint:

When working with conditional probabilities and Bayes' theorem, always ensure to calculate the total probability of the event first using the law of total probability. Then, use the relevant terms for the numerator and denominator in Bayes' theorem. This method allows you to break down the problem into smaller, manageable steps.

Answer 1

The Correct Option is C

Let \( A \) be the event the ball is not black, and \( B_2 \) be the event the ball was drawn from Bag-2. We want to find \( P(eg B_2 \mid A) \), the probability the ball was not from Bag-2 given it is not black.

Using Bayes' theorem:

\[ P(eg B_2 \mid A) = \frac{P(A \mid eg B_2)P(eg B_2)}{P(A)}. \]

First, calculate \( P(A) \), the probability of drawing a non-black ball:

• From Bag-1, the probability of a non-black ball is \( \frac{4}{10} = \frac{2}{5} \).
• From Bag-2, the probability of a non-black ball is \( \frac{5}{10} = \frac{1}{2} \).

Using the law of total probability to compute \( P(A) \):

\[ P(A) = P(A \mid B_1)P(B_1) + P(A \mid B_2)P(B_2). \]

The probability of drawing from Bag-1 is \( \frac{1}{3} \) (die shows a multiple of 3), and from Bag-2 is \( \frac{2}{3} \).

Therefore:

\[ P(A) = \frac{1}{3} \cdot \frac{2}{5} + \frac{2}{3} \cdot \frac{1}{2} = \frac{2}{15} + \frac{1}{3} = \frac{7}{15}. \]

Next, find \( P(A \mid eg B_2) \), the probability of a non-black ball from Bag-1:

\[ P(A \mid eg B_2) = \frac{2}{5}. \]

Applying Bayes' theorem:

\[ P(eg B_2 \mid A) = \frac{\frac{2}{5} \cdot \frac{1}{3}}{\frac{7}{15}} = \frac{\frac{2}{15}}{\frac{7}{15}} = \frac{2}{7}. \]

The final answer is: \[ \frac{2}{7}. \]