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. \(\frac{4}{19}\)

Answer 1

The Correct Option is C

Let \( A \) represent the event that the ball is not black, and \( B_2 \) represent the event that the ball was drawn from Bag-2. We need to find \( P(\neg B_2 \mid A) \), which is the probability that the ball was not drawn from Bag-2, given that it is not black.

We can use Bayes' theorem:

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

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

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

Now, we compute \( P(A) \) using the law of total probability:

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

The probability of drawing from Bag-1 is \( \frac{1}{3} \) (since the die shows a number divisible by 3), and the probability of drawing from Bag-2 is \( \frac{2}{3} \).

Thus:

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

Next, calculate \( P(A \mid \neg B_2) \), which is the probability of drawing a non-black ball from Bag-1:

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

Now, apply Bayes' theorem:

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

Thus, the correct answer is: \[ \frac{2}{7}. \]