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 \( A \) 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: \[ \frac{2}{7}. \]

Answer 2

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:

\[ \frac{2}{7}. \]