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

Answer: (C)

$\frac{2}{7}$

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: $\frac{2}{7}.$