A box contains `4` white and `3` black balls. Another box contains `3` white and `4` black balls. A die is thrown. If it exhibits a number greater than 3`, the ball is drawn from the first box. Otherwise, a ball is drawn from the second box. A ball drawn is found to be black. The probability that it has been drawn from the second box is

Options

A. `3//7`
B. `4//7`
C. `6//17`
D. `8//17`

Explanation

To solve the problem, we will use Bayes' theorem to find the probability that the black ball drawn came from the second box given that a black ball was drawn.

### Step-by-Step Solution:

1. **Define Events:**
- Let \( A_1 \): The event that the ball is drawn from the first box.
- Let \( A_2 \): The event that the ball is drawn from the second box.
- Let \( B \): The event that a black ball is drawn.

2. **Determine Probabilities of Drawing from Each Box:**
- The die can show numbers greater than 3 (4, 5, 6) or less than or equal to 3 (1, 2, 3).
- Probability of rolling a number greater than 3 (i.e., 4, 5, or 6):
\[ P(A_1) = \frac{3}{6} = \frac{1}{2} \]
- Probability of rolling a number less than or equal to 3 (i.e., 1, 2, or 3):
\[ P(A_2) = \frac{3}{6} = \frac{1}{2} \]

3. **Determine Probabilities of Drawing a Black Ball from Each Box:**
- From the first box (4 white and 3 black):
\[ P(B|A_1) = \frac{3}{7} \]
- From the second box (3 white and 4 black):
\[ P(B|A_2) = \frac{4}{7} \]

4. **Use Total Probability to Find \( P(B) \):**
- The total probability of drawing a black ball:
\[ P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) \]
- Substitute the values:
\[ P(B) = \left(\frac{3}{7} \cdot \frac{1}{2}\right) + \left(\frac{4}{7} \cdot \frac{1}{2}\right) \]
\[ P(B) = \frac{3}{14} + \frac{4}{14} = \frac{7}{14} = \frac{1}{2} \]

5. **Apply Bayes' Theorem to Find \( P(A_2|B) \):**
- We want to find the probability that the ball was drawn from the second box given that a black ball was drawn:
\[ P(A_2|B) = \frac{P(B|A_2)P(A_2)}{P(B)} \]
- Substitute the values:
\[ P(A_2|B) = \frac{\left(\frac{4}{7}\right) \left(\frac{1}{2}\right)}{\frac{1}{2}} \]
- Simplifying gives:
\[ P(A_2|B) = \frac{4}{7} \]

### Final Answer:
The probability that the black ball drawn came from the second box is \( \frac{4}{7} \).