There are 4 white and 3 black balls in a box. In another box, there are 3 white and 4 black balls. An unbiased die is rolled. If it shows a number less than or equal to 3. then a ball is drawn from the second box, otherwise from the first box. If the balls drawn is black, then the probability that the ball was drawn from the first box. is:

Options

A. `(1)/(2)`
B. `(6)/(7)`
C. `(4)/(7)`
D. `(3)/(7)`

Answer

To solve the problem step by step, we will use the concept of conditional probability.

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

### Step 2: Determine the Probabilities of Choosing Each Box
When a die is rolled:
- The probability of rolling a number less than or equal to 3 (choosing the second box) is \( P(A_2) = \frac{3}{6} = \frac{1}{2} \).
- The probability of rolling a number greater than 3 (choosing the first box) is \( P(A_1) = \frac{3}{6} = \frac{1}{2} \).

### Step 3: Calculate the Probability of Drawing a Black Ball from Each Box
- In the first box, there are 4 white and 3 black balls, so the total number of balls is 7. The probability of drawing a black ball from the first box is:
\[ P(B | A_1) = \frac{3}{7} \]

- In the second box, there are 3 white and 4 black balls, so the total number of balls is 7. The probability of drawing a black ball from the second box is:
\[ P(B | A_2) = \frac{4}{7} \]

### Step 4: Calculate the Total Probability of Drawing a Black Ball
Using the law of total probability:
\[ P(B) = P(A_1) \cdot P(B | A_1) + P(A_2) \cdot P(B | A_2) \]
Substituting the values:
\[ P(B) = \left(\frac{1}{2} \cdot \frac{3}{7}\right) + \left(\frac{1}{2} \cdot \frac{4}{7}\right) \]
\[ P(B) = \frac{3}{14} + \frac{4}{14} = \frac{7}{14} = \frac{1}{2} \]

### Step 5: Calculate the Conditional Probability
We need to find \( P(A_1 | B) \), the probability that the ball was drawn from the first box given that a black ball was drawn. Using Bayes' theorem:
\[ P(A_1 | B) = \frac{P(B | A_1) \cdot P(A_1)}{P(B)} \]
Substituting the values:
\[ P(A_1 | B) = \frac{\left(\frac{3}{7}\right) \cdot \left(\frac{1}{2}\right)}{\frac{1}{2}} \]
\[ P(A_1 | B) = \frac{3}{7} \]

### Final Answer
The probability that the ball drawn is from the first box given that it is black is:
\[ \frac{3}{7} \]