A bag P contains 5 distinct white and 3 distinct black balls. Four balls are drawn from P and put in an empty bag Q. A ball is drawn from Q and is found to be black. Let p denote the probability that all the three black balls are transferred to bag Q, then 7p = ______

Answer

To solve the problem, we need to find the probability \( p \) that all 3 black balls are transferred to bag Q given that a ball drawn from bag Q is black. We will use conditional probability to find this.

### Step-by-Step Solution:

1. **Define Events**:
- Let \( A \) be the event that all 3 black balls are transferred to bag Q.
- Let \( B \) be the event that a ball drawn from bag Q is black.

2. **Use Conditional Probability**:
We need to find \( P(A | B) \), which is given by the formula:
\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]

3. **Calculate \( P(A \cap B) \)**:
- If all 3 black balls are transferred to bag Q, then we must also transfer 1 white ball (since we are transferring a total of 4 balls).
- The probability of transferring all 3 black balls and 1 white ball can be calculated as follows:
- The number of ways to choose 3 black balls from 3 is \( \binom{3}{3} = 1 \).
- The number of ways to choose 1 white ball from 5 is \( \binom{5}{1} = 5 \).
- Therefore, the total number of favorable outcomes for transferring all 3 black balls and 1 white ball is \( 1 \times 5 = 5 \).

4. **Total Ways to Choose 4 Balls from 8**:
- The total number of ways to choose any 4 balls from the 8 balls (5 white + 3 black) is \( \binom{8}{4} = 70 \).

5. **Calculate \( P(A) \)**:
\[ P(A) = \frac{5}{70} = \frac{1}{14} \]

6. **Calculate \( P(B | A) \)**:
- If event \( A \) occurs (all 3 black balls are in Q), then the probability of drawing a black ball from Q (which contains 3 black and 1 white) is:
\[ P(B | A) = \frac{3}{4} \]

7. **Calculate \( P(B) \)**:
- We need to consider all scenarios where at least one black ball is drawn:
- **Case 1**: 3 black and 1 white: \( P(B | A) = \frac{3}{4} \) and \( P(A) = \frac{1}{14} \).
- **Case 2**: 2 black and 2 white: The probability is:
- Choose 2 black from 3: \( \binom{3}{2} = 3 \)
- Choose 2 white from 5: \( \binom{5}{2} = 10 \)
- Total ways = \( 3 \times 10 = 30 \)
- Probability = \( \frac{30}{70} = \frac{3}{7} \)
- Probability of drawing a black ball = \( \frac{2}{4} = \frac{1}{2} \)
- Contribution to \( P(B) \): \( P(B | 2B) \cdot P(2B) = \frac{1}{2} \cdot \frac{3}{7} = \frac{3}{14} \).
- **Case 3**: 1 black and 3 white: The probability is:
- Choose 1 black from 3: \( \binom{3}{1} = 3 \)
- Choose 3 white from 5: \( \binom{5}{3} = 10 \)
- Total ways = \( 3 \times 10 = 30 \)
- Probability = \( \frac{30}{70} = \frac{3}{7} \)
- Probability of drawing a black ball = \( \frac{1}{4} \)
- Contribution to \( P(B) \): \( P(B | 1B) \cdot P(1B) = \frac{1}{4} \cdot \frac{3}{7} = \frac{3}{28} \).

8. **Combine All Cases for \( P(B) \)**:
\[ P(B) = P(B | A) \cdot P(A) + P(B | 2B) \cdot P(2B) + P(B | 1B) \cdot P(1B) \]
\[ P(B) = \frac{3}{4} \cdot \frac{1}{14} + \frac{1}{2} \cdot \frac{3}{7} + \frac{1}{4} \cdot \frac{3}{7} \]
\[ P(B) = \frac{3}{56} + \frac{3}{14} + \frac{3}{28} \]
\[ P(B) = \frac{3}{56} + \frac{12}{56} + \frac{6}{56} = \frac{21}{56} = \frac{3}{8} \]

9. **Final Calculation of \( P(A | B) \)**:
\[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{3}{4} \cdot \frac{1}{14}}{\frac{3}{8}} = \frac{\frac{3}{56}}{\frac{3}{8}} = \frac{3}{56} \cdot \frac{8}{3} = \frac{8}{56} = \frac{1}{7} \]

10. **Calculate \( 7p \)**:
\[ 7p = 7 \cdot \frac{1}{7} = 1 \]

### Final Answer:
\[ 7p = 1 \]