There are 8 white and 7 black balls in a bag . Three balls are drawn twice from the bag and balls drawn first time are not being replaced when the balls are drawn second time. Find the probability that first three balls drawn are white and second three balls drawn are black.

Explanation

To solve the problem, we need to find the probability that the first three balls drawn from the bag are white and the second three balls drawn are black.

### Step-by-Step Solution:

1. **Identify the Total Number of Balls**:
- There are 8 white balls and 7 black balls in the bag.
- Total balls = 8 (white) + 7 (black) = 15 balls.

2. **Calculate the Probability of Drawing 3 White Balls First (P1)**:
- The number of ways to choose 3 white balls from 8 is given by the combination formula \( \binom{n}{r} \):
\[ \text{Ways to choose 3 white balls} = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]
- The total number of ways to choose any 3 balls from 15 is:
\[ \text{Ways to choose any 3 balls} = \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \]
- Therefore, the probability \( P1 \) of drawing 3 white balls first is:
\[ P1 = \frac{\text{Ways to choose 3 white balls}}{\text{Ways to choose any 3 balls}} = \frac{56}{455} \]

3. **Update the Count of Balls After the First Draw**:
- After drawing 3 white balls, the remaining balls are:
- White balls left = 8 - 3 = 5
- Black balls left = 7
- Total balls left = 5 (white) + 7 (black) = 12 balls.

4. **Calculate the Probability of Drawing 3 Black Balls Second (P2)**:
- The number of ways to choose 3 black balls from 7 is:
\[ \text{Ways to choose 3 black balls} = \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]
- The total number of ways to choose any 3 balls from the remaining 12 is:
\[ \text{Ways to choose any 3 balls} = \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \]
- Therefore, the probability \( P2 \) of drawing 3 black balls second is:
\[ P2 = \frac{\text{Ways to choose 3 black balls}}{\text{Ways to choose any 3 balls}} = \frac{35}{220} = \frac{7}{44} \]

5. **Calculate the Total Probability**:
- The total probability of the event (first 3 balls white and second 3 balls black) is the product of \( P1 \) and \( P2 \):
\[ \text{Total Probability} = P1 \times P2 = \frac{56}{455} \times \frac{7}{44} \]
- Simplifying this:
\[ = \frac{56 \times 7}{455 \times 44} = \frac{392}{20020} \]
- Now, simplifying \( \frac{392}{20020} \):
- The GCD of 392 and 20020 is 14.
- Thus, \( \frac{392 \div 14}{20020 \div 14} = \frac{28}{1430} \).

6. **Final Simplification**:
- Further simplifying \( \frac{28}{1430} \):
- The GCD of 28 and 1430 is 14.
- Thus, \( \frac{28 \div 14}{1430 \div 14} = \frac{2}{102.14} \approx \frac{1}{51.07} \).

### Final Answer:
The probability that the first three balls drawn are white and the second three balls drawn are black is:
\[ \frac{14}{715} \]