An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement The probability that the first draw gives all white balls and second draw gives all black balls is

Options

A. `2/335`
B. `1/495`
C. `5/812`
D. `3/715`

Answer

To solve the problem, we need to find the probability of two successive draws from an urn containing 6 white and 9 black balls. The first draw consists of 4 balls, all of which must be white, and the second draw also consists of 4 balls, all of which must be black.

### Step-by-step Solution:

1. **Identify the Total Number of Balls**:
- Total white balls = 6
- Total black balls = 9
- Total balls = 6 + 9 = 15

2. **Calculate the Probability of Drawing 4 White Balls in the First Draw**:
- The number of ways to choose 4 white balls from 6 is given by the combination formula \( \binom{n}{r} \), which is \( \binom{6}{4} \).
- The total number of ways to choose any 4 balls from 15 is \( \binom{15}{4} \).
- Therefore, the probability \( P(A) \) of drawing 4 white balls is:
\[ P(A) = \frac{\binom{6}{4}}{\binom{15}{4}} \]

3. **Calculate \( \binom{6}{4} \) and \( \binom{15}{4} \)**:
- \( \binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15 \)
- \( \binom{15}{4} = \frac{15!}{4!(15-4)!} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365 \)

4. **Calculate \( P(A) \)**:
\[ P(A) = \frac{15}{1365} = \frac{1}{91} \]

5. **Calculate the Probability of Drawing 4 Black Balls in the Second Draw**:
- After drawing 4 white balls, the remaining balls are:
- White balls = 6 - 4 = 2
- Black balls = 9
- Total remaining balls = 2 + 9 = 11
- The number of ways to choose 4 black balls from 9 is \( \binom{9}{4} \).
- The total number of ways to choose any 4 balls from the remaining 11 is \( \binom{11}{4} \).
- Therefore, the probability \( P(B|A) \) of drawing 4 black balls after having drawn 4 white balls is:
\[ P(B|A) = \frac{\binom{9}{4}}{\binom{11}{4}} \]

6. **Calculate \( \binom{9}{4} \) and \( \binom{11}{4} \)**:
- \( \binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \)
- \( \binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330 \)

7. **Calculate \( P(B|A) \)**:
\[ P(B|A) = \frac{126}{330} = \frac{21}{55} \]

8. **Calculate the Combined Probability**:
- The combined probability of both events occurring is given by:
\[ P(A \cap B) = P(A) \times P(B|A) = \frac{1}{91} \times \frac{21}{55} \]
- Simplifying this gives:
\[ P(A \cap B) = \frac{21}{5005} = \frac{3}{715} \]

### Final Answer:
The probability that the first draw gives all white balls and the second draw gives all black balls is \( \frac{3}{715} \).