Among 15 players, 8 are batsmen and 7 are bowlers. Find the probability that a team is chosen of 6 batsmen and 5 bowlers:

Options

A. `(""^(8)C_(6)xx""^(7)C_(5))/(""^(15)C_(11))`
B. `28/15`
C. `15/28`
D. None of these

Answer

To solve the problem of finding the probability of selecting a team of 6 batsmen and 5 bowlers from a total of 15 players (8 batsmen and 7 bowlers), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the total number of players and their categories**:
- Total players = 15
- Batsmen = 8
- Bowlers = 7

2. **Determine the number of players to be selected**:
- We need to select 6 batsmen and 5 bowlers.

3. **Calculate the number of ways to choose the batsmen**:
- The number of ways to choose 6 batsmen from 8 is given by the combination formula:
\[ \text{Number of ways to choose batsmen} = \binom{8}{6} \]

4. **Calculate the number of ways to choose the bowlers**:
- The number of ways to choose 5 bowlers from 7 is given by:
\[ \text{Number of ways to choose bowlers} = \binom{7}{5} \]

5. **Calculate the total number of ways to choose any 11 players from 15**:
- The total number of ways to choose 11 players from 15 is:
\[ \text{Total ways} = \binom{15}{11} \]

6. **Combine the results to find the probability**:
- The probability \( P \) of choosing 6 batsmen and 5 bowlers is given by:
\[ P = \frac{\text{Number of ways to choose batsmen} \times \text{Number of ways to choose bowlers}}{\text{Total ways}} \]
- Substituting the values:
\[ P = \frac{\binom{8}{6} \times \binom{7}{5}}{\binom{15}{11}} \]

7. **Calculate the combinations**:
- \(\binom{8}{6} = \frac{8!}{6!(8-6)!} = \frac{8 \times 7}{2 \times 1} = 28\)
- \(\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7 \times 6}{2 \times 1} = 21\)
- \(\binom{15}{11} = \binom{15}{4} = \frac{15!}{4!(15-4)!} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365\)

8. **Substitute the values into the probability formula**:
- Now we substitute the values we calculated:
\[ P = \frac{28 \times 21}{1365} \]
- Calculate \( 28 \times 21 = 588 \)
- Therefore,
\[ P = \frac{588}{1365} \]

9. **Final Result**:
- The probability that a team is chosen consisting of 6 batsmen and 5 bowlers is:
\[ P = \frac{588}{1365} \]