A cricket club has 16 members out of which 6 can bowl. If a team of 11 members is selected. Find the probability that the team will contain exactly four bowlers

Options

A. `(5)/(146)`
B. `(7)/(1456)`
C. `(5)/(1456)`
D. `(75)/(182)`

Answer

To find the probability that a cricket team of 11 members contains exactly 4 bowlers from a club of 16 members (of which 6 can bowl), we can follow these steps:

### Step 1: Determine the Total Ways to Select the Team
We need to calculate the total number of ways to select 11 members from 16. This can be calculated using the combination formula:

\[ \text{Total Ways} = \binom{16}{11} \]

### Step 2: Determine the Favorable Ways to Select the Team
Next, we need to find the number of favorable ways to select exactly 4 bowlers and 7 non-bowlers.

1. **Selecting 4 Bowlers**: We can choose 4 bowlers from the 6 available bowlers. This can be calculated as:

\[ \text{Ways to choose 4 bowlers} = \binom{6}{4} \]

2. **Selecting 7 Non-Bowlers**: After selecting 4 bowlers, we have 10 members left (16 total members - 6 bowlers = 10 non-bowlers). We need to select 7 members from these 10 non-bowlers:

\[ \text{Ways to choose 7 non-bowlers} = \binom{10}{7} \]

3. **Total Favorable Ways**: The total number of favorable ways to select the team with exactly 4 bowlers is the product of the two combinations calculated above:

\[ \text{Favorable Ways} = \binom{6}{4} \times \binom{10}{7} \]

### Step 3: Calculate the Probability
The probability \( P \) of selecting a team with exactly 4 bowlers is given by the ratio of favorable ways to total ways:

\[ P = \frac{\text{Favorable Ways}}{\text{Total Ways}} = \frac{\binom{6}{4} \times \binom{10}{7}}{\binom{16}{11}} \]

### Step 4: Calculate the Combinations
Now we will compute the values of the combinations:

1. \(\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15\)
2. \(\binom{10}{7} = \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\)
3. \(\binom{16}{11} = \binom{16}{5} = \frac{16!}{5!(16-5)!} = \frac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1} = 4368\)

### Step 5: Substitute the Values into the Probability Formula
Now substituting the values into the probability formula:

\[ P = \frac{15 \times 120}{4368} \]

Calculating the numerator:

\[ 15 \times 120 = 1800 \]

Thus,

\[ P = \frac{1800}{4368} \]

### Step 6: Simplify the Probability
Now, we simplify the fraction:

\[ P = \frac{75}{182} \]

### Final Answer
The probability that the team will contain exactly 4 bowlers is:

\[ \boxed{\frac{75}{182}} \]