IF the mean and the variance of a binomial distribution are 4 and 3 respectively , then the probability of six successes is

Options

A. `.^(16)C_(6)(1/4)^(6)(3/4)^(10)`
B. `.^(16)C_(6)(1/4)^(10)(3/4)^(6)`
C. `.^(12)C_(6)(1/4)^(6)(3/4)^(6)`
D. `.^(12)C_(6)(1/4)^(6)(3/4)^(10)`

Answer

To solve the problem, we need to determine the probability of getting exactly 6 successes in a binomial distribution where the mean is 4 and the variance is 3.

### Step-by-Step Solution:

1. **Identify the Mean and Variance**:
- Given: Mean (μ) = 4 and Variance (σ²) = 3.

2. **Use the Properties of Binomial Distribution**:
- For a binomial distribution with parameters \( N \) (number of trials) and \( P \) (probability of success):
- Mean: \( μ = N \cdot P \)
- Variance: \( σ² = N \cdot P \cdot Q \) where \( Q = 1 - P \) (probability of failure).

3. **Set Up the Equations**:
- From the mean:
\[ N \cdot P = 4 \quad \text{(1)} \]
- From the variance:
\[ N \cdot P \cdot Q = 3 \quad \text{(2)} \]

4. **Express Q in terms of P**:
- Since \( Q = 1 - P \), we can substitute \( Q \) in equation (2):
\[ N \cdot P \cdot (1 - P) = 3 \]

5. **Substitute N from Equation (1) into Equation (2)**:
- From equation (1), we can express \( N \) as:
\[ N = \frac{4}{P} \]
- Substitute \( N \) into equation (2):
\[ \left(\frac{4}{P}\right) \cdot P \cdot (1 - P) = 3 \]
- Simplifying gives:
\[ 4(1 - P) = 3 \]
\[ 4 - 4P = 3 \]
\[ 4P = 1 \quad \Rightarrow \quad P = \frac{1}{4} \]

6. **Calculate Q**:
- Now, since \( Q = 1 - P \):
\[ Q = 1 - \frac{1}{4} = \frac{3}{4} \]

7. **Find N**:
- Substitute \( P \) back into equation (1):
\[ N \cdot \frac{1}{4} = 4 \quad \Rightarrow \quad N = 16 \]

8. **Calculate the Probability of Exactly 6 Successes**:
- The probability of getting exactly \( k \) successes in a binomial distribution is given by:
\[ P(X = k) = \binom{N}{k} P^k Q^{N-k} \]
- For \( k = 6 \):
\[ P(X = 6) = \binom{16}{6} \left(\frac{1}{4}\right)^6 \left(\frac{3}{4}\right)^{10} \]

9. **Calculate the Binomial Coefficient**:
- Calculate \( \binom{16}{6} \):
\[ \binom{16}{6} = \frac{16!}{6!(16-6)!} = \frac{16!}{6! \cdot 10!} = 8008 \]

10. **Combine Everything**:
- Thus, the probability becomes:
\[ P(X = 6) = 8008 \cdot \left(\frac{1}{4}\right)^6 \cdot \left(\frac{3}{4}\right)^{10} \]

### Final Probability Calculation:
- Calculate \( \left(\frac{1}{4}\right)^6 = \frac{1}{4096} \) and \( \left(\frac{3}{4}\right)^{10} = \frac{59049}{1048576} \).
- Therefore:
\[ P(X = 6) = 8008 \cdot \frac{1}{4096} \cdot \frac{59049}{1048576} \]