The mean and the variance of a binomial distribution are 4 and 2 respectively.then, the probabitly of 2 , successes is

Options

A. `28//256`
B. `219//256`
C. `128//256`
D. `37//256`

Explanation

To solve the problem, we need to find the probability of getting exactly 2 successes in a binomial distribution where the mean is 4 and the variance is 2.

### Step-by-step Solution:

1. **Identify the Mean and Variance Formulas**:
For a binomial distribution, the mean (μ) and variance (σ²) are given by:
\[ \text{Mean} \, (\mu) = np \]
\[ \text{Variance} \, (\sigma^2) = npq \]
where \( q = 1 - p \).

2. **Set Up the Equations**:
From the problem, we know:
\[ np = 4 \quad \text{(1)} \]
\[ npq = 2 \quad \text{(2)} \]

3. **Express q in terms of p**:
Since \( q = 1 - p \), we can substitute \( q \) in equation (2):
\[ np(1 - p) = 2 \]
Substituting \( np = 4 \) from equation (1):
\[ 4(1 - p) = 2 \]

4. **Solve for p**:
\[ 4 - 4p = 2 \]
\[ 4p = 2 \]
\[ p = \frac{1}{2} \]

5. **Find q**:
Using \( q = 1 - p \):
\[ q = 1 - \frac{1}{2} = \frac{1}{2} \]

6. **Find n**:
Substitute \( p \) back into equation (1):
\[ n \cdot \frac{1}{2} = 4 \]
\[ n = 4 \cdot 2 = 8 \]

7. **Calculate the Probability of 2 Successes**:
The probability of getting exactly \( r \) successes in a binomial distribution is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n - r} \]
Here, we want \( P(X = 2) \):
\[ P(X = 2) = \binom{8}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{8 - 2} \]
\[ = \binom{8}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^6 \]
\[ = \binom{8}{2} \left(\frac{1}{2}\right)^8 \]

8. **Calculate \(\binom{8}{2}\)**:
\[ \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 \]

9. **Final Calculation**:
\[ P(X = 2) = 28 \cdot \left(\frac{1}{2}\right)^8 = \frac{28}{256} \]

Thus, the probability of getting exactly 2 successes is:
\[ \frac{28}{256} = \frac{7}{64} \]