The mean and variance of a binomial distribution are 6 and 4 respectively, then n is

Options

A. 9
B. 12
C. 18
D. 10

Answer

To solve the problem, we need to find the value of \( n \) given that the mean and variance of a binomial distribution are 6 and 4, respectively.

### Step-by-Step Solution:

1. **Understand the Mean and Variance of a Binomial Distribution:**
- The mean \( \mu \) of a binomial distribution is given by:
\[ \mu = n \cdot p \]
- The variance \( \sigma^2 \) of a binomial distribution is given by:
\[ \sigma^2 = n \cdot p \cdot q \]
where \( q = 1 - p \).

2. **Set Up the Equations:**
- From the problem, we know:
\[ n \cdot p = 6 \quad \text{(1)} \]
\[ n \cdot p \cdot q = 4 \quad \text{(2)} \]

3. **Express \( q \) in Terms of \( p \):**
- Since \( q = 1 - p \), we can substitute \( q \) in equation (2):
\[ n \cdot p \cdot (1 - p) = 4 \]

4. **Substitute \( n \cdot p \) from Equation (1) into Equation (2):**
- From equation (1), we have \( n \cdot p = 6 \). Substitute this into the modified equation (2):
\[ 6 \cdot (1 - p) = 4 \]

5. **Solve for \( p \):**
- Rearranging the equation gives:
\[ 6 - 6p = 4 \]
\[ 6p = 2 \]
\[ p = \frac{1}{3} \]

6. **Find \( q \):**
- Now that we have \( p \), we can find \( q \):
\[ q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \]

7. **Substitute \( p \) Back to Find \( n \):**
- Now substitute \( p \) back into equation (1):
\[ n \cdot \frac{1}{3} = 6 \]
\[ n = 6 \cdot 3 = 18 \]

### Final Answer:
Thus, the value of \( n \) is \( 18 \).