The mean and the variance in a binomial distribution are found to be 2 and I respectively. The probability P(X= 0)is:

Options

A. a) (1)/(2)
B. b) (1)/(4)
C. c) (1)/(8)
D. d) (1)/(16)

Answer

To solve the problem, we need to find the probability \( P(X = 0) \) in a binomial distribution where the mean \( np = 2 \) and the variance \( npq = 1 \).

### Step-by-Step Solution:

1. **Identify the Mean and Variance Equations:**
- The mean of a binomial distribution is given by:
\[ np = 2 \quad \text{(1)} \]
- The variance of a binomial distribution is given by:
\[ npq = 1 \quad \text{(2)} \]

2. **Express \( q \) in terms of \( p \):**
- Since \( p + q = 1 \), we can express \( q \) as:
\[ q = 1 - p \]

3. **Substitute \( q \) in the Variance Equation:**
- Substitute \( q \) in equation (2):
\[ np(1 - p) = 1 \]
- Now, substitute \( np = 2 \) from equation (1):
\[ 2(1 - p) = 1 \]

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

5. **Find \( q \):**
- Now, substitute \( p \) back to find \( q \):
\[ q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \]

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

7. **Calculate \( P(X = 0) \):**
- The probability \( P(X = 0) \) is given by:
\[ P(X = 0) = \binom{n}{0} p^0 q^n = \binom{4}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^4 \]
- Since \( \binom{4}{0} = 1 \) and \( \left(\frac{1}{2}\right)^0 = 1 \):
\[ P(X = 0) = 1 \cdot 1 \cdot \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]

### Final Answer:
\[ P(X = 0) = \frac{1}{16} \]