If the mean and the variance of a binomial distribution are 2 and 1 respectively , then the probability of atleast one success is

Options

A. `2/3`
B. `3/4`
C. `7/8`
D. `15/16`

Answer

To solve the problem, we need to find the probability of at least one success in a binomial distribution where the mean is 2 and the variance is 1.

### Step-by-Step Solution:

1. **Understand the parameters of the binomial distribution**:
- The mean (μ) of a binomial distribution is given by the formula:
\[ \mu = n \cdot p \]
- The variance (σ²) is given by:
\[ \sigma^2 = n \cdot p \cdot q \]
where \( q = 1 - p \).

2. **Set up the equations based on the given mean and variance**:
- From the mean, we have:
\[ n \cdot p = 2 \quad \text{(1)} \]
- From the variance, we have:
\[ n \cdot p \cdot q = 1 \quad \text{(2)} \]

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

4. **Substituting n from equation (1) into equation (2)**:
- From equation (1), we can express \( n \) as:
\[ n = \frac{2}{p} \]
- Substitute this into equation (2):
\[ \frac{2}{p} \cdot p \cdot (1 - p) = 1 \]
- Simplifying this gives:
\[ 2(1 - p) = 1 \]
- Therefore:
\[ 2 - 2p = 1 \implies 2p = 1 \implies p = \frac{1}{2} \]

5. **Finding q**:
- Now that we have \( p \):
\[ q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \]

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

7. **Calculate the probability of at least one success**:
- The probability of at least one success is given by:
\[ P(X \geq 1) = 1 - P(X = 0) \]
- The probability of no successes (X = 0) in a binomial distribution is:
\[ P(X = 0) = \binom{n}{0} p^0 q^n = 1 \cdot 1 \cdot \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]
- Therefore:
\[ P(X \geq 1) = 1 - P(X = 0) = 1 - \frac{1}{16} = \frac{15}{16} \]

### Final Answer:
The probability of at least one success is:
\[ \frac{15}{16} \]