The mean and variance of a binomial distribution are `(5)/(4) and (15)/(16)` respectively, then value of p, is

Options

A. `(1)/(2)`
B. `(15)/(16)`
C. `(1)/(4)`
D. `(3)/(4)`

Answer

To solve the problem, we need to use the properties of a binomial distribution. The mean and variance of a binomial distribution can be expressed as follows:

1. **Mean (μ)**: \( \mu = np \)
2. **Variance (σ²)**: \( \sigma^2 = npq \)

Where:
- \( n \) is the number of trials,
- \( p \) is the probability of success,
- \( q \) is the probability of failure, and \( q = 1 - p \).

Given:
- Mean \( \mu = \frac{5}{4} \)
- Variance \( \sigma^2 = \frac{15}{16} \)

### Step 1: Set up the equations

From the mean:
\[ np = \frac{5}{4} \quad \text{(1)} \]

From the variance:
\[ npq = \frac{15}{16} \quad \text{(2)} \]

### Step 2: Substitute \( q \)

Since \( q = 1 - p \), we can substitute \( q \) in equation (2):
\[ np(1 - p) = \frac{15}{16} \quad \text{(3)} \]

### Step 3: Substitute \( np \) from equation (1) into equation (3)

From equation (1), we can express \( n \) in terms of \( p \):
\[ n = \frac{5}{4p} \]

Substituting \( np \) from equation (1) into equation (3):
\[ \frac{5}{4}(1 - p) = \frac{15}{16} \]

### Step 4: Solve for \( p \)

Now we can solve for \( p \):
\[ \frac{5}{4} - \frac{5p}{4} = \frac{15}{16} \]

Multiply through by 16 to eliminate the fraction:
\[ 16 \cdot \frac{5}{4} - 16 \cdot \frac{5p}{4} = 15 \]
\[ 20 - 20p = 15 \]

Rearranging gives:
\[ 20 - 15 = 20p \]
\[ 5 = 20p \]
\[ p = \frac{5}{20} = \frac{1}{4} \]

### Step 5: Conclusion

Thus, the value of \( p \) is:
\[ \boxed{\frac{1}{4}} \]