In a bionomial distribution, mean is 18 and variance is 12 then p = .....

Options

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

Explanation

To solve the problem, we need to use the properties of a binomial distribution. The mean (μ) and variance (σ²) of a binomial distribution can be expressed in terms of the number of trials (n) and the probability of success (p) as follows:

1. **Mean (μ)**:
\[ \mu = n \cdot p \]
Given that the mean is 18, we can write:
\[ n \cdot p = 18 \quad \text{(1)} \]

2. **Variance (σ²)**:
\[ \sigma^2 = n \cdot p \cdot q \]
where \( q = 1 - p \). Given that the variance is 12, we can write:
\[ n \cdot p \cdot q = 12 \quad \text{(2)} \]

Now, we can express \( q \) in terms of \( p \):
\[ q = 1 - p \]

Substituting \( q \) into equation (2):
\[ n \cdot p \cdot (1 - p) = 12 \quad \text{(3)} \]

Now, we have two equations:
1. \( n \cdot p = 18 \) (from equation 1)
2. \( n \cdot p \cdot (1 - p) = 12 \) (from equation 3)

From equation (1), we can express \( n \) in terms of \( p \):
\[ n = \frac{18}{p} \quad \text{(4)} \]

Substituting equation (4) into equation (3):
\[ \left(\frac{18}{p}\right) \cdot p \cdot (1 - p) = 12 \]

This simplifies to:
\[ 18 \cdot (1 - p) = 12 \]

Now, we can solve for \( p \):
\[ 18 - 18p = 12 \]
\[ 18p = 18 - 12 \]
\[ 18p = 6 \]
\[ p = \frac{6}{18} = \frac{1}{3} \]

Thus, the value of \( p \) is:
\[ \boxed{\frac{1}{3}} \]