Question

A random variable $X \sim B(n, p)$. If the values of the mean and variance of $X$ are $18$ and $12$ respectively, then $n =$

• A. $54$
• B. $18$
• C. $12$
• D. $55$

Hint:

You can calculate this mentally in seconds! The ratio $\frac{\text{Variance}}{\text{Mean}}$ always equals $q$ (the probability of failure). Here, $q = \frac{12}{18} = \frac{2}{3}$. This means the probability of success is $p = 1 - \frac{2}{3} = \frac{1}{3}$. Since $\text{Mean} = 18$, simply multiply the mean by the reciprocal of $p$ to get $n$: $18 \times 3 = 54$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
For a binomial distribution, use the relationship between mean, variance, and parameters $n$ and $p$ to find $n$ quickly.

Step 2: Key Formula or Approach:
In a binomial distribution: Mean = $np$, Variance = $npq$, where $q = 1-p$. The ratio $\frac{\text{Variance}}{\text{Mean}} = \frac{npq}{np} = q$ gives the failure probability directly.

Step 3: Detailed Explanation:
Given Mean = 18 and Variance = 12, compute $q = \frac{12}{18} = \frac{2}{3}$. Then $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$. Using Mean = $np$, substitute: $18 = n \times \frac{1}{3}$, so $n = 18 \times 3 = 54$.

Step 4: Final Answer:
The number of trials is $n = 54$.