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:
We are given that a random variable $X$ follows a standard Binomial Distribution with parameters $n$ (total trials) and $p$ (probability of success). Given its statistical mean and variance, we need to solve for the parameter $n$.

Step 2: Key Formula or Approach:
For a binomial distribution $B(n, p)$: $\text{Mean} = np$ $\text{Variance} = npq = np(1 - p)$ where $q = 1 - p$ represents the probability of failure. By dividing the variance equation by the mean equation, we can quickly isolate $q$.

Step 3: Detailed Explanation:
From the problem parameters, we write our system of equations: $$\text{Equation 1: } np = 18$$ $$\text{Equation 2: } np(1 - p) = 12$$ Let's divide Equation 2 by Equation 1 to cancel out the $np$ terms: $$\frac{np(1 - p)}{np} = \frac{12}{18}$$ $$1 - p = \frac{2}{3}$$ Isolate $p$: $$p = 1 - \frac{2}{3} = \frac{1}{3}$$ Now substitute this value of $p = \frac{1}{3}$ back into Equation 1 to compute $n$: $$n \left(\frac{1}{3}\right) = 18$$ $$n = 18 \times 3 = 54$$

Step 4: Final Answer:
The total number of trials $n$ is $54$, which corresponds to option (A).