Question

The mean of a binomial variate \( X \sim B(n, p) \) is 1. If \( n>2 \) and \( P(X = 2) = \frac{27}{128} \), then the variance of the distribution is:}

• A. \( \frac{3}{4} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{4}{3} \)
• D. 4 \bigskip

Hint:

For a binomial distribution, calculate the variance using \( \text{Var}(X) = np(1-p) \). Here, the given mean \( np = 1 \) helps determine \( p \) in terms of \( n \), and then trial values are used to find the appropriate \( n \).

Answer 1

The Correct Option is A

We are given that the random variable \( X \) follows a binomial distribution \( X \sim B(n, p) \) with the following properties: The mean is \( E(X) = np = 1 \). The probability \( P(X = 2) = \frac{27}{128} \). 

Step 1: Determine \( p \) using the Mean 
Since the mean of a binomial distribution is \( E(X) = np \), we have: \[ np = 1 \quad \Rightarrow \quad p = \frac{1}{n}. \] 

Step 2: Use \( P(X = 2) \) 
For a binomial distribution, the probability of exactly 2 successes is: \[ P(X = 2) = \binom{n}{2} p^2 (1-p)^{n-2}. \] Substitute \( p = \frac{1}{n} \) and the given \( P(X = 2) = \frac{27}{128} \): \[ \binom{n}{2} \left(\frac{1}{n}\right)^2 \left(1 - \frac{1}{n}\right)^{n-2} = \frac{27}{128}. \] Using the binomial coefficient \( \binom{n}{2} = \frac{n(n-1)}{2} \), the equation becomes: \[ \frac{n(n-1)}{2} \cdot \frac{1}{n^2} \cdot \left(1 - \frac{1}{n}\right)^{n-2} = \frac{27}{128}. \] Simplify to obtain: \[ \frac{(n-1)}{2n} \cdot \left(1 - \frac{1}{n}\right)^{n-2} = \frac{27}{128}. \] 

Step 3: Solve for \( n \) 
Due to the complexity of the expression, we test possible values for \( n \). By substituting \( n = 4 \), we find that: \[ P(X = 2) = \frac{27}{128}, \] confirming that \( n = 4 \) satisfies the condition. 

Step 4: Compute the Variance 
The variance of a binomial distribution is given by: \[ \text{Var}(X) = np(1-p). \] Using \( n = 4 \) and \( p = \frac{1}{4} \) (since \( p = \frac{1}{n} \)), we have: \[ \text{Var}(X) = 4 \times \frac{1}{4} \times \left(1 - \frac{1}{4}\right) = 1 \times \frac{3}{4} = \frac{3}{4}. \] Thus, the variance of \( X \) is: \[ \boxed{\frac{3}{4}}. \]