Question

If the mean and variance of a Binomial variate \(X\) are 2 and 1 respectively, then the probability that \(X\) takes a value greater than 1 is:

• A. \(\frac{2}{3}\)
• B. \(\frac{4}{3}\)
• C. \(\frac{7}{8}\)
• D. \(\frac{11}{16}\)

Hint:

For Binomial distributions, use \(P(X > k) = 1 - P(X \leq k)\).

Answer 1

The Correct Option is D

Given that the mean \(\mu = 2\) and variance \(\sigma^2 = 1\) for a binomial distribution, we know:
\[ \mu = np = 2 \quad \text{and} \quad \sigma^2 = np(1-p) = 1. \]
From \(\mu = np = 2\), we can express \(p\) in terms of \(n\):
\[ p = \frac{2}{n}. \]
Substituting into the variance equation:
\[ np(1 - p) = 1 \quad \Rightarrow \quad n \cdot \frac{2}{n} \cdot \left( 1 - \frac{2}{n} \right) = 1. \]
Simplifying:
\[ 2 \left( 1 - \frac{2}{n} \right) = 1 \quad \Rightarrow \quad 2 - \frac{4}{n} = 1 \quad \Rightarrow \quad \frac{4}{n} = 1 \quad \Rightarrow \quad n = 4. \]
Substitute \(n = 4\) into \(p = \frac{2}{n}\):
\[ p = \frac{2}{4} = \frac{1}{2}. \]
Now, we need to calculate \(P(X > 1)\). We know that:
\[ P(X > 1) = 1 - P(X \leq 1). \]
For \(X \sim B(4, \frac{1}{2})\), we can compute \(P(X \leq 1)\) as:
\[ P(X \leq 1) = P(X = 0) + P(X = 1). \]
Using the binomial probability mass function:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}. \]
Thus,
\[ P(X = 0) = \binom{4}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^4 = \frac{1}{16}, \quad P(X = 1) = \binom{4}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^3 = \frac{4}{16}. \]
Therefore,
\[ P(X \leq 1) = \frac{1}{16} + \frac{4}{16} = \frac{5}{16}. \]
Finally,
\[ P(X > 1) = 1 - \frac{5}{16} = \frac{11}{16}. \]
Thus, the probability that \(X\) takes a value greater than 1 is \(\frac{11}{16}\).