Question

If the mean and varian of a binomial variable $X$ are 2 and 1 respectively, then $P(X>1) = $

• A. $\frac{11}{32}$
• B. $\frac{1}{8}$
• C. $\frac{11}{12}$
• D. $\frac{11}{16}$

Hint:

Binomial Setup from Moments. Use $np$ and $np(1-p)$ to find $n$ and $p$ when given mean and variance.

Answer 1

The Correct Option is D

We know for a binomial distribution $X \sim B(n, p)$: \[ np = 2, \quad np(1 - p) = 1 \Rightarrow p = \frac{2}{n} \] Substitute into variance: \[ np(1-p) = 2(1 - \frac{2}{n}) = 1 \Rightarrow \frac{4}{n} = 1 \Rightarrow n = 4, \; p = \frac{1}{2} \] So, $X \sim B(4, \frac{1}{2})$. \[ P(X>1) = 1 - P(X \leq 1) = 1 - [P(0) + P(1)] \] Using binomial formula: \[ P(0) = \binom{4}{0}\left(\frac{1}{2}\right)^4 = \frac{1}{16}, \quad P(1) = \binom{4}{1}\left(\frac{1}{2}\right)^4 = \frac{4}{16} \] \[ P(X>1) = 1 - (\frac{1}{16} + \frac{4}{16}) = \frac{11}{16} \]