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: \[ P(0) = \binom{4}{0}\left(\frac{1}{2}\right)^4 = \frac{1}{16}, \quad P(1) = \binom{4}{1}\left(\frac{1}{2}\right)^4 = \frac{4}{16} \] \[ P(X>1) = 1 - (\frac{1}{16} + \frac{4}{16}) = \frac{11}{16} \]

Answer 2

Step 1: Understand the problem
We have a binomial random variable \(X\) with mean \(\mu = 2\) and variance \(\sigma^2 = 1\). We need to find \(P(X > 1)\).

Step 2: Use properties of binomial distribution
For a binomial variable with parameters \(n\) and \(p\):
\[ \mu = np = 2, \quad \sigma^2 = np(1-p) = 1 \]

Step 3: Find \(n\) and \(p\)
From the mean: \(np = 2\)
From the variance: \(np(1-p) = 1\)
Divide variance by mean:
\[ \frac{np(1-p)}{np} = 1 - p = \frac{1}{2} \implies p = \frac{1}{2} \]
Using \(p = \frac{1}{2}\) in mean:
\[ n \times \frac{1}{2} = 2 \implies n = 4 \]

Step 4: Write the probability expression
We want \(P(X > 1) = 1 - P(X \leq 1) = 1 - [P(X=0) + P(X=1)]\).

Step 5: Calculate \(P(X=0)\) and \(P(X=1)\)
Using binomial formula:
\[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Calculate:
\[ P(X=0) = \binom{4}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^4 = 1 \times 1 \times \frac{1}{16} = \frac{1}{16} \]
\[ P(X=1) = \binom{4}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^3 = 4 \times \frac{1}{2} \times \frac{1}{8} = \frac{4}{16} = \frac{1}{4} \]

Step 6: Calculate \(P(X > 1)\)
\[ P(X > 1) = 1 - \left(\frac{1}{16} + \frac{1}{4}\right) = 1 - \frac{1 + 4}{16} = 1 - \frac{5}{16} = \frac{11}{16} \]

Final answer: \(\displaystyle P(X > 1) = \frac{11}{16}\)