Question

Let 𝑋 be a random variable such that the moment generating function of 𝑋 exists in a neighborhood of zero and
\(E(X^n)=(-1)^n\frac{2}{5}+\frac{2^{n+1}}{5}+\frac{1}{5},\) n=1,2,3,…..
Then, 𝑃 (|𝑋-\(\frac{1}{2}\)|>1) equals

• A. \(\frac{1}{5}\)
• B. \(\frac{2}{5}\)
• C. \(\frac{3}{5}\)
• D. \(\frac{4}{5}\)

Answer 1

The Correct Option is D

Comparing term-by-term shows the probabilities (weights): \[ P(X=-1)=\frac{2}{5},\qquad P(X=2)=\frac{2}{5},\qquad P(X=1)=\frac{1}{5}, \] (these sum to \(1\), so they are valid probabilities). 

Now evaluate the event \(|X-\tfrac{1}{2}|>1\). Check each atom:

• For \(X=-1\): \(|-1-\tfrac12|=1.5>1\) → contributes \(P(X=-1)=\tfrac{2}{5}\).
• For \(X=1\): \(|1-\tfrac12|=0.5\not>1\) → contributes \(0\).
• For \(X=2\): \(|2-\tfrac12|=1.5>1\) → contributes \(P(X=2)=\tfrac{2}{5}\).

Therefore \[ P\!\left(|X-\tfrac12|>1\right)=P(X=-1)+P(X=2)=\frac{2}{5}+\frac{2}{5}=\frac{4}{5}. \]