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
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
\(E(X^n)=(-1)^n\frac{2}{5}+\frac{2^{n+1}}{5}+\frac{1}{5},\) n=1,2,3,…..
Then, 𝑃 (|𝑋-\(\frac{1}{2}\)|>1) equals
Updated On: Nov 17, 2025
- \(\frac{1}{5}\)
- \(\frac{2}{5}\)
- \(\frac{3}{5}\)
- \(\frac{4}{5}\)
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The Correct Option is D
Solution and Explanation
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 \[ P\!\left(|X-\tfrac12|>1\right)=P(X=-1)+P(X=2)=\frac{2}{5}+\frac{2}{5}=\frac{4}{5}. \]
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