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

Updated On: Oct 1, 2024

$\frac{1}{5}$
$\frac{2}{5}$
$\frac{3}{5}$
$\frac{4}{5}$

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The Correct Option is D

Solution and Explanation

The correct option is (D):

\frac{4}{5}

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Let 𝑋 be a random variable such that the moment generating function of 𝑋 exists in a neighborhood of zero andE(Xn)=(−1)n25+2n+15+15,E(X^n)=(-1)^n\frac{2}{5}+\frac{2^{n+1}}{5}+\frac{1}{5},E(Xn)=(−1)n52​+52n+1​+51​, n=1,2,3,…..Then, 𝑃 (|𝑋-12\frac{1}{2}21​|>1) equals

The Correct Option is D

Solution and Explanation

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Questions Asked in IIT JAM MS exam

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