Question:
Let π be a random variable such that the moment generating function of π 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,β¦..
Then, π (|π-\(\frac{1}{2}\)|>1) equals
Let π be a random variable such that the moment generating function of π 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,β¦..
Then, π (|π-\(\frac{1}{2}\)|>1) equals
\(E(X^n)=(-1)^n\frac{2}{5}+\frac{2^{n+1}}{5}+\frac{1}{5},\) n=1,2,3,β¦..
Then, π (|π-\(\frac{1}{2}\)|>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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