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This question was given to me as a review for an upcoming exam.

Let X be a positive random variable with mean 3 and variance $\frac{1}{4}$. Use Chebyshev's and Markov's inequalities to obtain the probability that $X \geq 6$.

Also give an example of such an X

My work thus far:

Chebyshev's

$P(|X-\overline{x}| \geq a) \leq \frac{\sigma^{2}}{a^2} $

$P(X -3 \geq6-3) \leq \frac{\frac{1}{4}}{3^2} = \frac{1}{36}$

Markov's

$P(X \gt a) \leq \frac{E[X]}{a}$

$P(X \gt 6) \leq \frac{3}{6}= \frac{1}{2}$

I feel as though I've made an error somewhere, could someone help me figure out where?

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Your expression for Chebyshev's two sided inequality perhaps should have been $$P(|X-\mu| \geq a) \leq \frac{\sigma^{2}}{a^2}$$ and would have led to an upper bound of $\frac1{36}$, slightly higher than you have. What you have actually used looks like a one sided version $$P(X-\mu \geq a) \leq \frac{\sigma^{2}}{\sigma^{2}+a^2}$$ for $a \gt 0$, and your result of $\frac1{37}$ as an upper bound would then appear to be correct. There would be equality if $P(X=6)=\frac1{37}$ and $P\left(X=\frac{35}{12}\right)=\frac{36}{37}$

Your use of Markov's inequality for a non-negative random variable and $a \gt 0$ and your result of $\frac12$ as an upper bound both appear to be correct. There would be equality if $P(X=6)=\frac1{2}$ and $P\left(X=0\right)=\frac{1}{2}$, though this would have a variance of $9$, rather higher than that given in the question, and showing that taking the variance into account here gives a tighter bound through Chebyshev's one sided inequality

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