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So for this question how to find the tighter bound using markov's inequality. How do we do the question if $X \geq 2000$.

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If $X\geqslant b$ almost surely, then Markov inequality applied to $Y=X-b$ and $y=x-b$ with $x\gt b$ yields $P[Y\geqslant y]\leqslant E[Y]/y$, that is, $P[X\geqslant x]\leqslant(E[X]-b)/(x-b)$. Note that the upper bound is the usual Markov upper bound $E[X]/x$ when $b=0$.

For $E[X]=1000$, $b=500$ and $x=2000$, Markov inequality yields $P[X\geqslant x]\leqslant1000/2000=1/2$ while the modified version yields $P[X\geqslant x]\leqslant500/1500=1/3$.

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