Use Chebyshev's inequality to get an upper bound on the probability $\displaystyle P\Big[|X-E[X]|>2\sqrt{Var[X]}\Big]$, and also calculate the exact probability.

X is the result of a fair dice toss, 1,2,3,4,5,6, each with a probability of $\displaystyle \frac{1}{6}$

I'm not sure what it means by "get an upper bound on the probability."

This is what I did, and it matches the manual's solution of what the "exact probability" is:

$\displaystyle E[X]=\frac{1+2+3+4+5+6}{6}=\frac{7}{2}$

$\displaystyle E[X^2]=\frac{1+4+9+16+25+36}{6}=\frac{91}{6}$

$\displaystyle Var[X]=E[X^2]-E[X]^2=\frac{91}{6}-\frac{49}{4}=\frac{35}{12}$

$\displaystyle P\left[|X-\frac{7}{2}|>2\sqrt{\frac{35}{12}}\right]=$$\displaystyle 1-P[0.0843\leq X \leq 6.916]=1-P[1,2,3,4,5,6]=1-1=0$

$\displaystyle 0\leq \frac{1}{r^2}=0\leq\frac{1}{4}$