Let $ X\in {L}^2$ be a real-valued random variable with mean $ \mu=\mathbb{E}[X]$ and variance $ \sigma^2=\operatorname{Var} [X]$ Then for any standard of accuracy $t>0$ $$ \Prob{}{\left| X - \mu \right| \ge t} \le \frac{\sigma^2}{t^2}. $$

Note: There is another Chebyshev's inequality, which is unrelated.