Markov's Inequality

Markov’s Inequality uses Expected Value to find the upper bound of the Probability of a nonnegative Random Variable being at least $c$

$$\frac{X\text{ is RV}P(X\ge 0)=1c>0}{P(X\ge c)\le \frac{E[X]}{c}}$$

Proof

$$\begin{array}{rl}E[X]={\int }_0^{\infty }xf(x)dx\ge & {\int }_c^{\infty }xf(x)dx\ge c{\int }_c^{\infty }f(x)dx=cP(X\ge c)\\ E[X]& \ge cP(X\ge c)\\ \frac{E[X]}{c}& \ge P(X\ge c)\\ & \square \end{array}$$