If $X$ is a nonnegative random variable and $a > 0$, then \(\Pr[X \ge a] \le \frac{E[X]}{a}.\)
Proof. Since $X \ge a$ on the event ${X \ge a}$, we have $E[X \mid X \ge a] \ge a$. Since $X \ge 0$ on the event ${X < a}$, we have $E[X \mid X < a] \ge 0$. Therefore, \(E[X] = \Pr[X < a] \cdot E[X \mid X < a] + \Pr[X \ge a] \cdot E[X \mid X \ge a] \ge \Pr[X \ge a] \cdot a.\) Dividing both sides by $a$ completes the proof. $\square$