Let $X$ be a random variable that satisfies $\Pr(X \leq a) = 1$ for some constant $a$ Then, for $d < E[X]$ $$ \Pr( X > d) \geq \frac{E[X] - d}{a - d} $$

Proof: Apply the Markov's inequality to the random variable $\tilde{X} = a-X$ $$\Pr(X\leq d) = \Pr(\tilde{X}\geq a-d) \leq \frac{E[\tilde{X}]}{a-d} = \frac{a-E[X]}{a-d}. $$

Hence $$ \Pr(X> d) \geq 1 - \frac{a-E[X]}{a-d} = \frac{E[X]-d}{a-d}. $$