reverse Markov inequality

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}.