# How to conclude $P[X < 2/3] < 3\epsilon$ from $0 \leq X \leq 1$ and $E[X] \geq 1-\epsilon$

I am trying to show that if $$\epsilon > 0$$ and $$X$$ is a random variable and with $$0 \leq X \leq 1$$ and $$E[X] \geq 1-\epsilon$$, then I can estimate that $$P[X < 2/3] < 3 \epsilon$$

I tried using markov's inequality and also proving the converse, but I cannot get it done.

I would appreciate any help!

Consider $$Y \equiv 1 - X$$ such that $$0 \leq Y \leq 1$$ and $$E[Y] \leq \epsilon$$. By Markov inequality $$P\left[ X < \frac23 \right] = P\left[ Y > \frac13 \right] \leq \frac{ E[Y]}{1/3} \leq 3\epsilon$$