If I have a non-negative random variable $X$, using Markov inequality:
$$P\left(X\geq a\right)\leq\frac{E\left[X\right]}{a}$$
Knowing $E[X]=75$, I can ensure:
$$P\left(X\geq85\right)\leq\frac{15}{17}$$
What I'm having trouble is how does knowing $P(X\geq 61)=1$, let me make the inequality more strict? Specifically, I wish to show: $$P\left(X\geq85\right)\leq\frac{7}{12}$$
But I can't seem to figure out how to use the latter knowledge in order to tighten the ineqaulity. I tried using Chebyshev's inequality but that quite clearly doesn't give me what I wish for. I will be glad for any help/clue.
Best Answer
Let $Y=X-61.$ Then $\operatorname E(Y) = 75-61=14.$ So $$ \Pr(X\ge85) = \Pr(Y\ge 24) \le \frac{\operatorname E(Y)}{24} = \frac{14}{24}. $$