How is this bound being computed in a chebyshev probability question Problem: X has mean and variance of 20.
What can be said about $P(0<X<40)$?    
Chebyshev formula = $P(|X-\mu|\geq k)\leq\frac{\sigma^2}{k^2}$    
The first step has $P(|X-20|\geq 20)\leq\frac{\sigma^2}{20\times20}$   
My question is: How did they get this part $P(|X-20|\geq 20)$?  (Did $P(0<X<40)$ turn into that???).   
 A: Let $A$ be the event $0\lt X\lt 40$. Then $A$ fails to happen if either $X\ge 40$ or $X\le 0$. 
The event "$X\ge 40$ or $X\le 0$" happens precisely if $|X-20|\ge 20$.  This is clear from the geometry. We have $X\ge 40$ or $X\le 0$ if the distance of $X$ from $20$ is $\ge 20$.
In our case, the Chebyshev Inequality tells us that $\Pr(|X-20|\ge 20)$ is $\le  \frac{20}{20^2}$, that is, $0.05$.
Remark: Informally, the Chebyshev Inequality, as you stated it, says that the sum of the probabilities in the "tails" of the distribution is "small." That is equivalent to saying that the probability of landing somewhere in the middle is large.
Going back to the event $A$, we conclude that since
$$\Pr(0\lt X\lt 40)=1-\Pr(|X-20|\le 20),$$
we can conclude that $\Pr(0\lt X\lt 40)\ge 1-0.05=0.95$.
A: Chebyshev formula is $P(|X-\mu|\geq k)\leq\frac{\sigma^2}{k^2}$. Just replace $\mu$ with mean and $k$ (that can be an arbitrary number) with $20$.
Insight on this formula: It says that it is more probable to be around the mean. As much as you get far from the mean, the probability falls.
