Calculating $P(|X-E(X)|<\epsilon)$ using Chebyshev inequality for discrete random variable

Suppose that $X$ is number of failure until getting the $r$-th success in infinite sequence of Bernoulli trails with probability $p$ for success.
I have calculated the following results:
$EX=\frac{r(1-p)}{p}$
$Var(X)=\frac{r(1-p)}{p^2}$

Now I want to find a lower bound for the next probability $P(\frac{r(1-p)}{p}-\epsilon<X<\frac{r(1-p)}{p}+\epsilon)$ using Chebyshev inequality (which states that: $P(|X-E(X)|\geq\delta)\leq\frac{\operatorname{Var}(X)}{\delta^2}$)

From The results above and using the inequality I get that $P(\frac{r(1-p)}{p}-\epsilon<X<\frac{r(1-p)}{p}+\epsilon)=P(|X-E(X)|<\epsilon)=$
$1-P(|X-E(X)|\geq\epsilon)\geq1-\frac{r(1-p)}{p^2\epsilon^2}$
Here I got stuck and couldn't continue for finding a good lower bound.

I know that X is a discrete random variable so $P(X\leq x_i)=\sum_{x\leq x_i}P(X=x_i)$.

Can anyone give me an answer or hint how can I continue from here?

Thank you, Michael

• If you need a better lower bound, you may need other inequalities like Chernoff's. – poyea May 22 '18 at 14:13
• Your $X$ is negative binomial, so you may search for bounds of negative binomial distribution. – StubbornAtom May 22 '18 at 14:41