Understanding Bernoulli's Theorem in games of chance Statement :- Let $A$ denote an event whose probability of occurrence in a single trial is $p$. If $k$ denotes the number of occurrences of $A$ in $n$ independent trials, then 
$$P\left(\left|\frac kn - p\right|> \epsilon\right) \lt \frac{pq}{n \epsilon ^2}$$
Someone please help me understanding this
I understood the proof of the inequality but I am not able to infer things from it.
More info :
I am studying the subject by myself and I reached here.
I am studying from Probability, Random Variables and Stochastic processes (By Athanasios Papoulis and S. Unnikrishna Pillai) and I dont know any generalized theorems of above...
edit : What the textbook has said : The inequality above states that the frequency definition of probability of an event $\frac kn$ and its axiomatic definition p can be made compatible to any degree of accuracy with probability 1 or with almost certainity.
In other words, given two positive numbers $\epsilon$ and $\delta$ the probability of the inequality will be greater than $1-\delta$, provided the number of trials is above a certain limit
 A: "In $n$ independent trials" implies that this "$k$" is a binomial distribution, say $X\sim \text{Bin}(n, p)$. Recall that $\text{Var}(X) = np(1-p) = npq$, and so 
$$\text{Var}\left[\frac{X}{n}\right] = \frac{1}{n^2}\text{Var}(X) = \frac{pq}{n}.$$
Thus, applying Chebyshev's inequality, we get
$$P\left[\left|\frac{X}{n}-p\right|>\epsilon\right] \leq \frac{\text{Var}(X/n)}{\epsilon^2} = \frac{pq}{n\epsilon^2}.$$

Addendum: Don't know Chebyshev's inequility? Then let me be "clever" and call $Y = \frac{X}{n}$, and notice that
\begin{align*}
P(|Y - p| \geq \epsilon)&=
E[I\{|Y - p| \geq \epsilon\}]\\
&=E\left[I\left\{\left(\frac{Y-p}{\epsilon}\right)^2\geq 1\right\}\right]\\
&\leq E\left[\left(\frac{Y-p}{\epsilon}\right)^2\right]\\
&= \frac{E[(Y-p)^2]}{\epsilon^2}\\
&=\frac{\text{Var}(Y)}{\epsilon^2}\\
&= \frac{pq}{n\epsilon^2},
\end{align*}
where $I$ is an indicator. The reality is that I was not clever and I simply adapted the many proofs of Chebyshev's inequality to show the result.
What your textbook says is an interpretation of the inequality. This is the "proof" of the result, if you want to call it that. Were you supposed to simply accept this without proof or derive it yourself? I don't know, but here it is.
A: Chebychev's inequality says the upper bound is $\frac{1}{\epsilon^2}$ for $\epsilon$ standard deviations in  $P((|k-\mu|)\ge \epsilon\sigma) = P(|\frac{k}{n}-p|\ge\epsilon\sqrt{\frac{pq}{n}})$.  The standard deviation here is $\sqrt{npq}$ and $\mu = np$. For the following expression $\left(\epsilon\sqrt{\frac{pq}{n}}\right)$ standard deviations, the upper bound   still is  $\frac{1}{\epsilon^2}$. But what you want is $\epsilon$.  Now take the entire remainder of $\left(\frac{\sqrt{n}\epsilon}{\sqrt{pq}}\right)\sqrt{\frac{pq}{n})}$ besides $\sqrt{\frac{pq}{n}}$ as $\epsilon'$ and apply the normal chebyshev to find the upper bound as $\frac{1}{\epsilon'^2}$,  which when you manipulate gives$\frac{1}{\left(\frac{\sqrt{n}\epsilon}{\sqrt{pq}}\right)^2}$.  If you do some rearrangement , you get the  desired result.
