Coin binomial. Proportion p defective, (1-p) fair. We take samples of n coins: $X_n$ random RV equal to the number of defective coins.
Show that for any $\epsilon > 0$ $P(|\frac {X_n} {n} - p| \geq \epsilon) \leq \frac {1} {4n\epsilon^2}$
I tried showing it using a form of Chebyshev's inequality:
$P({|X - \mu| \geq k}) \leq \frac {\sigma^2} {k^2}$
By multiplying by n, and rewriting based on E[X_n]:
$P(|X_n - \mu| \geq \epsilon n) \leq \frac {(np(1-p))^2} {(\epsilon n)^2}$, simplified to:
$P(|X_n - \mu| \geq \epsilon n) \leq \frac {(p(1-p))^2} {\epsilon ^2}$, but from here I have no idea how to get to the following step.
Thought about CLT as well, but couldn't see the 1/4 coming up.
Variance of binomial is $np(1-p)$ which shouldn't be squared. That will give you the $1/n$ factor, and you can get the $1/4$ factor by maximizing $p(1-p)$ over $p\in (0,1).$