Let there be $0< \epsilon \leq 1$ and let there be $X$ a random variable that acts in the following way:
$P(x=1) = P(x = -1) = (\epsilon^2)/2$, $P(x=0) = 1-\epsilon^2$
Calculate: $P(|X-E[x]|<2(Var[X])^{0.5})$
So we can either calculate by hand or claim that the distribution is symmetric around $0$
So $E[X] = 0$.
$Var(X) = \sigma^2 =\sum X^2P(x)-0 = \epsilon^2$ and therefore $\sigma = \epsilon$
Now I'm kind of stuck. How to proceed from here? In other words I need to calculate $P(|X|<2\sigma)$
Whereas $0< \epsilon \leq 1$
There's a photo with a better-looking explanation PHOTO
$$|X| = \begin{cases} 1 & \text{ w. p. } \epsilon^2 \\ 0 & \text{w.p. } 1-\epsilon^2\end{cases}$$
$$P(|X| < 2 \epsilon)=\begin{cases} 1 & ,\epsilon > \frac12 \\ 1-\epsilon^2 & , \epsilon\leq \frac12\end{cases}$$
Explanation:
If $\epsilon > \frac12$, $2 \epsilon > 1 \geq |X|.$
If $0 <\epsilon \leq \frac12$, $0 < 2\epsilon < 1$, only $|X|=0$ satisfies the condition.