Probability of Absolute Difference Between X and E(X)

Question

I have to prove that for every real random variable $X$ with finite variance and expectation, $P \left( \lvert X - \mathbb{E}(X)\rvert \leq \sqrt{\mathbb{V}(X)}/10 \right) \neq 1$.

So far, I have tried supposing equality for contradiction, and manipulated the equation into: $P \left( \mathbb{E}(X) - \frac{\sqrt{\mathbb{E}(X^2) - \mathbb{E}(X)^2}}{10} \leq X \leq \mathbb{E}(X) + \frac{\sqrt{\mathbb{E}(X^2) - \mathbb{E}(X)^2}}{10}\right) = 1$ but I cannot figure out how to proceed from here.

Answer 1

$\mathbb{V}(X) \ge 0$ and $P \left( \lvert X - \mathbb{E}(X)\rvert \leq \dfrac{\sqrt{\mathbb{V}(X)}}{10} \right) = P \left( ( X - \mathbb{E}(X))^2 \leq \dfrac{{\mathbb{V}(X)}}{100} \right)$

and if this is $1$ then $\mathbb{V}(X) = \mathbb{E}( X - \mathbb{E}(X))^2 \leq \dfrac{{\mathbb{V}(X)}}{100} $ which is only possible if $\mathbb{V}(X) = 0$

Answer 2

It is not correct. A counterexample would be a constant random variable, e.g. $X = 1$ with probability $1$.