# How to use Chebyshev's inequality to find a lower bound?

I have that $$\text{var}(X) = \frac13v^2$$ and I want to find the lower bound for $$P(|X| \le v)$$

I tried doing the following: $$P(|X| \le v) = 1 - P(|X| \ge v)$$ so using Chebyshev's inequality I have that

$$1 - P(|X| \ge v) \le \frac13$$

$$\frac23 \le P(|X| \ge v)$$,

but I don't know what can I say about $$P(|X| \le v)$$ since everything is now on terms of $$P(|X| \ge v)$$.

You are making a mistake in using Chebyshev's inequality. What you get $$P(|X| \geq v) \leq 1/3$$ (assuming that the mean is $$0$$) and not $$1-P(|X| \geq v) \leq 1/3$$