Finding lower bound of probability with Chebyshev's inequality

Question

I know that Chebyshev's inequality is $$\Bbb P(|X-\mu|\ge c)\le\frac{\sigma^2}{c^2}$$ for all $c\ge0$, and in the given problem, $\Bbb E[X]=7$ and $\sigma^2=9$. The problem asks to find a lower bound for the probability $$\Bbb P(4<X<10)$$

I have just been introduced to Chebyshev's inequality and am wondering how exactly to apply it in this case?

Answer 1

Proceed like this: $$\Bbb{P}(4<X<10) = \Bbb{P}(|X-\mu|<3) = 1-\Bbb{P}(|X-\mu|\geq3)\geq1-\frac{\sigma^2}{c^2}=1-\frac99=0$$