Determining a probability of a random variable using its variance and expected value. Given a random variable $X : \mathbb R \to S$ where $S \subset \mathbb R$, its expected value $\mathbb E[X] = 10$ and its variance $var(X) = 20$ how can one prove that $\mathbb P(X > 1) > \frac{3}{4}$.
Trying to split the space $\mathbb R$ into two parts $]-\infty, 1]$ and $]1, \infty[$ doesn't seem to work and neither the inequality of Markov nor the inequality of Chebyshev seem to reduce the problem as $X$ isn't necessarily positive.

Chebyshev tells us that $$P\left(|X-10|≥k\sigma\right)≤\frac 1{k^2}$$  Here, $\sigma=\sqrt {20}\sim 4.472135955$ so, taking $k=\frac 9{\sigma}=2.01246118$ the relevant expression is $$P(|X-10|≥9)≤ 0.24691358$$  Your result follows from this.
