We can represent the probability that a random variable deviates from its mean by more than \(k \sigma\) like this:
$$ P(|X - \mu| \ge k\sigma) = P((X - \mu)^2 \ge k^2\sigma^2) $$
We get an upper bound using Markov's Inequality here:
\[ P((X - \mu)^2 \ge k^2\sigma^2) \le \frac{E[(X- \mu)^2]}{k^2\sigma^2} \]
Simplify:
$$ P(|X - \mu| \ge k\sigma) \le \frac{\sigma^2}{k^2\sigma^2} $$$$ P(|X - \mu| \ge k\sigma) \le \frac{1}{k^2} $$