Absolute Value and Inequality

Question

From lecture notes:

$P(X>8)$

$=P(X-5>3)$

$\le P(|X-5|\ge 3)$

How did the prof go from 2nd step to 3rd step?

Answer 1

The events $\{X-5>3\}$ and $\{X-5\leq -3\}$ are disjoint, and their unon is (almost) $\{\lvert X-5\rvert\geq 3\}$.

$\begin{split}\mathsf P(X>8) &= \mathsf P(X>8)+0 \\ &\leq \mathsf P(X-5>3) +\mathsf P(X-5\leq -3)+\mathsf P(X=8)\\&\leq \mathsf P(\lvert X-5\rvert \geq 3)\end{split}$

or more clearly

$\begin{split}\mathsf P(X>8) &\leq \mathsf P(X>8)+(\mathsf P(X\leq 2)+\mathsf P(X=8)) \\ &\leq \mathsf P(X-5>3) +(\mathsf P(X-5\leq -3)+\mathsf P(X-5=3))\\&\leq \mathsf P(\lvert X-5\rvert \geq 3)\\[2ex]\therefore ~\mathsf P(X>8) &\leq \mathsf P(\lvert X-5\rvert\geq 3)\end{split}$