Let $S$ be a sample space, and let $X$ and $X'$ be random variables such that $X(s) \ge X'(s)$ for all $s \in S$. Prove that for any real constant $t$,

$$ \Pr\{X \ge t\} \ge \Pr\{X' \ge t\} $$

Both of them, expanded, are:

$$ \Pr\{X \ge t\} = \sum_{s \in S:X(s) \ge t}\Pr\{s\} \\ \Pr\{X' \ge t\} = \sum_{s \in S:X'(s) \ge t}\Pr\{s\} $$

Each term of the second sum is present in the first sum, because $X(s) \ge X'(s)$. This makes them at least equal. There might be additional terms in the first sum (when $X(s) \ge t > X'(s)$. Thus it can also be greater.