Monotonicity of probability

Statement

Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $A, B \in \mathcal{F}$ be events with $A \subseteq B$. Then $P(A) \le P(B)$. More precisely, $P(B \setminus A) = P(B) - P(A)$.

Remarks

The proof: $B$ is the disjoint union of $A$ and $B \setminus A$, so $P(B) = P(A) + P(B \setminus A)$, and $P(B \setminus A) \ge 0$. With $B = \Omega$, monotonicity gives $P(A) \le 1$ for each event $A$. If $A$ implies $B$, then $B$ is at least as likely as $A$. For example, "the die shows $6$" is at most as likely as "the die shows an even number".

Explanations

L1 human

A bigger event contains more outcomes, so it cannot be less likely. The chance to draw a heart from a deck is at most the chance to draw a red card, because every heart is red.