Probability of the complement

Statement

Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $A \in \mathcal{F}$ be an event. Let $A^c = \Omega \setminus A$ be the complement of $A$. Then $$P(A^c) = 1 - P(A).$$

Remarks

The proof: $A$ and $A^c$ are disjoint and $A \cup A^c = \Omega$, so $P(A) + P(A^c) = P(\Omega) = 1$ by additivity. With $A = \Omega$, the rule gives $P(\varnothing) = 0$. The rule is useful when $A^c$ is simpler than $A$. For example, the probability of at least one six in four rolls of a fair die is $1 - (5/6)^4 \approx 0.518$.

Explanations

L1 human

Either $A$ happens or it does not, and the two chances add up to $1$. If the chance of rain is $0.3$, then the chance of no rain is $0.7$.