---
tag: 0PQEBTYDZM
kind: statement
role: proposition
name: Probability of the complement
aliases: [complement-rule]
msc: [60A05]
uses: [0ZYBZ34RT2, 00SQGJNRTB, 0B1FSS1VZW]
sources:
  - url: https://en.wikipedia.org/wiki/Complementary_event
    locator: Complement rule
---

## Statement

Let $(\Omega, \mathcal{F}, P)$ be a [[probability space|0ZYBZ34RT2]] and let $A \in \mathcal{F}$ be an [[event|00SQGJNRTB]].
Let $A^c = \Omega \setminus A$ be the [[complement|0B1FSS1VZW]] 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 $0.7$.