Index: The Book of Statistical Proofs ▷ General Theorems ▷ Probability theory ▷ Probability axioms ▷ Probability of the complement

Theorem: The probability of a complement of a set is one minus the probability of this set:

\[\label{eq:prob-comp} P(A^\mathrm{c}) = 1 - P(A)\]

where $A^\mathrm{c} = \Omega \setminus A$ and $\Omega$ is the sample space.

Proof: Since $A$ and $A^\mathrm{c}$ are mutually exclusive and $A \cup A^\mathrm{c} = \Omega$, the third axiom of probability implies:

\[\label{eq:pAAc} \begin{split} P(A \cup A^\mathrm{c}) &= P(A) + P(A^\mathrm{c}) \\ P(\Omega) &= P(A) + P(A^\mathrm{c}) \\ P(A^\mathrm{c}) &= P(\Omega) - P(A) \; . \end{split}\]

The second axiom of probability states that $P(\Omega) =1$, such that we obtain:

\[\label{eq:prob-comp-qed} P(A^\mathrm{c}) = 1 - P(A) \; .\]
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Metadata: ID: P245 | shortcut: prob-comp | author: JoramSoch | date: 2021-07-30, 12:14.