Index: The Book of Statistical Proofs ▷ General Theorems ▷ Probability theory ▷ Probability axioms ▷ Probability of exhaustive events

Theorem: Let $B_1, \ldots, B_n$ be mutually exclusive and collectively exhaustive subsets of a sample space $\Omega$. Then, their total probability is one:

\[\label{eq:prob-exh} \sum_i P(B_i) = 1 \; .\]

Proof: Because all $B_i$ are mutually exclusive, we have:

\[\label{eq:B-exclusive} B_i \cap B_j = \emptyset \quad \text{for all} \quad i \neq j \; .\]

Because the $B_i$ are collectively exhaustive, we have:

\[\label{eq:B-exhaustive} \cup_i \, B_i = \Omega \; .\]

Thus, the third axiom of probability implies that

\[\label{eq:prob-exh-s1} \sum_i P(B_i) = P(\Omega) \; .\]

and the second axiom of probability implies that

\[\label{eq:prob-exh-s2} \sum_i P(B_i) = 1 \; .\]
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Metadata: ID: P249 | shortcut: prob-exh | author: JoramSoch | date: 2021-08-08, 04:10.