Index: The Book of Statistical ProofsGeneral 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: The addition law of probability states that for two events $A$ and $B$, the probability of at least one of them occurring is:

$\label{eq:prob-add} P(A \cup B) = P(A) + P(B) - P(A \cap B) \; .$

Recursively applying this law to the events $B_1, \ldots, B_n$, we have:

$\label{eq:prob-all-s1} \begin{split} P(B_1 \cup \ldots \cup B_n) &= P(B_1) + P(B_2 \cup \ldots \cup B_n) - P(B_1 \cap [B_2 \cup \ldots \cup B_n]) \\ &= P(B_1) + P(B_2) + P(B_3 \cup \ldots \cup B_n) - P(B_2 \cap [B_3 \cup \ldots \cup B_n])- P(B_1 \cap [B_2 \cup \ldots \cup B_n]) \\ &\;\; \vdots \\ &= P(B_1) + \ldots + P(B_n) - P(B_1 \cap [B_2 \cup \ldots \cup B_n]) - \ldots - P(B_{n-1} \cap B_n) \\ P(\cup_i^n \, B_i) &= \sum_i^n P(B_i) - \sum_i^{n-1} P(B_i \cap [\cup_{j=i+1}^n B_j]) \\ &= \sum_i^n P(B_i) - \sum_i^{n-1} P(\cup_{j=i+1}^n [B_i \cap B_j]) \; . \end{split}$

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

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

Since the probability of the empty set is zero, this means that the second sum on the right-hand side of \eqref{eq:prob-all-s1} disappears:

$\label{eq:prob-all-s2} P(\cup_i^n \, B_i) = \sum_i^n P(B_i) \; .$

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

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

Since the probability of the sample space is one, this means that the left-hand side of \eqref{eq:prob-all-s2} becomes equal to one:

$\label{eq:prob-all-s3} 1 = \sum_i^n P(B_i) \; .$

This proofs the statement in \eqref{eq:prob-exh}.

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Metadata: ID: P319 | shortcut: prob-exh2 | author: JoramSoch | date: 2022-03-27, 23:14.