Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Probability axioms ▷ Law of total probability

Theorem: Let $A$ be a subset of sample space $\Omega$ and let $B_1, \ldots, B_n$ be finite or countably infinite partition of $\Omega$, such that $B_i \cap B_j = \emptyset$ for all $i \neq j$ and $\cup_i \, B_i = \Omega$. Then, the probability of the event $A$ is

$\label{eq:prob-tot} P(A) = \sum_i P(A \cap B_i) \; .$

Proof: Because all $B_i$ are disjoint, sets $(A \cap B_i)$ are also disjoint:

$\label{eq:B-disjoint} B_i \cap B_j = \emptyset \quad \Rightarrow \quad (A \cap B_i) \cap (A \cap B_j) = A \cap (B_i \cap B_j) = A \cap \emptyset = \emptyset \; .$

Because the $B_i$ are exhaustive, the sets $(A \cap B_i)$ are also exhaustive:

$\label{eq:B-exhaustive} \cup_i \, B_i = \Omega \quad \Rightarrow \quad \cup_i \, (A \cap B_i) = A \cap \left( \cup_i \, B_i \right) = A \cap \Omega = A \; .$

Thus, the third axiom of probability implies that

$\label{eq:prob-tot-qed} P(A) = \sum_i P(A \cap B_i) \; .$
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Metadata: ID: P248 | shortcut: prob-tot | author: JoramSoch | date: 2021-08-08, 03:56.