Probability of Union of Disjoint Events is Sum of Individual Probabilities

Theorem

Let $\EE$ be an experiment.

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability measure on $\EE$.

Then:

$\forall A, B \in \Sigma: A \cap B = \O \implies \map \Pr {A \cup B} = \map \Pr A + \map \Pr B$

Proof

From the Kolmogorov Axioms:

$\ds \map \Pr {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map \Pr {A_i}$

where $\set {A_1, A_2, \ldots}$ is a countable set of pairwise disjoint events of $\EE$.

This applies directly to $\map \Pr {A \cup B}$ where $A \cap B = \O$.

$\blacksquare$

Sources

• 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $1$: Events and probabilities: $1.3$: Probabilities