Proof: Probability under mutual exclusivity
Index: The Book of Statistical Proofs ▷ General Theorems ▷ Probability theory ▷ Probability ▷ Probability under exclusivity
Metadata: ID: P242 | shortcut: prob-exc | author: JoramSoch | date: 2021-07-23, 17:19.
Theorem: Let $A$ and $B$ be two statements about random variables. Then, if $A$ and $B$ are mutually exclusive, the probability of their disjunction is equal to the sum of the marginal probabilities:
\[\label{eq:prob-exc} p(A \vee B) = p(A) + p(B) \; .\]Proof: If $A$ and $B$ are mutually exclusive, then their joint probability is zero:
\[\label{eq:exc} p(A,B) = 0 \; .\]The addition law of probability states that
\[\label{eq:prob-add-set} p(A \cup B) = p(A) + p(B) - p(A \cap B)\]which, in logical rather than set-theoretic expression, becomes
\[\label{eq:prob-add-log} p(A \vee B) = p(A) + p(B) - p(A,B) \; .\]Because the union of mutually exclusive events is the empty set and the probability of the empty set is zero, the joint probability term cancels out:
\[\label{eq:prob-exc-qed} p(A \vee B) = p(A) + p(B) - p(A,B) \overset{\eqref{eq:exc}}{=} p(A) + p(B) \; .\]∎
Sources: - Wikipedia (2021): "Mutual exclusivity"; in: Wikipedia, the free encyclopedia, retrieved on 2021-07-23; URL: https://en.wikipedia.org/wiki/Mutual_exclusivity#Probability.
Metadata: ID: P242 | shortcut: prob-exc | author: JoramSoch | date: 2021-07-23, 17:19.