Index: The Book of Statistical Proofs ▷ General Theorems ▷ Probability theory ▷ Probability ▷ Probability under exclusivity

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:

\[\label{eq:prob-exc-qed} p(A \vee B) = p(A) + p(B) - p(A,B) \overset{\eqref{eq:exc}}{=} p(A) + p(B) \; .\]
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Metadata: ID: P242 | shortcut: prob-exc | author: JoramSoch | date: 2021-07-23, 17:19.