prob-exc.md

layout	mathjax	author	affiliation	e_mail	date	title	chapter	section	topic	theorem	sources	proof_id	shortcut	username
proof	true	Joram Soch	BCCN Berlin	joram.soch@bccn-berlin.de	2021-07-23 10:19:00 -0700	Probability under mutual exclusivity	General Theorems	Probability theory	Probability	Probability under exclusivity	authors year title in pages url Wikipedia 2021 Mutual exclusivity Wikipedia, the free encyclopedia retrieved on 2021-07-23 https://en.wikipedia.org/wiki/Mutual_exclusivity#Probability	P242	prob-exc	JoramSoch

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:

$$ \label{eq:prob-exc-qed} p(A \vee B) = p(A) + p(B) - p(A,B) \overset{\eqref{eq:exc}}{=} p(A) + p(B) ; . $$