Mutually exclusive and exhaustive

by Eliezer Yudkowsky Jan 26 2016 updated Apr 26 2016

The condition needed for probabilities to sum to 1

A set of propositions is "mutually exclusive and exhaustive" when exactly one of the propositions must be true. For example, of the two propositions "The sky is blue" and "It is not the case that the sky is blue", exactly one of those must be the case. Therefore, the probabilities of those propositions must sum to exactly 1.

If a set $~$X$~$ of propositions is "mutually exclusive", this states that for every two distinct propositions, the probability that both of them will be true simultaneously is zero:

$~$\forall i: \forall j: i \neq j \implies \mathbb{P}(X_i \wedge X_j) = 0.$~$

This implies that for every two distinct propositions, the probability of their union equals the sum of their probabilities:

$~$\mathbb{P}(X_i \vee X_j) = \mathbb{P}(X_i) + \mathbb{P}(X_j) - \mathbb{P}(X_j \wedge X_j) = \mathbb{P}(X_i) + \mathbb{P}(X_j).$~$

The "exhaustivity" condition states that the union of all propositions in $~$X,$~$ has probability $~$1$~$ (the probability of at least one $~$X_i$~$ happening is $~$1$~$):

$~$\mathbb{P}(X_1 \vee X_2 \vee \dots \vee X_N) = 1.$~$

Therefore mutual exclusivity and exhaustivity imply that the probabilities of the propositions sum to 1:

$~$\displaystyle \sum_i \mathbb{P}(X_i) = 1.$~$