## Two Useful Theorems from the Discrete Probability Axioms

Two useful implications follow fairly easily from the above probability axioms: the “complement” as well as the “inclusion and exclusion” statements.

## The Probability of the Complement of an Outcome

When ‘A’ is a subset of ‘S’, we say that ~A is the “complement of A”, or the set of all the outcomes in ‘S’ that were not found in ‘A’. Then p(~A) = 1 – p(A).

A sketch of the proof follows. We defined ~A by saying that the union of sets ‘A’ and ‘~A’ is the set ‘S’, and also the intersection of ‘A’ and ‘~A’ is the empty set, {}. So S = A U ~A, and Ø = A ∩ ~A.

Therefore axiom 4 applies: p(S) = p(A U ~A) = p(A) + p(~A) = 1. Therefore p(~A) = 1 – p(A).

## Inclusion or Exclusion of Outcomes

The second implication is the “principle of inclusion and exclusion”. Unlike the fourth axiom, this does not depend on mutually exclusive outcome sets. We want to know p(A_{1} U A_{2}) as a sum of their separate probabilities, without completely teasing apart each set of outcomes.

The answer is p(A_{1} U A_{2}) = p(A_{1}) + p(A_{2}) – p(A_{1} ∩ A_{2}).

Let’s sketch the proof, which is also based largely on the fourth axiom. If A_{1} ∩ A_{2}= {}, then by the first axiom, that p(Ø) = zero, the new statement is simply the fourth axiom.

The more interesting case has A_{1} ∩ A_{2} = A_{3} for some new, non-empty set A_{3}. If these sets are finite, it is easy to create mutually exclusive sets A_{a} that consists of A_{1} excluding all elements of A_{3}, and A_{b} that consists of A_{2} excluding all elements of A_{3}. So we now have mutually exclusive sets A_{a}, A_{b} and A_{3}, where A_{1} = A_{a} U A_{3} and A_{2} = A_{b} U A_{3}.

By the fourth axiom, p(A_{a} U A_{b} U A_{3}) = p(A_{a}) + p(A_{b}) + p(A_{3}), which is precisely the required probability for p(A_{1} U A_{2}).

But p(A_{1}) = p(A_{a}) + p(A_{3}), and p(A_{2}) = p(A_{b}) + p(A_{3}). Therefore p(A_{1}) = p(A_{a}) + p(A_{3}). Likewise p(A_{2}) = p(A_{b}) + p(A_{3}). So p(A_{1}) + p(A_{2}) = p(A_{a}) + p(A_{b}) + 2 * p(A_{3}).

Clearly that is one “p(A_{3})” more than the required answer. Subtracting this “p(A_{3})” from “p(A_{2}) + p(A_{2})” brings the correct result.

Therefore p(A_{1} U A_{2}) = p(A_{1}) + p(A_{2}) – p(A_{1} ∩ A_{2}), as shown for finite discrete outcome sets. The proof for infinite discrete outcome sets is similar, but depends on describing the method to determine A_{3}, A_{a} and A_{b} given the descriptions for A_{1} and A_{2}.

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