Axioms and Two Useful Theorems of Discrete Probability Functions

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Image of a Selection of Bingo Balls by sjsharktank

Bingo is an exercise in probability. Image by sjsharktank

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(A1 U A2) as a sum of their separate probabilities, without completely teasing apart each set of outcomes.

The answer is p(A1 U A2) = p(A1) + p(A2) – p(A1 ∩ A2).

Let’s sketch the proof, which is also based largely on the fourth axiom. If A1 ∩ A2= {}, then by the first axiom, that p(Ø) = zero, the new statement is simply the fourth axiom.

Picture of Four Random Dice by dullhunk

Use dice to experiment with probability. Image by dullhunk

The more interesting case has A1 ∩ A2 = A3 for some new, non-empty set A3. If these sets are finite, it is easy to create mutually exclusive sets Aa that consists of A1 excluding all elements of A3, and Ab that consists of A2 excluding all elements of A3. So we now have mutually exclusive sets Aa, Ab and A3, where A1 = Aa U A3 and A2 = Ab U A3.

By the fourth axiom, p(Aa U Ab U A3) = p(Aa) + p(Ab) + p(A3), which is precisely the required probability for p(A1 U A2).

But p(A1) = p(Aa) + p(A3), and p(A2) = p(Ab) + p(A3). Therefore p(A1) = p(Aa) + p(A3). Likewise p(A2) = p(Ab) + p(A3). So p(A1) + p(A2) = p(Aa) + p(Ab) + 2 * p(A3).

Clearly that is one “p(A3)” more than the required answer. Subtracting this “p(A3)” from “p(A2) + p(A2)” brings the correct result.

Therefore p(A1 U A2) = p(A1) + p(A2) – p(A1 ∩ A2), as shown for finite discrete outcome sets. The proof for infinite discrete outcome sets is similar, but depends on describing the method to determine A3, Aa and Ab given the descriptions for A1 and A2.

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