6  Axioms of Probability

In this section we will take a detailed look at a situation in which probabilities are not equally likely, and develop some of the key concepts which are used to analyze the probability in the general case.

Example: A large HMO is planning for future expenses. One component of their planning is a study of the percentage of births which involve more than one child twins, triplets or more. The study leads to the following table:

Number of children 1 2 3
Percent of all births 96.70% 3.11% 0.19%

How will the company assign probabilities to multiple births for future planning?

Solution: The table shows that the individual outcomes are not equally likely. We can use the numbers as probability of individual outcomes: \[P(\text{births} = 1) = .9670 \qquad P(\text{births} = 2) = .0311 \qquad P(\text{births} = 3) = .0019\] We can verify that the sum of probabilities of all individual outcomes is 1, and we can also calculate the probability of any event happening.

6.1 Assigning Probabilities to a Finite Sample Space

We can assign probabilities to events in any finite sample space with \(n\) individual outcomes denoted by \(O_1,O_2,...,O_n\). 1. Assign a probability \(P(O_i) \ge 0\) to each individual outcome \(O_i\), while ensuring that \(\sum_{i=1}^n P(O_i) = 1\) 2. Define the probability of any event \(E\) as: \[P(E) = \sum_{O_i \in E} P(O_i)\] ## The General Definition of Probability Not all sample spaces are finite or as easy to handle. If we define a way to assign a probability \(P(E)\) to any event \(E\), the following axioms should be satisfied: - \(P(E) \ge 0\) for any event E - \(P(\Omega) = 1\) - Suppose \(E_1, ..., E_n, ...\) is a (possibly infinite) sequence of events in which each pair of events is mutually exclusive. Then \[P\bigg(\bigcup_{i=1}^\infty E_i\bigg) = \sum_{i=1}^\infty P(E_i)\] Then, the set function \(P\) is called a probability measure. This set of axioms is called the Kolmogorov axioms of Probability.

One could verify that the rules derived for equally likely outcomes can be shown to hold for any probability assignment that satisfies the axioms. ## Properties of the Probabiliity Measure Proposition: Let \(P: \Omega \to [0,1]\) be a probability measure, \(A, B\) be events of the sample space \(\Omega\). Then: * \(P(A^c) = 1 - P(A)\) * \(P(A \cup B) = P(A) + P(B) - P(A\cap B)\) * \(P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A\cap B) - P(B\cap C) - P(A\cap C) + P(A \cap B \cap C)\)