Probability: Sample Spaces

I’ve been doing a few games lately (can be seen here, here and here) and, while I think those are very good ways to become interested in some of the avenues of math research, I also have had a few people come to me with questions regarding help with their classes. So I decided to write a script to try to help understand some elementary probability theory, focusing on discrete sample spaces.

Probability Image

In statistics, any process of observation is referred to as an experiment.
The set of all possible outcomes of an experiment is called the sample space and it is usually denoted by S. Each outcome in a sample space is called an element of the sample space. An event is a subset of the sample space or which the event occurs. Two events are said to be mutually exclusive if they have no elements in common.

Similar to set theory, we can form new events by performing operations like unions, intersections and compliments on other events. If A and B are any two subsets of a sample space S, then their union A ∪ B is the subset of S that contains all the elements that are in either A, in B, or in both; their intersection A ∩ B is the subset of S that contains all the elements that are in both A and B; the compliment A’ of A is the subset of S that contains all the elements of S that are not in A.

A probability is a function that assigns real numbers to events of a sample space. The following are the axioms of probability that apply when the sample space is discrete (finite or countable).

Axiom 1: The probability of an event is a non-negative real number; that is P(A) ≥ 0 for any subset A of S.
Axiom 2: The probability of the entire sample space is 1; that is P(S) = 1.
Axiom 3: If A1, A2, A3, … , is a finite or infinite sequence of mutually exclusive events of S, then
P(A1 ∪ A2 ∪ A3 ∪ …) = P(A1) + P(A2) + P(A3) + …
If A and B are any two events in a sample space S and P(A) ≠ 0, the conditional probability of B given A is

P(B | A) =
P(A ∩ B)


Two events A and B are independent if and only if P(A | B) = P(A) ∙ P(B).

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