When we talk about discrete probability we mean the set of all possible outcomes of an experiment, the sample space, is countable. We use the notation \(\Omega\) for the sample space, and \(\omega\) for an individual outcome. An example will help illustrate this.

Events

Example 1.
On a roll of a die the set of outcomes, the sample space, is the integers from one to six.

$$\Omega = \{1,2,3,4,5,6\}$$
We can create subsets of \(\Omega\) e.g. the set of odd numbers is $$A_0 = \{1,3,5\}$$
In probability these sets are known as events.

An event is a subset of the sample space, \( \Omega \)

Example 2.

\(A_1\) is the set of all even numbers (on a roll of a die) $$A_1 = \{2,4,6\}$$
\(A_2\) is the set of numbers greater than or equal to two, $$A_2 = \{2,3,4,5,6\}$$
\(A_3\) is the set of all even numbers greater than or equal to five, $$A_3 = \{6\}$$
We can combine events using set operations.

Union:
$$A_0 \cup A_2 = \left\{1,2,3,4,5,6\right\}$$

Intersection:
$$A_0 \cap A_2 = \left\{3,5\right\}$$

Probability

We use the notation \(P(A)\) to mean the probability that an individual outcome, \( \omega \), in the set A occurs. We will leave the formal definition of probability to the next course, but at an informal level if an event cannot occur it has probability zero, and if it is certain to occur it has probability one i.e. \(P(\Omega) = 1\) . Since any of the six possible outcomes from one to six is equally likely we can write \(P(\{i\}) = \frac{1}{|A|} = \frac{1}{6}\) where \(i\) is equal to any value \( \in \Omega \)

Example 3.
Let’s calculate the probabilities of the sets in the example above.
\(A_1\) is the set of all even numbers (on a roll of a die) $$A_1 = \{2,4,6\}, \; P(A_1) = \frac{3}{6} = \frac{1}{2}$$
\(A_2\) is the set of numbers greater than or equal to 2, $$A_2 = \{2,3,4,5,6\},\; P(A_2) = \frac{5}{6}$$
\(A_3\) is the set of all even numbers greater than or equal to 5, $$A_3 = \{6\}, \; P(A_3) = \frac{1}{6} $$
Union:
$$A_0 \cup A_2 = \left\{1,2,3,4,5,6\right\}, \; P(A_0 \cup A_2) = 1 $$
Intersection:
$$A_0 \cap A_2 = \left\{3,5\right\}, \; P(A_0 \cap A_2) =\frac{2}{6} = \frac{1}{3}$$