The set-up for probability should be undertaken with examples, such as presented in the preceding section, in mind.

Axioms

A probability space is a set   $S$   (the elements of   $S$   will be called outcomes) with a collection   $C$   of subsets of   $S$ .  
The set   $S$   itself needs to be in   $C$ .   For each subset   $A\subset S$   in   $C$ , we also have   $S\setminus A\in C$ .   The elements of   $C$   are called events.   To each event is assigned a number between   $0$   and   $1$   (inclusive).   This means that we have a function   $P:C\to [0,1]$ .   This function needs to satisfy several properties:

1. $P(S)=1$
2. If   ${\displaystyle A_1}$ , ${\displaystyle A_2}$ , … , is a countable
   collection of mutually disjoint elements of   $C$   (that is if   $i\ne j$   then   ${\displaystyle A_i\cap A_j=\mathrm{\varnothing }}$ ) then   ${\displaystyle P(A_1\cup A_2\cup \cdots )=P\left(A_1\right)+P\left(A_2\right)+\cdots }$ .   This has the consequence that   ${\displaystyle P(A)=1-P(S\setminus A)}$ , and thus that   ${\displaystyle P\left(\mathrm{\varnothing }\right)=0}$ .

Here is how we can interpret our introductory examples
through these rules.

Example 1: Our first example may have as the underlying experiment the flip of a quarter.   The collection of possible outcomes is   $S=\{H,T\}$ , and we take as the collection of events   $C=\{\mathrm{\varnothing },\{H\},\{T\},\{H,T\}\}$ .   Thus the event corresponding to the coin landing with Washington’s profile showing is   $\{H\}$ .   The event corresponding to the coin landing flat is   $\{H,T\}$ .   And the event corresponding to the coin landing on edge is   $\mathrm{\varnothing }$ .   For a fair coin, the events of   $C$   are assigned probabilities as follows.
$$\begin{array}{rl}P(\mathrm{\varnothing })& =0\\ P(\{H\})& =.5\\ P(\{T\})& =.5\\ P(\{H,T\})& =1\end{array}$$
We often will simplify this notation as below.
$$\begin{array}{rl}P(\mathrm{\varnothing })& =0\\ P(H)& =.5\\ P(T)& =.5\\ P(S)& =1\end{array}$$

Example 2: Our second example may have as the underlying experiment the roll of a standard die.   The collection of possible outcomes is   $S=\{1,2,3,4,5,6\}$ , and we take as the collection of events (with sixty-four elements)
$$\begin{array}{rl}C& =\{\mathrm{\varnothing },\\ & \{1\},\{2\},\{3\},\{4\},\{5\},\{6\},\\ & \{1,2\},\{1,3\},\{1,4\},\{1,5\},\{1,6\},\{2,3\},\{2,4\},\{2,5\},\{2,6\},\{3,4\},\{3,5\},\{3,6\},\{4,5\},\{4,6\},\{5,6\},\\ & \{1,2,3\},\{1,2,4\},\{1,2,5\},\{1,2,6\},\{1,3,4\},\{1,3,5\},\{1,3,6\},\{1,4,5\},\{1,4,6\},\{1,5,6\},\\ & \{2,3,4\},\{2,3,5\},\{2,3,6\},\{2,4,5\},\{2,4,6\},\{2,5,6\},\{3,4,5\},\{3,4,6\},\{3,5,6\},\{4,5,6\},\\ & \{1,2,3,4\},\{1,2,3,5\},\{1,2,3,6\},\{1,2,4,5\},\{1,2,4,6\},\{1,2,5,6\},\{1,3,4,5\},\{1,3,4,6\},\{1,3,5,6\},\{1,4,5,6\},\\ & \{2,3,4,5\},\{2,3,4,6\},\{2,3,5,6\},\{2,4,5,6\},\{3,4,5,6\},\\ & \{1,2,3,4,5\},\{1,2,3,4,6\},\{1,2,3,5,6\},\{1,2,4,5,6\},\{1,3,4,5,6\},\{2,3,4,5,6\},\\ & \{1,2,3,4,5,6\}\}\text{.}\end{array}$$
Thus the event corresponding to the die landing with an even number showing is   $\{2,4,6\}$ , and so on.   For a fair die, the events   $\{1\}$ , $\{2\}$ , $\{3\}$ , $\{4\}$ , $\{5\}$ , and   $\{6\}$   of   $C$   are considered equally likely and assigned probabilities
$$P(1)=P(2)=P(3)=P(4)=P(5)=P(6)=\frac{1}{6}\text{.}$$
(We have used the notational shortcut mentioned above.)   From these, we can use the rules to determine the probability of any event in   $C$ .   For example, the probability of getting an even number on a roll of a fair die is
$$P(\{2,4,6\})=P(\{2\}\cup \{4\}\cup \{6\})=P(2\text{ or }4\text{ or }6)=P(2)+P(4)+P(6)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}\text{.}$$
And the probability of getting an odd number on a roll is
$$P(1\text{ or }3\text{ or }5)=P(\{1,3,5\})=P(S\setminus \{2,4,6\})=1-\frac{1}{2}=\frac{1}{2}\text{.}$$

It should be clear here that even for relatively simply experiments (like, enumerating the possible events can be onerous.   It behooves us to develop a better way to count the number of outcomes and so determine probabilities.

Example 3: Our third example has a continuum of possible outcomes   –   all positive real numbers.   It would be impossible to try to list all possible outcomes, so we settle for a description like   ${\displaystyle S={\mathbb{R}}_{+}}$ .   It would be even more impossible to list all possible events (subsets of   ${\displaystyle {\mathbb{R}}_{+}}$ ).   It is also not at all clear how we might assign probabilities to events, although this may be important for various reasons.   In a future section we will see how one might go about determining the probabilities of events in such a case.