# Axioms of Probability and Immediate Consequences

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   $\displaystyle{ A_1 }\,$ , $\displaystyle{ A_2 }\,$ , â€¦ , is a countable
collection of mutually disjoint elements of   $C$   (that is if   $i \neq j$   then   $\displaystyle{ A_i \cap A_j =\emptyset }\,$ ) then   $\displaystyle{ P\left( A_1 \cup A_2 \cup \cdots \right) = P\left( A_1\right) +P\left( A_2\right) +\cdots }\;$ .   This has the consequence that   $\displaystyle{ P(A)=1-P\left( S\setminus A\right) }\,$ , and thus that   $\displaystyle{ P\left( \emptyset\right)=0 }\;$ .

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   $C =\{ \emptyset , \{ H \} , \{ T \} , \{ H , T \} \}\;$ .   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   $\emptyset\;$ .   For a fair coin, the events of   $C$   are assigned probabilities as follows.
$$\begin{array}{rl} P(\emptyset ) & =0 \\ P( \{ H \} ) & =.5 \\ P( \{ T \} ) & =.5 \\ P( \{ H , T \} ) & =1 \end{array}$$
We often will simplify this notation as below.
$$\begin{array}{rl} P(\emptyset ) & =0 \\ P( H ) & =.5 \\ P( T ) & =.5 \\ P( S ) & =1 \end{array}$$

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)
$$\begin{array}{rl} C & =\{ \emptyset , \\ & \quad\, \{ 1 \} , \{ 2 \} , \{ 3 \} , \{ 4 \} , \{ 5 \} , \{ 6 \} , \\ & \quad\, \{ 1, 2 \} , \{ 1, 3 \} , \{ 1, 4 \} , \{ 1, 5 \} , \{ 1, 6 \} , \{ 2, 3 \} , \{ 2, 4 \} , \{ 2, 5 \} , \{ 2, 6 \} , \{ 3, 4 \} , \{ 3, 5 \} , \{ 3, 6 \} , \{ 4, 5 \} , \{ 4, 6 \} , \{ 5, 6 \} , \\ & \quad\, \{ 1, 2, 3 \} , \{ 1, 2, 4 \} , \{ 1, 2, 5 \} , \{ 1, 2, 6 \} , \{ 1, 3, 4 \} , \{ 1, 3, 5 \} , \{ 1, 3, 6 \} , \{ 1, 4, 5 \} , \{ 1, 4, 6 \} , \{ 1, 5, 6 \} , \\ & \quad\, \{ 2, 3, 4 \} , \{ 2, 3, 5 \} , \{ 2, 3, 6 \} , \{ 2, 4, 5 \} , \{ 2, 4, 6 \} , \{ 2, 5, 6 \} , \{ 3, 4, 5 \} , \{ 3, 4, 6 \} , \{ 3, 5, 6 \} , \{ 4, 5, 6 \} , \\ & \quad\, \{ 1, 2, 3, 4 \} , \{ 1, 2, 3, 5 \} , \{ 1, 2, 3, 6 \} , \{ 1, 2, 4, 5 \} , \{ 1, 2, 4, 6 \} , \{ 1, 2, 5, 6 \} , \{ 1, 3, 4, 5 \} , \{ 1, 3, 4, 6 \} , \{ 1, 3, 5, 6 \} , \{ 1, 4, 5, 6 \} , \\ & \quad\, \{ 2, 3, 4, 5 \} , \{ 2, 3, 4, 6 \} , \{ 2, 3, 5, 6 \} , \{ 2, 4, 5, 6 \} , \{ 3, 4, 5, 6 \} , \\ & \quad\, \{ 1, 2, 3, 4, 5 \} , \{ 1, 2, 3, 4, 6 \} , \{ 1, 2, 3, 5, 6 \} , \{ 1, 2, 4, 5, 6 \} , \{ 1, 3, 4, 5, 6 \} , \{ 2, 3, 4, 5, 6 \} , \\ & \quad\, \{ 1, 2, 3, 4, 5 ,6 \} \}\; \text{.} \end{array}$$
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
$$P(1) =P(2) =P(3) =P(4) =P(5) =P(6) =\textstyle{\frac{1}{6}} \;\text{.}$$
(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
$$P(\{ 2, 4, 6\}) =P( \{ 2\} \cup \{ 4\} \cup \{ 6\} ) =P( 2 \text{ or } 4 \text{ or } 6 ) =P( 2 ) +P( 4 ) +P( 6 ) =\textstyle{ \frac{1}{6} +\frac{1}{6} +\frac{1}{6} } =\textstyle{ \frac{1}{2} } \;\text{.}$$
And the probability of getting an odd number on a roll is
$$P( 1 \text{ or } 3 \text{ or } 5 ) =P(\{ 1, 3, 5\}) =P( S \setminus \{ 2, 4, 6\} ) =1 -\textstyle{ \frac{1}{2} } = \textstyle{ \frac{1}{2} } \;\text{.}$$

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   â€“   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.