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:
- $P(S) =1$
- If $\displaystyle{ A_1 }\,$ , $\displaystyle{ A_2 }\,$ , … , 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 – 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.
