To gain some insight into the amount of freedom we have in defining probability functions, it is useful to consider finite sample description spaces. The sample description space \(S\) of a random observation or experiment is defined as finite if it is of finite size, which is to say that the random observation or experiment under consideration possesses only a finite number of possible outcomes.

Consider now a finite sample description space \(S\) , of size \(N\) . We may then list the descriptions in \(S\) . If we denote the descriptions in \(S\) by \(D_{1}\) , \(D_{2}, \ldots, D_{N}\) , then we may write \(S=\left\{D_{1}, D_{2}, \ldots, D_{N}\right\}\) . For example, let \(S\) be the sample description space of the random experiment of tossing two coins; if we define \(D_{1}=(H, H), D_{2}=(H, T), D_{3}=(T, H), D_{4}=\) \((T, T)\) , then \(S=\left\{D_{1}, D_{2}, D_{3}, D_{4}\right\}\) .

It is shown in section 1 of Chapter 2 that \(2^{N}\) possible events may be defined on a sample description space of finite size \(N\) . For example, if \(S=\left\{D_{1}, D_{2}, D_{3}, D_{4}\right\}\) , then there are sixteen possible events that may be defined; namely, \(S\) , \(\emptyset\) , \(\left\{D_{1}\right\}\) , \(\left\{D_{2}\right\}\) , \(\left\{D_{3}\right\}\) , \(\left\{D_{4}\right\}\) , \(\left\{D_{1}, D_{2}\right\}\) , \(\left\{D_{1}, D_{3}\right\}\) , \(\left\{D_{1}, D_{4}\right\}\) , \(\left\{D_{2}, D_{3}\right\}\) , \(\left\{D_{2}, D_{4}\right\}\) , \(\left\{D_{3}, D_{4}\right\}\) , \(\left\{D_{1}, D_{2}, D_{3}\right\}\) , \(\left\{D_{1}, D_{2}, D_{4}\right\}\) , \(\left\{D_{1}, D_{3}, D_{4}\right\}\) , \(\left\{D_{2}, D_{3}, D_{4}\right\}\) .

Consequently, to define a probability function \(P[\cdot]\) on the subsets of \(S\) , one needs to specify the \(2^{N}\) values that \(P[A]\) assumes as \(A\) varies over the events on \(S\) . However, the values of the probability function cannot be specified arbitrarily but must be such that axioms 1 to 3 are satisfied.

There are certain events of particularly simple structure, called the single-member events, on which it will suffice to specify the probability function \(P[\cdot]\) in order that it be specified for all events. A single-member event is an event that contains exactly one description . If an event \(E\) has as its only member the description \(D_{i}\) , this fact may be expressed in symbols by writing \(E=\left\{D_{i}\right\}\) . Thus \(\left\{D_{i}\right\}\) is the event that occurs if and only if the random situation being observed has description \(D_{i}\) . The reader should note the distinction between \(D_{i}\) and \(\left\{D_{i}\right\}\) ; the former is a description, the latter is an event (which because of its simple structure is called a single member event).

Example 6A. The distinction between a single-member event and a sample description. Suppose that we are drawing a ball from an urn containing six balls, numbered 1 to 6 (or, alternately, we may be observing the outcome of the toss of a die, bearing numbers 1 to 6 on its sides). As sample description space \(S\) , we take \(S=\{1,2,3,4,5,6\}\) . The event, denoted by \(\{2\}\) , that the outcome of the experiment is a 2 is a single-member event. The event, denoted by \(\{2,4,6\}\) , that the outcome of the experiment is an even number is not a single-member event. Note that 2 is a description, whereas \(\{2\}\) is an event.

A probability function \(P[\cdot]\) defined on \(S\) can be specified by giving its value \(P\left[\left\{D_{i}\right\}\right]\) on the single-member events \(\left\{D_{i}\right\}\) which correspond to the members of \(S\) . Its value \(P[E]\) on any event \(E\) may then be computed by the following formula:

Formula For Calculating the Probability of Events When the Sample Description Space Is Finite. Let \(E\) be any event on a finite sample description space \(S=\left\{D_{1}, D_{2}, \ldots, D_{N}\right\}\) . Then the probability \(P[E]\) of the event \(E\) is the sum, over all descriptions \(D_{i}\) that are members of \(E\) , of the probabilities \(P\left[\left\{D_{i}\right\}\right]\) ; we express this symbolically by writing that if \(E=\left\{D_{i_{1}}, D_{i_{2}}, \ldots, D_{i_{k}}\right\}\) then

\[P[E]=P\left[\left\{D_{i_{1}}\right\}\right]+P\left[\left\{D_{i_{2}}\right\}\right]+\cdots+P\left[\left\{D_{i_{k}}\right\}\right] . \tag{6.1}\] 

To prove (6.1), one need note only that if \(E\) consists of the descriptions \(D_{i_{1}}, D_{i_{2}}, \ldots, D_{i_{k}}\) then \(E\) can be written as the union of the mutually exclusive single-member events \(\left\{D_{i_{1}}\right\},\left\{D_{i_{2}}\right\}, \ldots,\left\{D_{i_{k}}\right\}\) . Equation (6.1) follows immediately from (5.8) .

Example 6B. Illustrating the use of (6.1). Suppose one is drawing a sample of size 2 from an urn containing white and red balls. Suppose that as the sample description space of the experiment one takes \(S=\{(W, W),(W, R),(R, W),(R, R)\}\) . To specify a probability function \(P[\cdot]\) on \(S\) , one may specify the values of \(P\left[{\cdot}\right]\) on the single-member events by a table:

\(x\) \((W, W)\) \((W, R)\) \((R, W)\) \((R, R)\) 
\(P[\{x\}]\) \(\frac{6}{15}\) \(\frac{4}{15}\) \(\frac{4}{15}\) \(\frac{1}{15}\) 

Let \(E\) be the event that the ball drawn on the first draw is white. The event \(E\) may be represented as a set of descriptions by \(E=\{(W, W),(W, R)\}\) . Then, by (6.1) , \(P[E]=P[\{(W, W)\}]+P[\{(W, R)\}]=\frac{2}{3}\) .