Elementary events

Now suppose that we divide $S$ uniformly into $N$ equally likely elementary outcomes, $e_1, e_2, \ldots, e_N$. For example, we could divide a year into 365 units for which rain is possible,

$$S = \{e_1, \ldots, e_{365} \} e_k - \mbox{rains on day } k$$

If $E$ consists of $r \leq N$ of these elementary outcomes,

$$\begin{eqnarray*} S & = & \{e_1,\ldots, e_N\} , \\ E & = & \{e_{(1)}, \ldots, e_{(r)} \} , \end{eqnarray*}$$

then the probability of $E$ is

$$P(E) = {{\mbox{number of elementary events in } E} \over {\... ...ntary events in } S}} = {{n(E)} \over {n(S)}} = { r \over N}$$

Strictly we should say

$$P(E) = {{\mbox{measure of } E} \over {\mbox{measure of } S}} \$$

but commonly the measure is just the count of elements. In Figure 2, $P(E)$ could be interpreted as the ratio of the area of $E$ to the area of $S$ and that area represents the number of elementary units.

It follows that if there are no elementary events in $E$, $P(E)=0$ and if $E$ is the same as $S$, $P(E)=1$, ie

$$0 \leq P(E) \leq 1 .$$ (1)

If $\bar E$ is the complement of $E$,

$$P(\bar E) =1 - P(E) .$$ (2)