The probability of an event

Consider a random experiment, and some event $E$ . Similar to the probability of an outcome, we can define the probability of an event as the long-run relative frequency of its occurrence:

Definition 1

The probability of event E, written $p(E)$ , is defined as

p(E)=\frac{n}{N}\quad (N \text{ large})

where $N$ is the number of times that the experiment is repeated under the same conditions, and $n$ is the number of times event $E$ occurred. Or expressed differently, $p(E)$ is the percentage of times that $E$ occurs if an experiment is repeated many times.

Clearly, there must be a close relation between outcome probabilities and event probabilities.

Theorem 1

The probability of event $E$ is the sum of its outcome probabilities. That is, if event $E$ contains the $r$ outcomes $o_1,...,o_r$ ,

E=\{o_1,...,o_r\}

then

\begin{array}{lll} p(E)&=&\sum_{i=1}^r p(o_i)\\ &=&p(o_1)+...+p(o_r) \end{array}

Exercise 1

Proof the statement above.

Solution

Let us repeat the experiment $N$ times, where $N$ is a big number. So $p(o_i)$ is the percentage of times the outcome $o_i$ occurs. The event $E$ occurs every time that one of the outcomes $o_1, ..., o_r$ occurs, and this percentage is $p(o_1)+...+p(o_r)$ . Thus $p(E)=p(o_1)+...+p(o_r)$ .

Exercise 2

A die has differently weighted faces, so that some numbers occur more often than others:

\begin{array}{lll} p(1)&=&0.25\\ p(2)&=&0.27\\ p(3)&=&0.12\\ p(4)&=&0.12\\ p(5)&=&0.12\\ p(6)&=&0.12\\ \end{array}

Determine the probability of the event "an even number occurred".

Solution

$E=\{2,4,6\}$ , thus $p(E)=p(2)+p(4)+p(6)=0.27+0.12+0.12=\underline{0.51}$