The elementary events are the building blocks (or atoms) of a probability model.
ELEMENTARY SETS AS EVENTS AND THEIR COMPLEMENTS
The elementary events are the building blocks (or
atoms) of a probability model. They are the events that cannot be decomposed
further into smaller sets of events. The set of elementary events is just the
collection of all the elementary events. In example 2, the event {1, 1} “snake
eyes” is an elementary event. The set [{1, 1}, {1, 2}, {1, 3}, {1, 4}, {1, 5},
{1, 6}, {2, 1}, {2, 2}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {3, 1}, {3, 2}, {3, 3},
{3, 4}, {3, 5}, {3, 6}, {4, 1}, {4, 2}, {4, 3}, {4, 4}, {4, 5}, {4, 6}, {5, 1},
{5, 2}, {5, 3}, {5, 4}, {5, 5}, {5, 6}, {6, 1}, {6, 2}, {6, 3}, {6, 4}, {6, 5},
and {6, 6}] is the set of elementary events.
It is customary to use Ω, the Greek letter omega, to represent the set containing all the
elementary events. This set is also called the universal set. For Ω we have P(Ω) = 1. The set containing no events is denoted by Ø and is called the null set, or empty
set. For the empty set Ø we have P(Ø) = 0.
For any set A,
Ac denotes the complement
of A. The complement of set A is just the set of all elementary
events not contained in A. From Example
2, if A = {sum of the faces on the
two dice is seven}, then A = [{1, 6},
{2, 5}, {3, 4}, {4, 3}, {5, 2}, {6, 1}] and the set Ac is the set [{1, 1}, {1, 2}, {1, 3}, {1, 4}, {1, 5},
{2, 1}, {2, 2}, {2, 3}, {2, 4}, {2, 6}, {3, 1}, {3, 2}, {3, 3}, {3, 5}, {3, 6},
{4, 1}, {4, 2}, {4, 4}, {4, 5}, {4, 6}, {5, 1}, {5, 3}, {5, 4}, {5, 5}, {5, 6},
{6, 2}, {6, 3}, {6, 4}, {6, 5}, and {6, 6}] .
By simply counting the elementary events in the set
and dividing by the total number of elementary events in Ω, we obtain the probability for the event. In problems with a large
number of elementary events, this method for finding a probability can be
tedious; it also requires that the elementary events are equally likely.
Formulas that we derive in later sections will allow us to compute more easily
the probabilities of certain events.
Consider the probability of A = {sum of the faces on the two dice is seven}. As we saw in the
previous section, P(A) = 6/36 = 1/6 0.167. Since there are
30 elementary events in Ac,
P(Ac)
= 30/36 = 5/6 0.833. We see that P(Ac) = 1 – P(A),
which is always the case, as demonstrated in the next section.
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