Introduction to Probability and Statistics for Engineers and Scientists

60 Chapter 3:Elements of Probability

the outcome will be a member of the sample spaceS. Axiom 3 states that for any set of
mutually exclusive events the probability that at least one of these events occurs is equal to
the sum of their respective probabilities.
It should be noted that if we interpretP(E) as the relative frequency of the eventE
when a large number of repetitions of the experiment are performed, thenP(E) would
indeed satisfy the above axioms. For instance, the proportion (or frequency) of time that
the outcome is inEis clearly between 0 and 1, and the proportion of time that it is inS
is 1 (since all outcomes are inS). Also, ifEandFhave no outcomes in common, then
the proportion of time that the outcome is in eitherEorFis the sum of their respective
frequencies. As an illustration of this last statement, suppose the experiment consists of
the rolling of a pair of dice and suppose thatEis the event that the sum is 2, 3, or 12 and
Fis the event that the sum is 7 or 11. Then if outcomeEoccurs 11 percent of the time
and outcomeF22 percent of the time, then 33 percent of the time the outcome will be
either 2, 3, 12, 7, or 11.
These axioms will now be used to prove two simple propositions concerning prob-
abilities. We first note thatEandEcare always mutually exclusive, and sinceE∪Ec=S,
we have by Axioms 2 and 3 that

1 =P(S)=P(E∪Ec)=P(E)+P(Ec)

Or equivalently, we have the following:

PROPOSITION 3.4.1

P(Ec)= 1 −P(E)

In other words, Proposition 3.4.1 states that the probability that an event does not occur
is 1 minus the probability that it does occur. For instance, if the probability of obtaining
a head on the toss of a coin is^38 , the probability of obtaining a tail must be^58.
Our second proposition gives the relationship between the probability of the union of
two events in terms of the individual probabilities and the probability of the intersection.

PROPOSITION 3.4.2

P(E∪F)=P(E)+P(F)−P(EF)

Proof

This proposition is most easily proven by the use of a Venn diagram as shown in Figure
3.4. As the regions I, II, and III are mutually exclusive, it follows that

P(E∪F)=P(I)+P(II)+P(III)

P(E)=P(I)+P(II)

P(F)=P(II)+P(III)