Page 476 - Discrete Mathematics and Its Applications

7.2 Probability Theory 455

Solution: We want to find the probability of the event E ={1, 3, 5}. By Exercise 2, we have

p(1) = p(2) = p(4) = p(5) = p(6) = 1/7; p(3) = 2/7.

It follows that

p(E) = p(1) + p(3) + p(5) = 1/7 + 2/7 + 1/7 = 4/7. ▲

When possible outcomes are equally likely and there are a finite number of possible out-
comes, the definition of the probability of an event given in this section (Definition 2) agrees
with Laplace’s definition (Definition 1 of Section 7.1). To see this, suppose that there are n
equally likely outcomes; each possible outcome has probability 1/n, because the sum of their
probabilities is 1. Suppose the event E contains m outcomes. According to Definition 2,

m
1 m
p(E) = = .
n n
i=1
Because |E|= m and |S|= n, it follows that

m |E|
p(E) = = .
n |S|
This is Laplace’s definition of the probability of the event E.

Probabilities of Complements and Unions of Events

The formulae for probabilities of combinations of events in Section 7.1 continue to hold
when we use Definition 2 to define the probability of an event. For example, Theorem 1 of
Section 7.1 asserts that

p(E) = 1 − p(E),

where E is the complementary event of the event E. This equality also holds when Definition 2
is used. To see this, note that because the sum of the probabilities of the n possible outcomes
is 1, and each outcome is either in E or in E, but not in both, we have

p(s) = 1 = p(E) + p(E).
s S

Hence, p(E) = 1 − p(E).
Under Laplace’s definition, by Theorem 2 in Section 7.1, we have

p(E 1 ∪ E 2 ) = p(E 1 ) + p(E 2 ) − p(E 1 ∩ E 2 )

whenever E 1 and E 2 are events in a sample space S. This also holds when we define the prob-
ability of an event as we do in this section. To see this, note that p(E 1 ∪ E 2 ) is the sum of
the probabilities of the outcomes in E 1 ∪ E 2 . When an outcome x is in one, but not both,
of E 1 and E 2 , p(x) occurs in exactly one of the sums for p(E 1 ) and p(E 2 ). When an
outcome x is in both E 1 and E 2 , p(x) occurs in the sum for p(E 1 ), in the sum for p(E 2 ),
and in the sum for p(E 1 ∩ E 2 ), so it occurs 1 + 1 − 1 = 1 time on the right-hand side. Conse-
quently, the left-hand side and right-hand side are equal.