Discrete Mathematics for Computer Science

516 CHAPTER 8 Discrete Probability

Solution. Each outcome (o = (i, j) c Q can be regarded as an event {o0} g 2, and the

Theorem of Total Probability can be used. The sample space Q2 is the disjoint union of two

events of positive probability-namely, the event E, = {(1, 0), (1, 1)1 of sending a 1 and

the event E 2 = {(0, 0), (0, 1)1 of sending a 0. According to the data, P(E 1 ) = 0.6, and

P(E 2 ) = 0.4. Hence, the event {(0, 0)) has probability

P({(0, 0)}) = P({(0, 0)} 1 E) .P(E 1 ) + P({(0, 0)} I E 2 ). P(E 2 )

= (0)(0.6) + (0.8)(0.4)

= 0.32

Event {(l, 0)) has probability

P({(1, 0)1) ) P({(1, 0)) 1 EI) • P(E 1 ) + P({(1, 0)}] E 2 )D P(E 2 )

= (0.1)(0.6) + (0)(0.4)

= 0.06

Similarly,

P({(0, 1)1) = (0)(0.6) + (0.2)(0.4) = 0.08

and

P({(1, 1)1) = (0.9)(0.6) + (0)(0.4) = 0.54

The probabilities of these singleton events must agree with the probability density p as-

signed to Q2, so there is only one choice for p consistent with the data. U

Situations that consist of stages depending on the results of preceding stages, such

as the channel reliability examples, are called dependent trial processes. In designing

sample spaces and probability distributions, it is important to consider whether trials are

independent or dependent. Sometimes, the dependency of one trial on a preceding trial is

not immediately obvious (see Exercise 9 in Section 8.6).

We conclude this section with one more application of Bayes' Rule and the Theorem

of Total Probability.

Example 8. (Identifying the Source of a Bad Apple) Suppose apples are shipped to

a grocery store by three different orchards O1, 02, and 03. Suppose the percentages of

each shipment that are bad are 10%, 8%, and 3%, respectively. Suppose further that the

percentages of the apple supply from these orchards are 20%, 30%, and 50%, respectively.

Now, suppose a customer selects an apple at random and finds that it is bad. What is the

probability the apple came from orchard 02?

Solution. Let Ai denote the event that an apple (good or bad) selected at random comes

from orchard Oi for i = 1, 2, and 3. Hence,

P(A 1 ) = 0.2 P(A 2 ) = 0.3 P(A 3 ) = 0.5

Let B denote the event that an apple selected at random is bad. We are to determine

P (A 2 I B). We are given the conditional probabilities

P(B IA1) = O.1 P(B IA 2 ) = 0.08 P(B IA 3 ) = 0.03