ch4

CHAPTER 4
PROBABILITY
Prem Mann, Introductory Statistics, 7/E

Copyright © 2010 John Wiley & Sons. All right reserved
EXPERIMENT, OUTCOMES, AND SAMPLE SPACE
● Definition
● An experiment is a process that, when
performed, results in one and only one of
many observations. These observations are
called that outcomes of the experiment.
The collection of all possible outcomes for an
experiment is called a sample space.

Table 4.1 Examples of Experiments, Outcomes,
and Sample Spaces

Example 4-1
● Draw the Venn and tree diagrams for the

experiment of tossing a coin once.

Figure 4.1 (a) Venn Diagram and (b) tree diagram
for one toss of a coin.

Example 4-2
● Draw the Venn and tree diagrams for the

experiment of tossing a coin twice.

Figure 4.2 (a) Venn diagram and (b) tree diagram
for two tosses of a coin.

Example 4-3
● Suppose we randomly select two workers

from a company and observe whether the
worker selected each time is a man or a
woman. Write all the outcomes for this
experiment. Draw the Venn and tree
diagrams for this experiment.

Figure 4.3 (a) Venn diagram and (b) tree diagram
for selecting two workers.

Simple and Compound Events
● Definition
● An event is a collection of one or more of
the outcomes of an experiment.

● Definition
● An event that includes one and only one of
the (final) outcomes for an experiment is
called a simple event and is denoted by Ei.

Example 4-4
● Reconsider Example 4-3 on selecting two workers

from a company and observing whether the worker
selected each time is a man or a woman. Each of the
final four outcomes (MM, MW, WM, and WW) for this
experiment is a simple event. These four events can
be denoted by E1, E2, E3, and E4, respectively. Thus,
● E1 = (MM), E2 = (MW), E3 = (WM), and E4 = (WW)

● Definition
● A compound event is a collection of more
than one outcome for an experiment.

Example 4-5
● Reconsider Example 4-3 on selecting two workers
from a company and observing whether the worker
selected each time is a man or a woman. Let A be
the event that at most one man is selected. Event A
will occur if either no man or one man is selected.
Hence, the event A is given by
A = {MW, WM, WW}
● Because event A contains more than one outcome,

it is a compound event. The Venn diagram in Figure
4.4 gives a graphic presentation of compound event
A.
Figure 4.4 Venn diagram for event A.

Example 4-6
● In a group of a people, some are in favor of genetic
engineering and others are against it. Two persons are
selected at random from this group and asked whether they
are in favor of or against genetic engineering. How many
distinct outcomes are possible? Draw a Venn diagram and a
tree diagram for this experiment. List all the outcomes
included in each of the following events and mention whether
they are simple or compound events.
(a) Both persons are in favor of the genetic engineering.

(b) At most one person is against genetic engineering.
(c) Exactly one person is in favor of genetic engineering.

Example 4-6: Solution
● Let
■ F = a person is in favor of genetic engineering
■ A = a person is against genetic engineering
■ FF = both persons are in favor of genetic engineering
■ FA = the first person is in favor and the second is
against
■ AF = the first is against and the second is in favor
■ AA = both persons are against genetic engineering

Figure 4.5 Venn and tree diagrams.

a) Both persons are in favor of genetic engineering

= { FF }
Because this event includes only one of the final
four outcomes, it is a simple event.
b) At most one person is against genetic
engineering = { FF, FA, AF }
Because this event includes more than one
outcome, it is a compound event.
c) Exactly one person is in favor of genetic
engineering = { FA, AF }
It is a compound event.
CALCULATING PROBABLITY
● Definition
● Probability is a numerical measure of the
likelihood that a specific event will occur.

Two Properties of Probability
The probability of an event always lies in the range 0 to 1.
The sum of the probabilities of all possible outcomes of an

experiment is always 1.

Three Conceptual Approaches to Probability
● Classical Probability
● Definition
● Two or more outcomes (or events) that
have the same probability of occurrence
are said to be equally likely outcomes
(or events).

Classical Probability
Classical Probability Rule to Find Probability

Example 4-7
● Find the probability of obtaining a head and

the probability of obtaining a tail for one
toss of a coin.

Similarly,

Example 4-8
● Find the probability of obtaining an even

number in one roll of a die.

● A = {2, 4, 6}. If any one of these three
numbers is obtained, event A is said to occur.
Hence,

Example 4-9
● In a group of 500 women, 120 have played

golf at least once. Suppose one of these 500
women is randomly selected. What is the
probability that she has played golf at least
once?

● One hundred twenty of these 500 outcomes are
included in the event that the selected woman
has played golf at least once. Hence,

COUNTING RULE
● Counting Rule to Find Total Outcomes
● If an experiment consists of three steps and if

the first step can result in m outcomes, the
second step in n outcomes, and the third in k
outcomes, then
● Total outcomes for the experiment = m · n · k

Example 4-12
● Suppose we toss a coin three times. This
experiment has three steps: the first toss, the
second toss and the third toss. Each step has two
outcomes: a head and a tail. Thus,
● Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
● The eight outcomes for this experiment are

● HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT

Example 4-13
● A prospective car buyer can choose between

a fixed and a variable interest rate and can
also choose a payment period of 36 months,
48 months, or 60 months. How many total
outcomes are possible?

There are two outcomes (a fixed or a variable

interest rate) for the first step and three outcomes
(a payment period of 36 months, 48 months, or
60 months) for the second step. Hence,
Total outcomes = 2 x 3 = 6

Example 4-14
A National Football League team will play 16 games
during a regular season. Each game can result in
one of three outcomes: a win, a lose, or a tie. The
total possible outcomes for 16 games are calculated
as follows:
Total outcomes = 3·3·3·3·3·3·3·3·3·3·3·3 ·3·3·3·3
= 316 = 43,046,721
One of the 43,046,721 possible outcomes is all 16
wins.

MARGINAL AND CONDITIONAL
PROBABILITIES
Suppose all 100 employees of a company were

asked whether they are in favor of or against
paying high salaries to CEOs of U.S. companies.
Table 4.3 gives a two way classification of the
responses of these 100 employees.

Table 4.3 Two-Way Classification of Employee
Responses

Table 4.4 Two-Way Classification of Employee
Responses with Totals

PROBABILITIES
● Definition
● Marginal probability is the probability of a
single event without consideration of any
other event. Marginal probability is also
called simple probability.

Table 4.5 Listing the Marginal Probabilities
P (M ) = 60/100 = .60
P (F ) = 40/100 = .40
P (A ) = 19/100 P (B ) = 81/100
= .19 = .81

PROBABILITIES

PROBABILITIES
● Definition
● Conditional probability is the probability that an
event will occur given that another has already
occurred. If A and B are two events, then the
conditional probability A given B is written as
● P ( A | B )
● and read as “the probability of A given that B has
already occurred.”

Example 4-15
● Compute the conditional probability

P ( in favor | male) for the data on 100
employees given in Table 4.4.


Figure 4.6 Tree Diagram.

Example 4-16
● For the data of Table 4.4, calculate the

conditional probability that a randomly
selected employee is a female given that
this employee is in favor of paying high
salaries to CEOs.


Figure 4.7 Tree diagram.

MUTUALLY EXCLUSIVE EVENTS
● Definition
● Events that cannot occur together are said
to be mutually exclusive events.

Example 4-17
● Consider the following events for one roll of a

die:
● A= an even number is observed= {2, 4, 6}
● B= an odd number is observed= {1, 3, 5}
● C= a number less than 5 is observed= {1, 2, 3, 4}
● Are events A and B mutually exclusive? Are
events A and C mutually exclusive?


Example 4-18
● Consider the following two events for a

randomly selected adult:
● Y = this adult has shopped on the Internet at
least once
● N = this adult has never shopped on the Internet
● Are events Y and N mutually exclusive?

As we can observe from the definitions of events Y and N
and from Figure 4.10, events Y and N have no common
outcome. Hence, these two events are mutually exclusive.

INDEPENDENT VERSUS DEPENDENT EVENTS
● Definition
● Two events are said to be independent if the
occurrence of one does not affect the
probability of the occurrence of the other. In
other words, A and B are independent
events if
● either P(A | B) = P(A) or P(B | A) = P(B)

Example 4-19
● Refer to the information on 100 employees

given in Table 4.4 in Section 4.4. Are
events “female (F)” and “in favor (A)”
independent?

● Events F and A will be independent if
● P (F) = P (F | A)
●
Otherwise they will be dependent.
Using the information given in Table 4.4, compute
●
P (F) = 40/100 = .40 and
P (F | A) = 4/19 = .2105
Because these two probabilities are not equal, the
two events are dependent.

Example 4-20
● A box contains a total of 100 CDs that were
manufactured on two machines. Of them, 60 were
manufactured on Machine I. Of the total CDs, 15
are defective. Of the 60 CDs that were
manufactured on Machine I, 9 are defective.
Let D be the event that a randomly selected CD is
defective, and let A be the event that a randomly
selected CD was manufactured on Machine I. Are
events D and A independent?

● From the given information,

● P (D) = 15/100 = .15 and
P (D | A) = 9/60 = .15
Hence,
P (D) = P (D | A)
● Consequently, the two events, D and A, are
independent.

Table 4.6 Two-Way Classification Table

Multiplication Rule
● P(A and B) = P(A) P(B|A) or P(B) P(A|B)

● Multiplication Rule to Calculate the
Probability of Independent Events
● The probability of the intersection of two

independent events A and B is
● P(A and B) = P(A) P(B)

Example 4-26
● An office building has two fire detectors. The

probability is .02 that any fire detector of
this type will fail to go off during a fire. Find
the probability that both of these fire
detectors will fail to go off in case of a fire.

We define the following two events:
A = the first fire detector fails to go off during

a fire
B = the second fire detector fails to go off
during a fire
Then, the joint probability of A and B is

P(A and B) = P(A) P(B) = (.02)(.02) = .0004

Example 4-27
● The probability that a patient is allergic to
penicillin is .20. Suppose this drug is
administered to three patients.
a) Find the probability that all three of them are

allergic to it.
b) Find the probability that at least one of the them
is not allergic to it.

a) Let A, B, and C denote the events the first,

second and third patients, respectively, are
allergic to penicillin. Hence,
P (A and B and C) = P(A) P(B) P(C)

= (.20) (.20) (.20) = .008

b) Let us define the following events:

G = all three patients are allergic
H = at least one patient is not allergic
▪ P(G) = P(A and B and C) = .008
▪ Therefore, using the complementary event
rule, we obtain
▪ P(H) = 1 – P(G) = 1 - .008 = .992

Figure 4.18 Tree diagram for joint probabilities.

Multiplication Rule for Independent Events
● Joint Probability of Mutually Exclusive Events
● The joint probability of two mutually

exclusive events is always zero. If A and B
are two mutually exclusive events, then
● P(A and B) = 0

Example 4-28
● Consider the following two events for an

application filed by a person to obtain a car
loan:
● A = event that the loan application is approved
● R = event that the loan application is rejected
● What is the joint probability of A and R?

● The two events A and R are mutually

exclusive. Either the loan application will be
approved or it will be rejected. Hence,
P(A and R) = 0

UNION OF EVENTS AND THE ADDITION RULE
● Definition
● Let A and B be two events defined in a
sample space. The union of events A and B
is the collection of all outcomes that belong
to either A or B or to both A and B and is
denoted by
● A or B

Example 4-29
● A senior citizen center has 300 members.
Of them, 140 are male, 210 take at least
one medicine on a permanent basis, and 95
are male and take at least one medicine on
a permanent basis. Describe the union of
the events “male” and “take at least one
medicine on a permanent basis.”

Let us define the following events:

M = a senior citizen is a male
F = a senior citizen is a female
A = a senior citizen takes at least one medicine
B = a senior citizen does not take any medicine
The union of the events “male” and “take at least

one medicine” includes those senior citizens who are
either male or take at least one medicine or both.
The number of such senior citizen is
140 + 210 – 95 = 255

Table 4.8

Figure 4.19 Union of events M and A.

UNION OF EVENTS AND THE ADDITION
RULE
● Addition Rule
● Addition Rule to Find the Probability of Union

of Events
● The portability of the union of two events A
and B is
● P(A or B) = P(A) + P(B) – P(A and B)

Example 4-30
● A university president has proposed that all

students must take a course in ethics as a
requirement for graduation. Three hundred faculty
members and students from this university were
asked about their opinion on this issue. Table 4.9
gives a two-way classification of the responses of
these faculty members and students.
● Find the probability that one person selected at

random from these 300 persons is a faculty
member or is in favor of this proposal.

Table 4.9 Two-Way Classification of Responses

A = the person selected is a faculty member
B = the person selected is in favor of the proposal
From the information in the Table 4.9,

P(A) = 70/300 = .2333
P(B) = 135/300 = .4500
P(A and B) = P(A) P(B | A) = (70/300)(45/70) = .1500
Using the addition rule, we obtain

P(A or B) = P(A) + P(B) – P(A and B)
= .2333 + .4500 – .1500 = .5333

Example 4-31
● In a group of 2500 persons, 1400 are

female, 600 are vegetarian, and 400 are
female and vegetarian. What is the
probability that a randomly selected person
from this group is a male or vegetarian?

● Let us define the following events:
● F = the randomly selected person is a female
● M = the randomly selected person is a male
● V = the randomly selected person is a vegetarian
● N = the randomly selected person is a non-vegetarian.

Table 4.10 Two-Way Classification Table

Addition Rule for Mutually Exclusive Events
● Addition Rule to Find the Probability of the

Union of Mutually Exclusive Events
● The probability of the union of two mutually

exclusive events A and B is
● P(A or B) = P(A) + P(B)

Example 4-32
● A university president has proposed that all
students must take a course in ethics as a
requirement for graduation. Three hundred faculty
members and students from this university were
asked about their opinion on this issue. The
following table, reproduced from Table 4.9 in
Example 4-30, gives a two-way classification of the
responses of these faculty members and students.
● What is the probability that a randomly selected
person from these 300 faculty members and
students is in favor of the proposal or is neutral?



F = the person selected is in favor of the proposal
N = the person selected is neutral
From the given information,

P(F) = 135/300 = .4500
P(N) = 40/300 = .1333
Hence,
P(F or N) = P(F) + P(N) = .4500 + .1333 = .5833

Figure 4.20 Venn diagram of mutually exclusive
events.

Example 4-33
● Consider the experiment of rolling a die

twice. Find the probability that the sum of
the numbers obtained on two rolls is 5, 7,
or 10.

Table 4.11 Two Rolls of a Die

P(sum is 5 or 7 or 10)
= P(sum is 5) + P(sum is 7) + P(sum is 10)
= 4/36 + 6/36 + 3/36 = 13/36 = .3611

Example 4-34
The probability that a person is in favor of

genetic engineering is .55 and that a
person is against it is .45. Two persons are
randomly selected, and it is observed
whether they favor or oppose genetic
engineering.
a) Draw a tree diagram for this experiment
b) Find the probability that at least one of the two
persons favors genetic engineering.

a) Let
F = a person is in favor of genetic engineering
A = a person is against genetic engineering
This experiment has four outcomes. The tree

diagram in Figure 4.21 shows these four
outcomes and their probabilities.

Figure 4.21 Tree diagram.

b) P(at least one person favors)

= P(FF or FA or AF)
= P(FF) + P(FA) + P(AF)
= .3025 + .2475 + .2475 = .7975


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CHAPTER 4

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

● Draw the Venn and tree diagrams for the

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

● Draw the Venn and tree diagrams for the

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

● Suppose we randomly select two workers

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

● Reconsider Example 4-3 on selecting two workers

● E1 = (MM), E2 = (MW), E3 = (WM), and E4 = (WW)

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

● Because event A contains more than one outcome,

Prem Mann, Introductory Statistics, 7/E

(a) Both persons are in favor of the genetic engineering.

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

a) Both persons are in favor of genetic engineering

Prem Mann, Introductory Statistics, 7/E

The sum of the probabilities of all possible outcomes of an

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

● Find the probability of obtaining a head and

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

● Find the probability of obtaining an even

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

● In a group of 500 women, 120 have played

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

● Counting Rule to Find Total Outcomes

● If an experiment consists of three steps and if

Prem Mann, Introductory Statistics, 7/E

● Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8

● The eight outcomes for this experiment are

Prem Mann, Introductory Statistics, 7/E

● A prospective car buyer can choose between

Prem Mann, Introductory Statistics, 7/E

There are two outcomes (a fixed or a variable

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

Suppose all 100 employees of a company were

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E

Prem Mann, Introductory Statistics, 7/E