DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION SYSTEMS
Introduction to Business Statistics
QM 120
Chapter 4
Experiment, outcomes, and sample space
2

Probability and statistics are related in an important way. It
is used to allow us to evaluate the reliability of our
conclusions about the population when we have only
sample information.

Data are obtained by observing either uncontrolled events
in nature or controlled situation in laboratory. We use the
term experiment to describe either method of data
collection.

The observation or measurement generated by an
experiment may or may not produce a numerical value.
Here are some examples:

Recording a test grade

Interviewing a householder to obtain his or her opinion in certain
issue.
Experiment, outcomes, and sample space
3
Table 1: Examples of experiments, outcomes, and sample spaces

Experiment
Outcomes
Sample Space
Toss a coin once
Head, Tail
S = {Head, Tail}
Roll a die once
1, 2, 3, 4, 5, 6
S = {1, 2, 3, 4, 5, 6}
Play a lottery
Win, Lose
S = {Win, Lose}
Select a student
M, F
S = {M,F}
Toss a coin twice
HH, HT, TH, TT
S = {HH, HT, TH, TT}
Venn diagram is a picture that depicts all possible outcomes
for an experiment while in tree diagram, each outcome is
represented a branch of a tree.
H
●H
●T
Venn
Diagram
Tree
Diagram
T
Experiment, outcomes, and sample space
4
Example: Draw the Venn and tree diagrams for the
experiment of tossing a coin twice.
Solution:
H
● HH
● TH
● HT
● TT
Venn
Diagram
H
T
T
H
HH
HT
TH
Tree
Diagram
T
TT
Example: Draw a tree diagram for three tosses of a coin. List
all outcomes for this experiment in a sample space S.
Solution
Experiment, outcomes, and sample space
5
A simple event is the outcome that is observed on a single
repetition of the experiment. It is often denoted by E with a
subscript.
Example: Toss a die and observe the number that appears on
the upper face. List the simple event in the experiment.
Solution: When the die is tossed once, there are six possible
outcomes. There are the simple events, listed below.
Event E1: Observe a 1
Event E2: Observe a 2
Event E3: Observe a 3
Event E4: Observe a 4
Event E5: Observe a 5
Event E6: Observe a 6
Experiment, outcomes, and sample space
6
We can now define an event (or compound event) as a
collection of simple events, often denoted a capital letter.
Example: Tossing a die (continued)
We can define the events A and B as follow,
A: Observe an odd number
B: Observe a number less than 4
Solution:
Experiment, outcomes, and sample space
7
Example: Suppose we randomly select two persons from
members of a club and observe whether the person selected is
a man or woman. Write all the outcomes from experiment.
Draw the Venn and tree diagrams for this experiment.
Solution:

Two events are mutually exclusive if, when one event
occurs, the other cannot, and vice versa.

The set of all simple events is called the sample space, S.
Calculating Probability
8
Probability is a numerical measure of the likelihood that an
event will occur.

Two properties of probability
 The probability of an even always lies in the range 0 to 1.
0 ≤ P(Ei) ≤ 1
0 ≤ P(A) ≤ 1

The sum of the probabilities of all simple events for an experiment,
denoted by ΣP(Ei), is always 1.
∑ P(Ei) = P(E1) + P(E2) + P(E3) + . . . .
=1
Calculating Probability
9
Three conceptual approaches to probability
1. Classical probability

Two or more events that have the same probability of occurrence are
said to be equally likely events.

The probability of a simple event is equal to 1 divided by the total
number of all final outcomes for an equally likely experiment.

Classical probability rule to find probability
1
Total number of outcomes
Number of outcomes favorable to A
P (A ) 
Total number of outcomes
P (E i ) 
Calculating Probability
10
Example: Find the probability of obtaining a head and the
probability of obtaining a tail for one toss of a coin.
Solution
Calculating Probability
11
Example: Find the probability of obtaining an even number in
one roll of a die.
Solution
Calculating Probability
12
Calculating the probability of an event:

List all the simple events in the sample space.

Assign an appropriate probability to each simple event.

Determine which simple events result in the event of interest.

Sum the probabilities of the simple events that result in the event of
interest.
Always
1. Include all simple events in
the sample space.
2. Assign realistic probabilities
to the simple events.
Calculating Probability
13
Example: A six years boy has a safe box that contains four
banknotes: One-Dinar, Five-Dinar, Ten-Dinar, Twenty-Dinar.
His sister which is a three years old girl randomly grabbed
three banknotes from the safe box to buy a 30 KD toy. Find the
odds (probability) that this girl can buy the toy.
Solution
Calculating Probability
14
2. Relative frequency concept of probability

Suppose we want to know the following probabilities:

The next car coming out of an auto factory is a “lemon”

A randomly selected family owns a home

A randomly selected woman has never smoked

The outcomes above are neither equally likely nor fixed for each
sample.

The variation goes to zero as n becomes larger and larger

If an experiment is repeated n times and an event A is observed f
times, then
P (A ) 
f
n
Calculating Probability
15
Example: In a group of 500 women, 80 have played golf at
least once. Suppose one of these 500 woman is selected. What
is the probability that she played golf at least once
Solution
Calculating Probability
16
Example: Ten of the 500 randomly selected cars manufactured
at a certain auto factory are found to be defective. What is the
probability that the next car manufactured at that factory is a
defective?
Solution:
Calculating Probability
17
Example: Lucca Tool Rental would like to assign probabilities
to the number of car polishers it rents each day. Office records
show the following frequencies of daily rentals for the last 40
days.
Number of Polishers Rented
Number of Days
0
4
1
6
2
18
3
10
4
2
Calculating Probability
18
Solution
Calculating Probability
19
Law of large numbers: If an experiment is repeated again and
again, the probability of an event obtained from the relative
frequency approaches the actual probability.
3. Subjective probability


Suppose we want to know the following probabilities:

A student who is taking a statistics class will get an A grade.

KSE price index will be higher at the end of the day.

China will dominate the gold medal list in the 2008 Olympics.
Subjective probability is the probability assigned to an event based on
subjective judgment, experience, information, and belief.
Counting Rule
20
Suppose that an experiment involves a large number N of
simple events and you know that all the simple events are
equally likely. Then each simple event has probability 1/N and
the probability of an event A can be calculated as
nA
N
Where n A is the number of simple events that result in event A
P (A ) 

The mn rule

Consider an experiment that is performed in two stages. If the first
stage can be performed in m ways and for each of these ways, the
second stage can be accomplished in n ways, then there mn ways to
accomplish the experiment
Counting Rule
21
Example: Suppose you want to
order a car in one of three
styles and in one of four paint
colors. To find out how many
options are available, you can
think of first picking one of the
m = 3 styles and then one of n =
4 colors.
Style
Color
1
2
1
3
4
1
2
Using the mn rule, as shown in the
figure below, you have mn= (3)(4) = 12
possible options.
2
3
4
1
2
3
3
4
Counting Rule
22
The extended mn rule

If an experiment is performed in k stages, with n1 ways to accomplish
the first stage, n2 to accomplish the second stage…, and nk ways to
accomplish the kth stage, the number of ways to accomplish the
experiment is
n1n 2 n3 ...n k
Example: A bus driver can take three routes from city A to city
B, four routes from city B to city C, and three routes from city
C to city D. For traveling from A to D, the driver must drive
from A to B to C to D, how many possible routes from A to D
are available
Counting Rule
23
Example: A couple is planning their weeding reception. The
bird's parents have given thema choice of four reception
facilities, three caterers, five DJs, and two limo services. He
the couple randomly selects one reception facility, one caterer,
one DJ, and one limo service, how many different outcomes
are possible?.
Counting Rule
24
A counting rule for permutations (orderings)

The number of ways we can arrange n distinct objects, taking them r
at a time, is
n!
Pr  n Pr 
(n  r )!
where n !  n (n  1)(n  2).....(3)(2)(1) and 0!  1
n
A counting rule for combination

The number of distinct combinations of n distinct objects that can be
formed, taking them r at a time, is
n!
( )
nC r  C
r !(n  r )!
n
r
n
r
Counting Rule
25

Permutations: Given that position (order) is important, if
one has 5 different objects (e.g. A, B, C, D, and E), how
many unique ways can they be placed in 3 positions (e.g.
ADE, AED, DEA, DAE, EAD, EDA, ABC, ACB, BCA, BAC
etc.)
P35 

5!
(5)(4)(3)(2)(1)

 60
(5  3)!
(2)(1)
Combinations: If one has 5 different objects (e.g. A, B, C, D,
and E), how many ways can they be grouped as 3 objects
when position does not matter (e.g. ABC, ABD, ABE, ACD,
ACE, ADE are correct but CBA is not ok as is equal to ABC)
C 35 
5!
(5)(4)(3)(2)(1)

 10
3!(5  3)! (3)(2)(1) (2)(1) 
Marginal & Conditional Probabilities
26

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. The following table gives a twoway classification of their responses.
In favor
Against
Male
15
45
60
Female
4
36
40
19
81
100
Total

Total
Marginal probability is the probability of a single event
without consideration of any other event
Marginal & Conditional Probabilities
27

Suppose one employee is selected, he/she maybe classified
either on the bases of gender alone or on the bases of
opinion.

The probability of each of the following event is called
marginal probability


P(male) = 60/100

P(female) = 40/100


In favor
Against
Total
Male
15
45
60
P(in favor) = 19/100
Female
4
36
40
P(against) = 81/100
Total
19
81
100
Suppose the employee selected is known to be male. What
is the probability that he is in favor?
Marginal & Conditional Probabilities
28

This probability is called conditional probability and is
written as “the probability that the employee selected is in
favor given that he is a male.”
P( in favor | male )
The event whose probability
is to be determined

This event has
already occurred
Read as “given”
Conditional probability is the probability that an even will
occur given that another event has already occurred. If A
and B are two events, then the conditional probability of A
given B is written as
P(A | B )
Marginal & Conditional Probabilities
29
Example: Find the conditional probability P(in favor | male) for
the data on 100 employees
Solution
Marginal & Conditional Probabilities
30
Example: Find the conditional probability P(female|in favor) for
the data on 100 employees
Solution
Marginal & Conditional Probabilities
31
Example: Consider the experiment of tossing a fair die. Denote
by A and B the following events:
A={Observing an even number}, B={Observing a number of
dots less than or equal to 3}. Find the probability of the event
A, given the event B.
Solution
Mutually Exclusive Events
32

Events that cannot occur together are called mutually
exclusive events.
Example: 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}
- A and B are mutually exclusive events but A and C are not.
- How about B and C?
- Simple events are mutually exclusive always.
Independent vs. Dependent events
33

Two events are said to be independent if the occurrence of
one does not affect the probability of the other one.

A and B are said to be independent events if either
P(A| B) = P(A) or P(B | A) = P(B)
Example: Refer to the information on 100 employees. Are
events “female (F)” and “in favor (A)” independent?
Solution:
Independent vs. Dependent events
34

What is the difference between mutually exclusive and
independent events?

It is common to get confused or not to tell the difference between
these two terminologies.

When two events are mutually exclusive, they cannot both happen.
Once the event B has occurred, event A cannot occur, so that P(A|B) =
0, or vice versa. The occurrence of event B certainly affects the
probability that event A can occur. Therefore,


Mutually exclusive events must be dependent.

Independent events are never mutually exclusive.
But, dependent events may or may not be mutually exclusive
Independent vs. Dependent events
35
Example: A sample of 420 people were asked if they smoke or not
and whether they are graduate or not. The following two-way
classification table gives their responses
College graduate
Not a college graduate
Smoker
35
80
Nonsmoker
130
175
If an person is selected at random from this sample, find the
probability that this person is a
College graduate (G).
Nonsmoker (NS).
Smoker (S) given the person is not a college graduate (NG).
College graduate (G) given the person is a nonsmoker (NS).
Are the events Smoker and college graduate independent?
Why?

Independent vs. Dependent events
Independent vs. Dependent events
37
Complementary Events
38

Two mutually exclusive events that taken together include
all the outcome for an experiment (sample space, S) are
called complementary events.

Consider the following Venn diagram:
The complement of event A, denoted by
A and read as “A bar” or “A
A
complement,” is the event that includes
B
all the outcomes for an experiment that
Venn diagram of two complementary evens are not in A
 Since two complementary events, taken together, include all
the sample space S, the sum of probabilities of all outcomes
is 1
A
P(A) + P(A) = 1  P(A) = 1 - P(A)
Complementary Events
39
Example: In a lot of five machines, two are defective. If one of
machines is randomly selected, what are the complementary
events for this experiment and what are their probabilities?
Solution: The two complementary events for this experiment are
A = the machine selected is defective
A = the machine selected is not defective
Since there are 2 defective and three non defective machines, the
probabilities of each event is
P(A) = 2/5 = 0.4
P(A) = 3/5 = 0.6
or simply P(A) = 1 = 0.4 = 0.6
A
A
Intersection of Events & Multiplication Rule
40
Intersection of events

Let A and B be two events defined in a sample space. The
intersection of A and B represents the collection of all
outcomes that are common to both A and B is denoted by
any of the followings
A and B, A ∩ B, or AB
Example:
Let A = the event that a
person owns a PC
Let B = the event that a
person owns a mobile
A
A and B
B
Intersection of events A and B
Intersection of Events & Multiplication Rule
41
Multiplication rule

Sometimes we may need to find the probability of two
events happening together.

The probability of the intersection of two events A and B is
called their joint probability. It is written
P(AB) or P(A ∩ B)

It can be obtained by multiplying the marginal probability
of one event with the conditional probability of the second
one.

Multiplication rule: The probability of the intersection of
two events A and B is
P(AB) = P(A ∩ B) = P(A)P(B|A) = P(B)P(A|B)
Intersection of Events & Multiplication Rule
42
Example: The following table gives the classification of all
employees of a company by gender and college degree.
College
graduate (G)
Not a college graduate
(N)
Total
Male (M)
7
20
27
Female (F)
4
9
13
Total
11
29
40
If one employee is selected at random, what is the probability
that the employee is a female and a college graduate?
Solution
Intersection of Events & Multiplication Rule
Intersection of Events & Multiplication Rule
44
Example: A box contains 20 DVD, 4 of which are defective. If
two DVDs are selected at random (without replacement),
what is the probability that both are defective?
Solution:
Intersection of Events & Multiplication Rule
45

If events A and B are independent, their joint probability
simplifies from
P(A ∩ B) = P(A)P(B |A)

TO
P(A ∩ B) = P(A)P(B )
Sometimes we know the joint probability of two events A
and B, in this case, the conditional probability of B given A
or A given B is
P (A  B )
P (A | B ) 
P (B )
P (A  B )
P (B | A ) 
P (A )
given that P(A )  0 and P(B )  0
Intersection of Events & Multiplication Rule
46
Example: According to a survey, 60% of all homeowners owe
money on home mortgages. 36% owe money on both home
mortgages and car loans. Find the conditional probability that
a homeowner selected at random owes money on a car loan
given that he owes money on a home mortgage.
Solution:
Intersection of Events & Multiplication Rule
47
Example: A computer company has two quality control
inspectors, Mr. Smith and Mr. Robertson, who independently
inspect each computer before it is shipped to a client. The
probability that Mr. Smith fails to detect a defective PC is .02
while it is .01 for Mr. Robertson. Find the probability that both
inspectors will fail to detect a defective PC.
Solution:
Intersection of Events & Multiplication Rule
48
Example: The probability that a patient is allergic to Penicillin
is .2. Suppose this drug is administrated to three patients.
Find
a) The probability that all three of them are allergic to it
b) At least one of them is not allergic
Solution: Let
A = 1st patient is allergic to penicillin
B = 2nd patient is allergic to penicillin
C = 3rd patient is allergic to penicillin
Union of Events & Addition Rule
49
Union of events

Let A and B be two events defined in a sample space S. The
union of events A and B is the collection of all outcomes that
belong either to A and B or to both A and B and is denoted
by “ A or B” or “A ∪ B”
S
A
B
Union of Events & Addition Rule
50
Example: A company has 1000 employees. Of them, 400 are
females and 740 are labor union members. Of the 400 females,
250 are union members. Describe the union of events “female”
and “union member”
Solution
Union of Events & Addition Rule
51
Addition rule
The probability of the union of two events A and B is
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Example: 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 and it shown in the
following table.
Opinion
Favor
Oppose
Neutral
Total
Faculty
45
15
10
70
Student
90
110
30
230
Total
135
125
40
300
Find the probability that a person selected is a faculty member or in
favor.
Union of Events & Addition Rule
52
Solution:
Union of Events & Addition Rule
53
Example: There are a total of 7225 thousand persons with
multiple jobs in the US. Of them, 4115 thousand are male, 1742
thousand are single, and 905 thousand are male and single.
What is the probability that a selected person is a male or
single?
Solution:
Union of Events & Addition Rule
54

The probability of the union of two mutually exclusive
events A and B is
P(A ∪ B) = P(A) + P(B)
Example: 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 and it shown in the following table.
Opinion
Favor
Oppose
Neutral
Total
Faculty
45
15
10
70
Student
90
110
30
230
Total
135
125
40
300
Find the probability that a person selected is in favor or is
neutral
Union of Events & Addition Rule
55
Solution:
Union of Events & Addition Rule
56
Example: Eighteen percent of the working lawyers in the
United States are female. Two lawyers are selected at random
and it is observed whether they are male or female.
a) Draw a tree diagram for this experiment
b) Find the probability that at least one of the two lawyers is a
female.
Solution:
Law of Total Probability & Bayes’ Rule
Consider the following Venn diagram
B
B1
A1
B2
Bn-1
A2
An-1
Bn
A
An
Can we find the area (probability) of A assuming that we
know the probability of each Bi & P(A|Bi) ?
Law of Total Probability & Bayes’ Rule
58
Addition Law
P (A  B )  P (A )  P (B )  P (A  B )
Multiplication Law
P (A  B )  P (B )P (A | B )
P (A  B )  P (A )P (B | A )
Conditional Probability
P (A  B )
P (A | B ) 
P (B )
P (A  B )
P (B | A ) 
P (A )
Law of Total Probability & Bayes’ Rule
59
Bayes’ theorem (two-event case)
P (A1 )P (B | A1 )
P (A1 | B ) 
P (A1 )P (B | A1 )  P (A 2 )P (B | A 2 )
P (A 2 )P (B | A 2 )
P (A 2 | B ) 
P (A1 )P (B | A1 )  P (A 2 )P (B | A 2 )
Bayes’ theorem
P (A i )P (B | A i )
P (A i | B ) 
P (A1 )P (B | A1 )  P (A 2 )P (B | A 2 )  ...  P (A n )P (B | A n )
Law of Total Probability & Bayes’ Rule
60
Example: Manufacturing firm that receives shipment of parts
from two different suppliers. Currently, 65 percent of the
parts purchased by the company are from supplier 1 and the
remaining 35 percent are from supplier 2. Historical Data
suggest the quality rating of the two supplier are shown in the
table:
Good Parts
Bad Parts
Supplier 1
98
2
Supplier 2
95
5
a) Draw a tree diagram for this experiment with the
probability of all outcomes
b) Given the information the part is bad, What is the
probability the part came from supplier 1?
Law of Total Probability & Bayes’ Rule
61
Solution:
Law of Total Probability & Bayes’ Rule
62
Example: An insurance company rents 35% of the cars for its
customers from Avis and the rest from Hertz. From past
records they know that 8% of Avis cars break down and 5% of
Hertz cars break down. A customer calls and complains that
his rental car broke down. What is the probability that his car
was rented from Avis?
Law of Total Probability & Bayes’ Rule
63
Solution:
Example
64
A study was recently done in which 500 people were asked to indicate their
preferences for one of three products. The following table shows the
breakdown of the responses by gender of the respondents.
Gender
Male
Female
Product Preference
A
B
C
80
20
10
200
70
120
Suppose one person is randomly chosen. Based on this data, what is the
probability that the person chose is a female who prefers product C?
A) .31
B) .92
C) .86
D) .24
Example
65
The method of probability assessment that relies on an
examination of historical data from similar situations is
A) classical assessment
B) relative frequency of occurrence
C) subjective assessment
D) historical assessment