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Chapter 3 Probability
 The Concept of Probability
 Sample Spaces and Events
 Some Elementary Probability
Rules
 Conditional Probability and
Independence
Chapter 3 Probability
Section 3.1 The Concept of
Probability
An experiment is any process of observation with an uncertain
outcome.
--- On any single trial of the experiment, one and only one of the
possible outcomes will occur.
The possible outcomes for an experiment are called the
experimental outcomes
Probability is a measure of the chance that an experimental
outcome will occur when an experiment is carried out
Chapter 3 Probability
Example 3.1
Roll a die. The experimental outcomes are 1, 2, 3, 4, 5,
and 6.
An Outcome is the
particular result of
an experiment.
An Event is the
collection of one or
more outcomes of
an experiment.
Possible outcomes: The
numbers 1, 2, 3, 4, 5, 6
One possible event: The
occurrence of an even
number. That is, we collect
the outcomes 2, 4, and 6.
Chapter 3 Probability
 Regardless of the method used, probabilities must be
assigned to the experimental outcomes so that two
conditions are met:
Conditions
1. 0  P(E)  1
such that:
If E can never occur, then P(E) = 0
If E is certain to occur, then P(E) = 1
2. The probabilities of all the experimental outcomes must
sum to 1
Chapter 3 Probability
Section 3.2 Sample Spaces
and Events
Sample space (S):
The sample space is defined as the set of all possible
outcomes of an experiment.
e.g. All 6 faces of a die:
e.g. All 52 cards of a bridge deck:
Chapter 3 Probability
Example 3.2
Genders of Two Children
Let: B be the outcome that child is boy.
G be the outcome that child is girl.
Sample space S:
S = {BB, BG, GB, GG}
If B and G are equally
likely , then
P(B) = P(G) = ½
and
P(BB) = P(BG) = P(GB) =
P(GG) = ¼
Chapter 3 Probability
Recall example 3.2: Genders of Two Children
An event is a set of sample space outcomes.
Events
P(one boy and one girl) =
P(BG) + P(GB) = ¼ + ¼ = ½.
P(at least one girl) =P(BG) +
P(GB) + P(GG) = ¼ + ¼ + ¼
= ¾.
Note: Experimental Outcomes: BB, BG, GB, GG
All outcomes equally likely: P(BB) = … = P(GG) = ¼
Chapter 3 Probability
Example 3.3
Answering Three True-False Questions
A student takes a quiz that consists of three true-false
questions. Let C and I denote answering a question correctly
and incorrectly, respectively.
The graph on the next slide shows the sample space outcomes
for the experiment. The sample space consists of 8 outcomes:
CCC CCI CIC CII ICC ICI IIC III
Suppose the student is totally unprepared for the quiz and has
to blindly guess the answers. That is, the student has a 50-50
chance of correctly answering each question.
Chapter 3 Probability
So, each of the 8 outcomes is equally likely to occur.
P(CCC)=P(CCI)= ... = P(III)=1/8.
Chapter 3 Probability
Probabilities: Equally Likely Outcomes
If the sample space outcomes (or experimental
outcomes) are all equally likely, then the
probability that an event will occur is equal to
the ratio:
the number of outcomes that correspond to the event
The total number of outcomes
Chapter 3 Probability
Basic Computation of Probabilities
The probability of an event is also equal the sum of the
probabilities of the sample space outcomes that correspond to
the event.
Example 3.4
The probability that the student will get exactly two questions
correct is
P(CCI) + P(CIC) + P(ICC) = 1/8 + 1/8 + 1/8 = 3/8.
The probability that the student will get at least two questions
correct is
P(CCC) + P(CCI) + P(CIC) + P(ICC) = 1/8 + 1/8 + 1/8 + 1/8
= 1/2.
Chapter 3 Probability
Relative Frequency Method
Let E be an outcome of an experiment.
If the experiment is performed many times, P(E) is the
relative frequency of E.
P(E) is the percentage of times E occurs in many repetitions
of the experiment.
Use sampled or historical data to calculate probabilities.
Example 3.5
Suppose that of 1000 randomly selected consumers, 140
preferred brand X.
The probability of randomly picking a person who prefers
brand X is
140/1000 = 0.14 or 14%.
Chapter 3 Probability
Section 3.3 Some Elementary Probability Rules
The complement A of an event A is
the set of all sample space outcomes
not in A. Further, P(A) = 1 - P(A).
These figures are “Venn diagrams”.
Chapter 3 Probability
Union of A and B, A  B
Is an event consisting of the outcomes that
belong to either A or B (or both).
Intersection of A and B, A  B
Is an event consisting of the outcomes that
belong to both A and B.
Chapter 3 Probability
The Addition Rule
The probability that A or B (the union of A and B) will
occur is P(A  B) = P(A) + P(B) - P(A  B)
where P(A  B) is the “joint” probability of A and B,
i.e., both occurring.
A and B are mutually exclusive if they have
no sample space outcomes in common, or
P(A  B) = 0.
equivalently, if
If A and B are mutually exclusive, then
P(A  B)=P(A)+P(B).
Chapter 3 Probability
Example 3.6


Newspaper Subscribers #1
Define events:
 A = event that a randomly selected household subscribes
to the Atlantic Journal.
 B = event that a randomly selected household subscribes
to the Beacon News.
Given:
 total number in city, N = 1,000,000
 number subscribing to A, N(A) = 650,000
 number subscribing to B, N(B) = 500,000
 number subscribing to both, N(A∩B) = 250,000
Chapter 3 Probability
Newspaper Subscribers #2

Use the relative frequency method to assign
probabilities
650,000
P  A 
 0.65
1,000,000
500,000
P B  
 0.50
1,000,000
250,000
P A  B  
 0.25
1,000,000
Chapter 3 Probability
Table3.1 A Contingency Table Subscription Data for the
Atlantic Journal and the Beacon News
Events
Subscribes to Does Not
Beacon News, Subscribe to
B
Beacon News,
Total
Subscribes to
Atlantic Journal, A
250,000
400,000
650,000
Does not
Subscribes to
Atlantic Journal,
250,000
100,000
350,000
Total
500,000
500,000
1,000,000
Chapter 3 Probability
Newspaper Subscribers #3
Refer to the contingency table in Table 3.1 for
all probabilities
For example, the chance that a household does
not subscribe to either newspaper
Calculate PA  B  , so from middle row
and middle column of Table 3.1,
100,000
P A  B  
 0.10.
1,000,000
Chapter 3 Probability
Newspaper Subscribers #4
The
chance that a household subscribes to either newspaper:
P(A  B)=P(A)+P ( B)  P( A  B)
 0.65  0.50  0.25
 0.90.
Note that if the joint probability was not subtracted, then
we would have gotten 1.15, greater than 1, which is
absurd.
Note: The subtraction avoids double counting the joint
probability.
Chapter 3 Probability
A Mutually Exclusive Case
If A and B are mutually exclusive, then
P(A  B)=P(A)+P(B).
Example 3.7: Consider randomly selecting a card from a standard
deck of 52 playing cards, and define the events
J= the randomly selected card is jack.
Q=the randomly selected card is queen.
P(J∪Q)=?
Since there are four jacks, four queens, we have P(J)=4/52 and
P(Q)=4/52. Furthermore, since there is no card that is both a jack
And a queen, the events J and Q are mutually exclusive and thus
P(J∩Q)=0. So we have
8
2

P(J ∪Q)=P(J)+P(Q)=
52 13
Chapter 3 Probability
Section 3.4 Conditional Probability and
Independence
The probability of an event A, given that the event B
has occurred, is called the “conditional probability
of A given B” and is denoted as
Further,
P(A  B)
P(A|B) =
P(B)
Assume that P(B) is greater than 0.
Interpretation: Restrict the sample space to just event
B. The conditional probability P(A|B) is the chance
of event A occurring in this new sample space.
Chapter 3 Probability
Similarly, if A occurred, then what is the chance of
B occurring?
To answer this question, we need to introduce the
probability of event B, given that the event A has
occurred, i.e., the conditional probability of B
given A, denoted by P(B|A).
P(A  B)
P(B | A) =
P(A)
Assume that P(A) is greater than 0.
Chapter 3 Probability
Newspaper Subscribers
Given that the households that subscribe to the
Atlantic Journal, what is the chance that they also
subscribe to the Beacon News?
Calculate P(B|A), where
P A  B 
P  B | A 
P  A
0.25

 0.3846.
0.65
Chapter 3 Probability
Example: The Dean of the School of Business at Owens
University collected the following information about
undergraduate students in her college:
Major
Accounting
Male
170
Female
110
Total
280
Finance
120
100
220
Marketing
160
70
230
Management
150
120
270
Total
600
400
1000
Example 7
If a student is selected at random, what is the
probability that the student is a female (F)
accounting major (A)?
110
P(A and F) =.
1000
Given that the student is a
female, what is the
probability that she is an
accounting major?
110
P ( A and F )
 1000  0.275
P(A|F)=
400
P( F )
1000
Independence of Events
Two events A and B are said to be independent if and
only if P(A|B) = P(A) or, equivalently, P(B|A) = P(B).
That is, if the chance of event A occurring is not
influenced by whether the event B occurs and
vice versa; or if the occurrences of the events A
and B have nothing to do with each other, then A
and B are independent.
In fact if one of the above two equations holds,
so does the other, why?
Chapter 3 Probability
Newspaper Subscribers



Given that the households that subscribe to the Atlantic
Journal subscribers, what is the chance that they also
subscribe to the Beacon News?
 If independent, the P(B|A) = P(B).
Is P(B|A) = P(B)?
 Know that P(B) = 0.50.
 Just calculated that P(B|A) = 0.3846.
 0.50 ≠ 0.3846, so P(B|A) ≠ P(B).
B is not independent of A.
 A and B are said to be dependent.
Chapter 3 Probability
The Multiplication Rule
The joint probability that A and B (the intersection
of A and B) will occur is
P(A  B) = P(A)  P(B|A)
= P(B)  P(A|B).
If A and B are independent, then the probability
that A and B (the intersection of A and B) will
occur is
P(A  B) = P(A) P(B)  P(B)  P(A).
Chapter 3 Probability
A Question
Suppose in the following contingency table, where the
numbers represent probabilities, some data are lost.
1.Can you recover the missing data?
2.Are events R and C independent?
R
C
.4
C
.3
.5
R
Total
Chapter 3 Probability
Total
.6
1.00
Contingency Tables
P(R )
P(R  C)
R
R
Total
C
.4
.1
.5
P(R  C)
As P( R  C )  0.4
C
.2
.3
.5
Total
.6
.4
1.00
P(C)
P ( R)  P (C )  0.6  0.5  0.3
P ( R  C )  P ( R )  P(C )
the events R and C are dependent.
Chapter 3 Probability
General Addition Rule:
P(A or B) = P(A) + P(B) - P(A and B)
Special Rule of Addition: If A and B are mutually exclusive
P(A or B) = P(A) + P(B)
Complement Rule
P(A) + P(~A) = 1 or P(A) = 1 - P(~A).
General Rule of Multiplication
P(A and B) = P(A)P(B|A) or P(A and B) = P(B)P(A|B)
Special Rule of Multiplication: If A and B are independent
P(A and B) = P(A)P(B)
Chapter 3 Probability