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Lecture Slides
Elementary Statistics
Twelfth Edition
and the Triola Statistics Series
by Mario F. Triola
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Chapter 4
Probability
4-1 Review and Preview
4-2 Basic Concepts of Probability
4-3 Addition Rule
4-4 Multiplication Rule: Basics
4-5 Multiplication Rule: Complements and Conditional
Probability
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Experiments, Outcomes, and Sample
Spaces
Example #3:
Draw a tree diagram for three tosses of a coin. List all outcomes for this
experiment in a sample space S.
Solution:
Let “H” represent head and “T” represent tail.
Therefore, for each experiment, the outcome is either a “H” or “T”.
1st
Selectio
n
H
T
3rd
2nd
Selectio Selectio
n
n
HHH
H
HHT
T
H
HTH
H
HTT
T
THH
T
H
H
T
H
T
T
S = {HHH, HHT, HTH, HTT, THH, THT, TTH,
TTT}
THT
TTH
TTT
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
3
Section 4.3-‹#›
Simple and Compound Events
a.
b.
Event is a collection of one or more of the outcomes of an
experiment. An event could be the entire or portion of a
sample space. Therefore, an event can be classified as:
 Simple event
 Compound event
A simple event, Ei, consists of one and only one of the final
outcomes of an experiment. In Example #3,
E1=(HHH), E2=(HHT), E3=(HTH), E4=(HTT), E5=(THH),
E6=(THT), E7=(TTH), and E8=(TTT)
c.
A compound event consists of more than one outcome of an
experiment. It is represented by
A, B, C, D,..., or A1, A2, A3,..., B1, B2, B3,,...
Reconsider Example #3, let A be the event that two of the three tosses
will result in heads. Then, event A is given by
A = {HHH, HHT, HTH, THH}
4
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Simple and Compound Events
Example #4:
A box contains a certain number
of computer parts, a few of
which are defective. Two parts
are selected at random from this
box and inspected to determine
if they are good or defective.
List all the outcomes included in
each of the following events.
Indicate which are simple and
which are compound events.
a)At
least one part is good.
b)Exactly one part is defective.
c)The first part is good and the
second is defective.
d)At most one part is good.
Solution:
Let,
D = a defective part
G = a good part
The experiment has the following outcomes:
DD = both parts are defective
DG = the 1st part is defective and the 2nd is good
GG = both parts are good
GD = the 1st part is good and the 2nd is defective
a)At
least one part is good = {DG, GG, GD}
compound event
b)Exactly one part is defective = {DG, GD}
compound event
c)The 1st is good and 2nd defective = {GD}
simple event
d)At most one part is good = {DD, DG, GD}
compound event.
5
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
MARGINAL AND CONDITIONAL
PROBABILITIES
The following table gives the responses of 2,000 randomly selected adults who
were asked whether or not they have shopped on internet.
Have shopped
Have never shopped
Male
500
700
Female
300
500
Discussion
1.
The table is a two-way classification of 2,000 adults.
2.
The table is called contingency table and each box with a
numeric entry is called a cell.
3.
Each cell gives the frequency of two characteristics:
a)
Gender (male or female) and
b)
Opinion (have shopped or have never shopped).
6
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Marginal and Conditional
Probabilities
Discussion
3.
By adding the row totals and column totals, we obtain a new
table.
4.
Have shopped
Have never
shopped
Total
Male
500
700
1200
Female
300
500
800
Total
800
1200
2000
If only one characteristic, “have shopped”, “have never
shopped”, “male”, or “female”, is being considered at a time,
the probability of each event is called marginal probability or
simple probability.
7
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Section 4.3-‹#›
Marginal and Conditional
Probabilities
Have shopped
Have never
shopped
Total
Male
500
700
1200
Female
300
500
800
Total
800
1200
2000
The marginal probability or simple probability is a probability of a
single event without consideration of any other event.
From the table, the marginal probabilities of the characteristics are as
follows:
P (has shopped) 
# of adults who have shopped 800
# of males
1200

 0.4 P (male) 

 0.6
total number of adults
2000
total number of adults 2000
# of adults who have never shopped 1200

 0.6
total number of adults
2000
# of females
800
P (Female) 

 0.4
total number of adults 2000
P (has never shopped) 
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8
Section 4.3-‹#›
Marginal and Conditional
Probabilities
Now suppose we want to find the probability that the randomly
selected adult has shopped on the internet, assuming that the
adult is female.
In other words, the event that the adult is female has already
occurred. This probability is called conditional probability, and it
is written,
P(has shopped | female) and is read as
The probability that the selected adult has shopped on the
internet given that the event “female” has already occurred.
9
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Marginal and Conditional
Probabilities
General Statement
Suppose A and B are two events, then the conditional
probability of A given B is written as ,
P(A|B).
Again from the contingent table,
P Male | has never shopped  
Number of males who have never shopped
700

 0.58
Total number of adults who have never shopped 1200
10
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Section 4.3-‹#›
Marginal and Conditional Probabilities
Tree Diagram
HS | M
M | HS
HS
M
HNS | M
M
HS
HS
F
800/2000
M | HS
500/800
HS | F
M
HNS | F
HNS
1200/2000
500/1200
M | HNS
F | HNS
11
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
MUTUALLY EXCLUSIVE EVENTS
Definition
 Mutually exclusive events are events that do not have any outcome in
common.





Ex. Events for rolling a die: A = an even number is observed
B = an odd number is observe
C = a number less than five is observed
Mutually Exclusive
Mutually nonexclusive event
Most importantly, the occurrence of one event prevents the occurrence of
the other mutually exclusive events.
12
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Mutually Exclusive Events
• For example, the outcomes of tossing a coin are mutually exclusive
because both Head and Tail outcomes could not occur at the same
time. The occurrence of Head prevents occurrence of Tail to occur.
HH
H
T
HT
H
TH
H
T
T
TT
13
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Mutually Exclusive Events
Solution
Example #12
There are 160 practicing physicians
in a city. Of them, 75 are female
and 25 are pediatricians. Of the 75
female, 20 are pediatricians. Are
the events “female” and
“pediatrician” mutually exclusive?
Explain why or why not.
Example #13
Male
Ped
Non-Ped
Totals
5
80
85
Female
20
55
75
The events “female” and “pediatrician” are
not mutually exclusive because a
physician could be a female and a
pediatrician as shown above.
Define the following two events for two tosses of a coin:
A = at least one head
B = both tails are obtained
Are A and B mutually exclusive? Explain why or why not.
Solution:
The experiment involves tossing a coin twice. The sample space
S = {HH, HT, TT, TH}
where H = Head and T = Tail.
The events are:
A = {HH, HT, TH} & B = {TT}
A and B are mutually exclusive. They do not have any outcome in common.
14
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Section 4.3-‹#›
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).



If P(A | B) = P(A) is true, then P(B | A) = P(B) is also true.
If P(A | B) = P(A) is false, then P(B | A) = P(B) is also false.
If the occurrence of one event affects the probability of the other,
then we say that the events are dependent. In other words, two
events are dependent if
either P(A | B) ≠ P(A) or P(B | A) ≠ P(B).
15
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Independent Versus Dependent
Events
General Statement
1. Two events are either mutually exclusive or independent.
a. Mutually exclusive events are dependent
b. Independent events are never mutually exclusive
2.
Dependent events may or may not be mutually exclusive.
Solution
Example #14
There are 160 practicing physicians
in a city. Of them, 75 are female
and 25 are pediatricians. Of the 75
female, 20 are pediatricians. Are
the events “female” and
“pediatrician” independent? Explain
why or why not.
Since P(female | Pediatrician) ≠ P(female),
then the events are not independent.
Ped
Non-Ped
Totals
Male
5
80
85
Female
20
55
75
Totals
25
135
160
P  female  
75
 0.47
160
P  female | Pediatrician  
20
 0.80
25
16
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Section 4.3-‹#›
Independent Versus Dependent
Events
Solution
Example #15:
Two donut bakers baked 1000 donut
holes. Baker A baked 600 donuts, of
which 450 were sold and the
remaining were discarded. Baker B
baked the remaining donuts, of which
100 were discarded. The events are
“Baker A”, “Baker B”, “sold donuts”,
and “discarded donuts”. Prepare a
contingent table for this experiment.
Are the events “Baker A” and “sold
donuts” independent? Explain why or
S|A
why not.
A
S|B
S
D |A
A
D
250/1000
100/250
D|B
Sold
Donut
Discarded
Donuts
Totals
Baker A
450
150
600
Baker B
300
100
400
Totals
750
250
1000
P  Baker A  
600
 0.60
1000
P  Baker A| Sold Donuts  
450
 0.60
750
Since P(Baker A | sold donut) =
P(Baker A), then the events are
independent because the occurrence of
event “sold donuts” does affect the
probability of event “Baker A”.
17
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Independent Versus Dependent
Events
Example #15.1:
A statistical experiment has 10 equally likely outcomes that are
denoted by 10, 11, 12, 13, 14, 15, 16, 17, 18, 19.
Let event A = { 10, 12, 14, 16} and event B = {11, 13, 15}
a. Are events A and B mutually exclusive?
yes
b. Are event A and B independent events?
P ( A) 
4
 0.4
10
P( A / B)  0
Because the two probabilities are not the same, the two events are not
independent. Also, we know that mutually exclusive events are always
dependent.
18
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Section 4.3-‹#›
COMPLEMENTARY EVENTS
Definition
1. The complement of event A, denoted by Ā and is read as “A bar” or
“A complement,” is the event that includes all the outcomes for an
experiment that are not in A.
2. Therefore, complementary events are always mutually exclusive.
3. Two complementary events, combined together, includes all the
outcomes of the experiment.
P  A   P(A)  1
P  A   1  P(A)
and
 
P A  1  P(A)
19
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Section 4.3-‹#›
Complementary Events
Example #15 – Problem
Let A be the event that a number less than 3 is obtained
if we roll a die once. What is the probability of A? What is
the complementary event of A, and what is its
probability?
Solution
2
P  A    0.33
6
A ={a number  3}
4
P  A   1  P(A)  0.67 or P  A    0.67
6
20
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Section 4.3-‹#›
Complementary Events
Example #15.1:
A statistical experiment has 10 equally likely outcomes that are
denoted by 10, 11, 12, 13, 14, 15, 16, 17, 18, 19.
Let event A = { 10, 12, 14, 16} and event B = {11, 13, 15}
What are the complements of event A and
B, respectively, and their probabilities?
Solution
A  {11,13,15,17,18,19},
6
P ( A) 
10
B  {10,12,14,16,17,18,19},
7
P( B) 
 .7
10
21
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Section 4.3-‹#›
INTERSECTION OF EVENTS AND THE
MULTIPLICATION RULE
Suppose an experiment resulted in a sample space described as,
S = {1, 2, 3, 4, 5, 6, 7, 8}
Also, three events from the experiment are define as consisting of the following
outcomes:
A = {1, 2, 3, 4}
A and B
B = {3, 4, 5, 6}
A  B, or simply AB
C = {5, 6, 7, 8}
1.
2.
3.
You can see that the Events A and B are not
mutually exclusive because they have two
common outcomes, 3 and 4.
Likewise, Events B and C are not mutually
exclusive because they have two outcomes, 5
and 6, in common.
Given two Events, A and B, we can say that
the intersection of Events A and B is the
collection of all outcomes that are common
to both A and B. It can be written as,
22
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Section 4.3-‹#›
Multiplication Rule
The probability of the intersection of Events A and B is called the
joint probability and is define as the product of the marginal and
conditional probabilities. Joint probability is written as,
P(A  B) = P(A)  P(B|A) or
P(A  B) = P(B)  P(A|B)
From the two formulas, we can calculate the conditional
probabilities as,
P(A  B)
P(A  B)
P(B|A) =
and P(A|B) =
P(A)
P(B)
23
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Multiplication Rule for Independent
Events
Recall, P(A  B) = P(A)  P(B|A) or P(A  B) = P(B)  P(A|B)
This is true for dependent events if the probability of one is affected by
the occurrence of the other event. In other words,
P(A|B) ≠ P(A) and P(B|A)≠ P(B).
For independent events, the occurrence of one event does not affect the
probability of the other. Therefore,
P(A|B) = P(A) and P(B|A) = P(B).
Hence, we can rewrite the formula for calculating probability of
intersection of two independent events as,
P(A  B) = P(A)  P(B)
Note:
You can extend the multiplication rule to calculate the joint probability of
as many events as you want.
24
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Joint Probability of Mutually Exclusive
Events
We know from discussion of mutually exclusive events that mutually
exclusive events have no common outcomes. Therefore, they do not
have an intersection. In this case, we write the intersection of two or
more mutually exclusive events as
P(AB) = 0
Solution
Example #17
Find the joint probability of A and B for the a.
following:
b.
a.
P(B) = 0.59 and P(A|B) = 0.77
b.
P(A) = 0.28 and P(B|A) = 0.35
P(AB)=
=
P(AB)=
=
P(B) P(A|B)
(0.59)(0.77) = 0.4543
P(A) P(B|A)
(0.28)(0.35) = 0.098
Solution
Example #18
Find the joint probability for the following
three independent events:
a. P(A) = 0.49, P(B) = 0.67, P(C) = .75
b. P(A) = 0.71, P(B) = 0.34, P(C) = 0.45
a.
b.
P(ABC)= P(A)P(B)P(C)
= (.49)(.67)(.75) = 0.2462
P(ABC)= P(A)P(B)P(C)
= (.71)(.34)(.45) = 0.1086
25
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Intersection of Events and Multiplication
Rule
Solution
Example #19
The following table gives two way
classification of all basketball players at a
state university who began their college
careers between 2001 and 2005, based on
gender and whether or not they graduate.
Graduate
Did not
Graduate
Totals
Male
126
55
181
Female
133
32
165
Totals
259
87
346
a. If one of these players is selected at
random, find the following probabilities:
i. P(female and graduate)
ii. P(male and did not graduate)
b. Find P(graduate and did not graduate).
Is this probability zero? If yes, why?
133
 0.3844
346
P(F  G) = P(F)P(G|F)
165 133


 0.3844
346 165
55
P(male and did not graduate) =
 0.1590
346
P(M  not G) = P(M)P( not G|M)
P(female and graduate) =
181 55

346 181
 0.1590

P(graduate and did not graduate) = 0,
because these events are mutually
exclusive and you could not have
someone that is both “a graduate” and
“a non graduate”.
26
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Intersection of Events and Multiplication
Rule
Example #20
The following table gives two way
classification of the responses based
on the education levels of the persons
included in the survey and whether
they are financially better off, the
same as, or worse off than their
parents.
< High
Sch (D)
High
Sch (E)
> High
Sch (F)
Better off
(A)
140
450
420
Same as
(B)
60
250
110
Worse off
(C)
200
300
70
a. Suppose one adult is selected at random
from these 2000 adults. Find the
following probabilities:
i. P(better off and high school)
ii. P(more than high school and
worse off)
b. Find the joint probability of the events
“worse off” and “better off.” Is this
probability zero? Explain why or why
not.
27
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Intersection of Events and Multiplication
Rule
Solution
< High
Sch (D)
High
Sch (E)
> High
Sch (F)
Totals
Better
off (A)
140
450
420
1010
Same
as (B)
60
250
110
420
Worse
off (C)
200
300
70
570
Totals
400
1000
600
2000
450
 0.225
2000
P(A  E) = P(A)P(E|A)
1010 450


 0.225
2000 1010
P(A  E) =
70
 0.035
2000
P(F  C) = P(F)P(C|F)
600 70


 0.035
2000 600
P(F  C ) =
P(C  A = 0
Mutually exclusive events.
28
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Section 4.3-‹#›
UNION OF EVENTS AND THE ADDITION RULE
Definition
Let a sample space, S, consist of all
outcomes in Events A and B. Then the union
of the two events is the collection of all
outcomes that belong to either A or B or to
both A and B. This is denoted by
A  B or just A or B
S
A
B
A
B
S
Addition Rule
Addition rule is the method for calculating the
probability of the union of events. It is defined as,
P(A  B) = P(A)+P(B)-P(A  B)
Essentially, we calculate the probability of union of events by:
1. Adding the probability of each event and
2. Subtract the probability of the intersection of the events from result
in (1).
29
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Addition Rule for Mutually Exclusive Events
Let re-examine the formula for calculating the probability of union of
events.
P(A  B) = P(A)+P(B)-P(A  B)
However, we have said that for mutually exclusive events,
P(A  B) = 0
Then, for mutually exclusive events, the union of two events is,
P(A  B) = P(A)+P(B)
30
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Section 4.3-‹#›
Union of Events and the Addition Rule
Example #24 - Solution
Example #24
The following table gives two way
classification of all basketball players at a
state university who began their college
careers between 2001 and 2005, based on
gender and whether or not they graduate.
Graduate
(C)
Did not
Graduate (D)
Totals
Male
(A)
126
55
181
Female
(B)
133
32
P(B D) = P(B) + P(D) - P(B  D)

P(C
165
165 87 32


 0.6358
346 346 346
A) = P(C) + P(A) - P(C  A)

259 181 126


 0.9075
346 346 346
Totals
259
87
346
If one of these players is selected at
random, find the following probabilities:
a.
P(female or did not graduate)
b.
P(graduate or male)
31
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Union of Events and the Addition Rule
Example #25
The following table gives two way
classification of the responses based
on the education levels of the persons
included in the survey and whether
they are financially better off, the
same as, or worse off than their
parents.
< High
Sch (D)
High
Sch (E)
> High
Sch (F)
Better off
(A)
140
450
420
Same as
(B)
60
250
110
Worse off
(C)
200
300
70
Suppose one adult is selected at random from
these 2000 adults. Find the following
probabilities:
i. P(better off or high school)
ii. P(more than high school or worse off)
iii. P(better off or worse off)
32
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Section 4.3-‹#›
Union of Events and the Addition Rule
Solution
< High
Sch (D)
High
Sch (E)
> High
Sch (F)
Totals
Better
off (A)
140
450
420
1010
Same
as (B)
60
250
110
420
Worse
off (C)
200
300
70
570
Totals
400
1000
600
2000
a.
P(A E) = P(A) + P(E) - P(A E)
=
b.
P(F C ) = P(F) + P(C) - P(F C )
=
c.
1010 1000 450


 0.78
2000 2000 2000
600 570
70


 0.55
2000 2000 2000
P(A C ) = P(A) + P(C)
=
1010 570

 0.79
2000 2000
33
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 4.3-‹#›
Union of Events and the Addition
Rule
Example #26
The probability of a student getting an A grade in an economics class is
0.24 and that of getting a B grade is 0.28. What is the probability that a
randomly selected student from this class will get an A or a B in this
class? Explain why the probability is not equal to 1.0.
Solution
P(A) = 0.24
P(B) = 0.28
Then, the probability of getting an A or B is,
P(A or B) = P(A) + P(B) = 0.24 + 0.28 = 0.52
The probability is not equal to 1.0 because the student can get a grade of
C, D, or F.
34
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Section 4.3-‹#›