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© 2012 McGraw-Hill
Ryerson Limited
© 2009 McGraw-Hill Ryerson Limited
1
© 2012 McGraw-Hill Ryerson Limited
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Learning Objectives
LO 1 Define probability.
LO 2 Explain the terms experiment, event, outcome,
LO 3
LO 4
LO 5
LO 6
permutations, and combinations.
Describe the classical, empirical, and subjective
approaches to probability.
Define the terms conditional probability and joint
probability
Calculate probabilities using the rules of addition and
rules of multiplication.
Use a tree diagram to organize and compute
probabilities.
© 2012 McGraw-Hill Ryerson Limited
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Statistical Inference
We now turn to the second facet of statistics, namely,
computing the chance that something will occur in the
future.
This facet of statistics is called statistical inference or
inferential statistics.
• Deals with conclusions about a population based on a
sample taken from that population.
• Probability theory, which has often been referred to
as the science of uncertainty.
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LO
1
WHAT IS A PROBABILITY?
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Probability
A probability is a measure of the likelihood that an event
in the future will happen.
It can only have a value between 0 and 1.
• An event with a value near 0 is not likely to happen.
• An event with a value near 1 is likely to happen.
LO
1
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LO
2
EXPLAIN THE TERMS
EXPERIMENT, EVENT, OUTCOME,
PERMUTATIONS, AND
COMBINATIONS
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Three Key Words
An experiment is a process that leads to the occurrence of
one, and only one, of several possible observations.
An outcome is a particular result of an experiment.
An event is a collection of one or more outcomes of an
experiment.
LO
2
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An Illustration
LO
2
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You Try It Out!
A lamp manufacturer is launching a new emergency
lamp. Its durability is to be tested by 90 consumers.
(a) What is the experiment?
(b) What is one possible outcome?
(c) Suppose 75 consumers used the new emergency
lamp and said that it is durable. Is 75 a probability?
(d) The probability that the new emergency lamp will be
a success is computed to be –1.0. Comment.
(e) Specify one possible event.
LO
2
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LO
3
APPROACHES TO PROBABILITY
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Approaches To Probability
Approaches to assigning probabilities:
• Objective Probability
• Classical
• Empirical
• Subjective Probability
LO
3
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Classical Probability
Classical probability is based on the assumption that the
outcomes of an experiment are equally likely.
Using the classical viewpoint, the probability of an event
happening is computed by dividing the number of
favourable outcomes by the number of possible outcomes:
LO
3
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Example – Classical Probability
Consider an experiment of rolling a six-sided die. What is
the probability of the event “an odd number of spots appear
face up”?
LO
3
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Solution – Classical Probability
The possible outcomes are:
There are three “favourable” outcomes (a one, a three, and
a five) in the collection of six equally likely possible
outcomes, Therefore:
Probability of an odd number 
3
Number of favorable outcomes
6 Total number of possible outcomes
 0.5
LO
3
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Two More Terms
Events are mutually exclusive if occurrence of one event
means that none of the other events can occur at the same
time.
Events are collectively exhaustive if at least one of the
events must occur when an experiment is conducted.
If the set of events is collectively exhaustive and the events
are mutually exclusive, then the sum of the probabilities
equals 1.
LO
3
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Empirical Probability
The probability of an event happening is the fraction of the
time similar events happened in the past.
The empirical approach to probability is based on what is
called the law of large numbers.
Over a large number of trials, the empirical probability of an
event will approach its true probability.
number of times event occurred in the past
Probability of event happening =
Total number of observations
LO
3
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Law of Large Numbers
Suppose we toss a fair coin. The result of each toss is
either a head or a tail. If we toss the coin a great number of
times, the probability of the outcome of heads will approach
0.5.
The following table reports the results of an experiment of
flipping a fair coin 1, 10, 50, 100, 500, 1000, and 10 000
times and then computing the relative frequency of heads.
LO
3
Number of
Trials
Number of
Heads
Relative Frequency
of Heads
1
0
.00
10
3
.30
50
26
.52
100
52
.52
500
236
.472
10 00
494
.494
10 000
5027
.5027
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Example – Empirical Probability
A tree plantation survey agency found that out of 175 trees
planted at a hillside area,140 trees survived without water
for a week. What is the probability that a tree will survive
without water for a week?
LO
3
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Solution – Empirical Probability
Using formula:
Number of times event occured in past
Total number of observations
140
P( A) 
 0.80
175
Probability of event happening 
We can use this as an estimate of probability. In other words,
based on past experience, the probability is .80 or 80% that a
tree will survive without water for a week.
LO
3
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Subjective Probability
The likelihood (probability) of a particular event happening
that is assigned by an individual based on whatever
information is available.
Illustrations of subjective probability are:
1. Estimating when the next provincial election will be.
2. Estimating the likelihood you will be married before
the age of 30.
3. Estimating the chance of the Maple Leafs winning
the Stanley Cup.
LO
3
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Probability Approaches Summary
Approaches to Probability
Objective Probability
Classical Probability
Empirical Probability
Based on equally
likely outcomes
Based on relative
frequencies
Subjective Probability
Based on available
information
CHART 4-1 Summary of Approaches to Probability
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You Try It Out!
1. A city club organized a rose fair. There were 100 red
roses, 75 pink roses, and 80 white roses. What is the
probability that a particular rose chosen at random is
white? Which approach did you use?
2. What is the probability that house prices in the Halifax
area will exceed $2000 per square foot during next 6
months? Which approach to probability did you use to
answer this question?
3. One card will be randomly selected from a standard 52card deck. What is the probability the card will be a
king? Which approach to probability did you use to
answer this question?
LO
3
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Principles of Counting
There are three principles for counting the number of
outcomes:
• The multiplication formula
• The permutation formula
• The combination formula
LO
3
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The Multiplication Formula
If there are m ways of doing one thing and n ways of doing
another thing, then there are (m)(n) ways of doing both.
For three events m, n, o:
The total number of arrangements = (m)(n)(o)
LO
3
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Example – The Multiplication Formula
An automobile dealer wants to advertise that for $29 999
you can buy a convertible, a two-door sedan, or a four-door
sedan, with your choice of either wire wheel covers or solid
wheel covers. How many different arrangements of models
and wheel covers can the dealer offer?
LO
3
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Solution – The Multiplication Form
Of course the dealer could determine the total number of
arrangements by picturing and counting them. There are
arrangements.
( )( ) ( )( )
Total possible arrangements = m n = 3 2 = 6
LO
3
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You Try It Out!
1. Six men and five women are participating in a pairs
dance competition. In how many ways can a pair be
formed?
2. A sewing kit assembler has 3 scissors, 4 thread
sets, and 5 button packets. A sewing kit contains
each of these items. How many sewing kits can be
made available?
LO
3
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The Permutation Formula
Any arrangement of r objects selected from a single group
of n possible objects when order is considered.
The arrangements “a b c” and “b a c” are different
permutations. The formula to count the total number of
different permutations is:
where:
n is the total number of objects.
r is the number of objects selected.
LO
2
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Example – The Permutation Formula
Jack had decided to give return gift to his 9 friends on his
birthday party. He bought 9 gifts. Only 4 friends could
attend the birthday party. In how many different ways can 9
gifts be distributed among 4 friends?
LO
2
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Solution – The Permutation Formula
This could also be found using formula as follows :
n Pr 
n!
 n  r !
If n = 9 gifts, and r = 4 friends available, the formula gives:
n Pr 
n!
9!
9! (9)(8)(7)(6)5!

 
 3024
5!
 n  r !  9  4! 5!
Therefore, 9 return gifts can be distributed among 4 friends
in 3024 different ways.
LO
2
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The Permutation Formula In Excel
LO
2
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The Combination Formula
A combination is the number of ways to choose r objects
from a group of n objects without regard to order.
The formula to count the number of r object combinations
from a set of n objects is:
LO
2
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Example – The Combination Formula
A group singing competition has been announced. Each
group should contain 4 students. A student can only be a
part of one group at a time. There should be at least 76
participants in total. If 8 schools participated with 4 students
at a time then how many possible combinations would
there be and would it be equal to or greater than 76? If not,
then how many schools should participate in the
competition?
LO
2
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Solution – The Combination Formula
Using formula :
n Cr 
n!
r ! n  r !
There are 70 combinations, found by:
n Cr 

n!
8!

r ! n  r  ! 4! 8  4  !
8!
 70
4!4!
Which is less than 76. Combination of 9 schools with 4
students at a time will give 126 different combinations. This
would be more than 76. Therefore 9 schools should
participate in the competition.
LO
2
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The Combination Formula In Excel
LO
2
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You Try It Out!
1. Five balls are to be drawn randomly from an urn. In how
many different ways can these balls can be drawn?
2. To be certified in a certain computer course you need to
pass different three tests. It does not matter in which
order you take these tests.
(a) How many permutations are possible?
(b) How many combinations are possible?
LO
2
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LO
5
SOME RULES FOR COMPUTING
PROBABILITIES
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Rules for Computing Probabilities
1.
2.
3.
4.
5.
LO
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Special rule of addition
General rule of addition
Complement rule
Special rule of multiplication
General rule of multiplication
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The Rules of Addition
Special Rule of Addition
If two events A and B are mutually exclusive, the probability
of one or the other events occurring equals the sum of their
probabilities.
Event
A
LO
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Event
B
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The Rules of Addition
The General Rule of Addition
If A and B are two events that are not mutually exclusive,
then P(A or B) is given by the following formula:
A
A
and
B
B
Both
LO
5
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Example – Special Rule of Addition
500 people were asked in which city they would like to buy
their own house. The number of preferences for three cities
are as follows:
City
Event
Number of
preferences
Probability of
Occurrence
Halifax
A
208
0.42
Yellowknife
B
137
0.27
Calgary
C
155
0.31
500
1.00

208
500
What is the probability that a person will prefer either
Halifax or Calgary?
LO
5
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Solution – Special Rule of Addition
The outcome “Halifax” is the event A. The outcome
“Calgary” is the event C. Applying the special rule of
addition:
P  A or C   P  A  P  C   0.42  0.31  .73
LO
5
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Venn Diagrams
A Venn diagram is a useful tool to depict addition or
multiplication rules.
To construct a Venn diagram:
• Draw a rectangle. The space inside represents the
total of all possible outcomes.
• Draw circles inside the rectangle to represent each
event.
Event
A
LO
5
Event
B
Event
C
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Example – General Rule of Addition
A music band consists of 55 musicians. Out of 55, 32 are
qualified and 15 are individuals working in different fields.
What is the probability that a musician selected at random
from this band is both qualified and working in a different
field?
LO
5
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Solution – General Rule of Addition
Musician
Probability
Explanation
Qualified
P(A)
= 32/55 32 qualified in a band of 55
musicians.
Working
Individual
P(B)
= 15/55 15 working individuals in a band of 55
musicians.
Qualified
Working
Individual
P(A and B) = 1/55
1 qualified working individual in a
band of 55 musicians.
P  A or B   P  A   P  B   P  A and B 
 32 / 55  15 / 55  1 / 55
 46 / 55, or 0.8364
LO
5
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LO
4
JOINT PROBABILITY
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JOINT PROBABILITY
A probability that measures the likelihood two or more
events will happen concurrently is called a joint
probability.
LO
5
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The Complement Rule
Used to determine the probability of an event occurring by
subtracting the probability of the event not occurring from 1.
Useful because sometimes it is easier to calculate the
probability of an event happening by determining the
probability of it not happening and subtracting the result
from 1.
Event
A
~A
LO
5
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The Complement Rule
Notice that the events A and ~A are mutually exclusive and
collectively exhaustive. Therefore, the probabilities of A and
~A sum to 1.
LO
5
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Example – The Complement Rule
500 people were asked in which city they would like to
purchase their own house. The number of preferences for
three cities are as follows :
City
Halifax
Yellowknife
Calgary
Event
Number of
preferences
A
B
C
Probability of
Occurrence
208
137
155
500
0.42
0.27
0.31
1.00
Use the complement rule to show the probability of
preference for Yellowknife city is 0.27. Show the solution
using a Venn diagram.
LO
5
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Solution – The Complement Rule
P  A or C   P  A   P  C 
 0.42  0.31  0.73
P  B  1 
 P  A  P  C 

 1   0.42  0.31  0.27
The Venn diagram of this situation is:
A
0.42
C
0.31
~(A or C) 0.27
LO
5
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You Try It Out!
A sample of candidates who came for an interview is
selected. Candidates are from four different educational
Backgrounds. They are classified as follows:
LO
5
Background
Event
Number of
Candidates
Engineering
A
100
Management
B
40
HR
C
130
Science
D
110
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You Try It Out!
(a) What is the probability that the first candidate
selected is:
(i) either from management or science background?
(ii) not from HR?
(b) Draw a Venn diagram illustrating your answers to
part (a).
(c) Are the events in part (a)(i) complementary, or
mutually exclusive, or both?
LO
5
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You Try It Out!
A sample of working women are surveyed regarding
how they go to work. Forty percent use their own
vehicle, 50 percent use public transport, and 10 percent
walk.
(a) What is the probability that a woman selected at
random either uses public transport or walks?
(b) Show this situation in the form of a Venn diagram.
LO
5
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The Rules of Multiplication
Special Rule of Multiplication
The special rule of multiplication requires that two events, A
and B, be independent.
Two events are independent if the occurrence of one does
not alter the probability of the occurrence of the other
event.
LO
5
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The Rules of Multiplication
The General Rule of Multiplication
Used to find the joint probability that two events will occur
when the events are not independent (i.e., they are
dependent).
It states that for two events, A and B, the joint probability
that both events will happen is found by multiplying the
probability that event A will occurring by the conditional
probability of event B occurring given that A has occurred.
LO
5
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Example – General Rule of Multiplication
A store has 12 umbrellas on sale. Nine are white and the
others are blue. Two umbrellas are sold. What is the
probability that both of them were blue?
LO
5
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Solution – General Rule of Multiplication
The event that the first umbrella sold was blue is B1. The
probability is P  B1   3 12.
The event that the second umbrella sold was also blue is
B2. The conditional probability that the second umbrella
sold was blue, given that the first umbrella sold was also
blue, is
P  B2 | B1   2 11.
 3  2 
P  B1 and B2   P  B1  P  B2 | B1        0.05
 12   11 
LO
5
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You Try It Out!
On Valentines Day it has been decided that all the
employees will dress in either red or white.
a) What is the probability that all employees will dress
in white?
b) What is the probability that all employees will dress
in red?
c) Does the sum of the probabilities for the events
described in parts (a) and (b) equal 1? Explain.
LO
5
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Example – Special Rule of Multiplication
In a marathon, 67 percent of the participants reached the
end within 4 hours. Three participants are selected at
random. What is the probability that all three participants
finished the race within 4 hours?
LO
5
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Solution – Special Rule of Multiplication
The individual probability that the first, second, and third
participant finished the race within 4 hours is 0.67.
P(S1) = 0.67 P(S2) = 0.67 P(S3) = 0.67
P(S1 and S2 and S3 )  P  S1  P  S2  P(S3 )   0.67  0.67  (0.67)  0.30
LO
5
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You Try It Out!
The probability that a scoop of ice cream will not melt at
room temperature for 15 minutes is 0.70. What is the
probability that all three scoops will not melt at room
temperature before 15 minutes?
LO
5
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Contingency Tables
A table used to classify sample observations according to
two or more identifiable characteristics
Level of measurement can be nominal.
Makes it easy to figure out probabilities.
E.g., Classification of the marks, out of 100, of 165 students
on a statistics test are given below :
Marks Scored
50–60
60–70
70 or more
Total
LO
5
Gender
Girls
Boys
30
35
35
45
15
5
80
85
Total
65
80
20
165
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Example – Contingency Tables
A sample of 160 consumers who use cell phones were
asked, “If another service provider offered more “talk time”
for less money, would you continue with the same provider
or go with the new one?” The responses were crosstabulated with their duration of use with the current service
provider.
Duration of service
Continuity
LO
5
Less than 1
month B1
1–5 months B2
6–10 months B3
More than 10
months B4
Total
Would
continue, A1
10
12
14
74
110
Would not
continue, ~A1
12
17
8
13
50
22
29
22
87
160
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Example – Contingency Tables
What is the probability of randomly selecting a consumer
who
1. Would continue and who has used the current provider
for more than 10 months of service?
2. Would continue or has used less than one month of
service?
LO
5
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Solution – Contingency Tables
P  A1 and B4   P  A1  P  B4 | A1 
74
 110   74 


 0.463


 160   110  160
P  A1 or B1   P  A1   P  B1   P  A1 and B1 
 0.69  0.14  0.06  0.77
LO
5
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You Try It Out!
Refer the table below to find the following probabilities.
Length of Training
Continuity
Would continue,
A1
Would not
continue, ~A1
Less than 1
month B1
1–5 months B2 6–10 months B3
More than 10
months B4
Total
9
12
6
74
101
21
17
8
32
78
30
29
14
106
179
(a) What is the probability of selecting candidate with more than 10
months of training?
(b) What is the probability of selecting a candidate with more than
10 months of training who would not continue the training?
(c ) What is the probability of selecting a candidate with more than
10 months of training or one who would not continue with the
training?
LO
5
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LO
6
TREE DIAGRAMS
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Tree Diagrams
The tree diagram is a graph that is helpful in organizing
calculations that involve several stages.
Each segment in the tree is one stage of the problem.
The branches of a tree diagram are weighted by
probabilities.
LO
6
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Construction of a Tree Diagram
1. To construct a tree diagram, draw a heavy dot on the
left to represent the root of the tree.
2. Two main branches go out from the root, the upper one
representing “would continue” and the lower one “would
not continue.” These probabilities can also be denoted
as P(A1) and P(~A1).
3. Four branches “grow” out of each of the two main
branches. These branches represent the duration of
service.
4. Finally, joint probabilities, that the events A1 and Bi, or
the events A2 and Bi will occur together.
LO
6
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Example – Tree Diagrams
LO
6
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You Try It Out!
A random sample of the students of a university was
chosen to determine their interest in the management
and engineering field.
Opinion
Interested
Not Interested
Area of
Interest
Total
Management
12
6
18
Engineering
16
13
29
47
(a) What is the table called?
(b) Draw a tree diagram, and determine the joint probabilities.
(c) Do the joint probabilities total 1.00? Why?
LO
6
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Chapter Summary
I. A probability is a value between 0 and 1 inclusive that
represents the likelihood a particular event will happen.
A. An experiment is the observation of some activity or
the act of taking some measurement.
B. An outcome is a particular result of an experiment.
C. An event is the collection of one or more outcomes
of an experiment.
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Chapter Summary
II. There are three definitions of probability.
A. The classical definition applies when there are n
equally likely outcomes to an experiment.
B. The empirical definition occurs when the number
of times an event happens is divided by the
number of observations.
C. A subjective probability is based on whatever
information is available.
III. Two events are mutually exclusive if by virtue of one
event happening the other cannot happen.
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Chapter Summary
IV. Two events are independent if the occurrence of one
event does not affect the occurrence of the other event.
V. There are three counting rules that are useful in
determining the number of outcomes in an experiment.
A. The multiplication rule states that if one event can
happen in m ways and another event can happen in
n ways, then the two events can happen in mn
ways.
Number of arrangements  (m)(n)
[4–2]
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Chapter Summary
B. A permutation is an arrangement in which the order
of the objects selected from a specific pool of
objects is important.
n!
n Pr 
(n  r )!
[4–3]
C. A combination is an arrangement in which the order
of the objects selected from a specific pool of
objects is not important.
n!
n Cr 
r!(n  r )!
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[4–4]
77
Chapter Summary
VI. The rules of addition refer to the union of events.
A. The special rule of addition is used when events are
mutually exclusive.
P ( A or B)  P ( A)  P ( B)
[4–5]
B. The general rule of addition is used when the events
are not mutually exclusive.
[4–7]
P ( A or B)  P ( A)  P ( B)  P ( A and B)
C. The complement rule is used to determine the
probability of an event happening by subtracting the
probability of the event not happening from 1.
P ( A)  1  P ( A)
[4–6]
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Chapter Summary
VII.The rules of multiplication refer to the product of events.
A. The special rule of multiplication refers to events that
are independent.
P ( A and B)  P ( A) P ( B)
[4–8]
B. The general rule of multiplication refers to events
that are not independent.
[4–9]
P ( A and B)  P ( A) P ( B | A)
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Chapter Summary
C. A joint probability is the likelihood that two or more
events will happen at the same time.
D. A conditional probability is the likelihood that an
event will happen, given that another event has
already happened.
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