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Chapter Five
Elementary
Probability Theory
Understanding Basic Statistics
Fourth Edition
By Brase and Brase
Prepared by: Lynn Smith
Gloucester County College
Probability
• Probability is a numerical
measurement of likelihood of an event.
• The probability of any event is a
number between zero and one.
• Events with probability close to one are
more likely to occur.
• Events with probability close to zero
are less likely to occur.
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Probability Notation
If A represents an event,
P(A)
represents the probability of A.
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If P(A) = 1
Event A is certain to occur
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If P(A) = 0
Event A is certain not to occur
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Three methods to find probabilities:
• Intuition
• Relative frequency
• Equally likely outcomes
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Intuition Method of Determining Probability
• Incorporates past experience,
judgment, or opinion.
• Is based upon level of confidence in the
result
• Example: “I am 95% sure that I will
attend the party.”
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Probability as Relative Frequency
Probability of an event =
the fraction of the time that the event
occurred in the past =
f
n
where f = frequency of an event
n = sample size
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Example of Probability
as Relative Frequency
If you note that 57 of the last 100
applicants for a job have been female,
the probability that the next applicant is
female would be:
57
100
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Equally Likely Outcomes
• No one result is expected to occur
more frequently than any other.
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When outcomes are equally likely
Probabilit y of an event 
Number of outcomes favorable to event
Total number of outcomes
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Example of Equally Likely Outcome Method
When rolling a die, the probability of
getting a number less than three =
2 1

6 3
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Law of Large Numbers
In the long run, as the sample size
increases and increases, the relative
frequencies of outcomes get closer and
closer to the theoretical (or actual)
probability value.
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Statistical Experiment or
Statistical Observation
• Any random activity that results in a
definite outcome
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Event
• A collection of one or more outcomes of
a statistical experiment or observation
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Simple Event
• An outcome of a statistical experiment
that consists of one and only one of the
outcomes of the experiment
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Sample Space
• The set of all possible distinct
outcomes of an experiment
• The sum of all probabilities of all simple
events in a sample space must equal
one.
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Sample Space for the
rolling of an ordinary die:
1, 2, 3, 4, 5, 6
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For the experiment of
rolling an ordinary die:
3 1
• P(even number) = 
6 2
• P(result less than five) = 4  2
6
3
5
• P(not getting a two) =
6
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Complement of Event A
• the event that A does not occur
• Notation for the complement of event A:
A
c
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Event A and its complement Ac
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Probability of a Complement
• P(event A does not occur) =
P(Ac) = 1 – P(A)
• So, P(A) + P(Ac) = 1
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Probability of a Complement
If the probability that it will snow today
is 30%,
P(It will not snow) = 1 – P(snow) =
1 – 0.30 = 0.70
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Probability Related to Statistics
• Probability makes statements about
what will occur when samples are
drawn from a known population.
• Statistics describes how samples are to
be obtained and how inferences are to
be made about unknown populations.
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Independent Events
• The occurrence (or non-occurrence) of
one event does not change the
probability that the other event will
occur.
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If events A and B are independent,
• P(A and B) = P(A) P(B)
•
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Conditional Probability
• If events are dependent, the fact that
one occurs affects the probability of the
other.
• P(A, given B) equals the probability that
event A occurs, assuming that B has
already occurred.
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General Multiplication Rule for Any Events:
• P(A and B) = P(A) P(B, given A)
•
• P(A and B) = P(B) P(A, given B)
•
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The Multiplication Rules:
• For independent events:
P(A and B) = P(A) P(B)
•
• For any events:
P(A and B) = P(A) P(B, given A)
•
P(A and B) = P(B) P(A, given B)
•
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For independent events:
P(A and B) = P(A) P(B)
•
When choosing two cards from two separate
decks of cards, find the probability of getting
two fives.
P(two fives) =
P(5 from first deck and 5 from second) =
1 1
1
 
13 13 169
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For dependent events:
P(A and B) = P(A) · P(B, given A)
When choosing two cards from a deck
without replacement, find the
probability of getting two fives.
P(two fives) =
P(5 on first draw and 5 on second) =
4 3
12
1



52 51 2652 221
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“And” versus “or”
• And means both events occur
together.
• Or means that at least one of the
events occur.
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The Event A and B
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The Event A or B
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General Addition Rule
For any events A and B,
P(A or B) =
P(A) + P(B) – P(A and B)
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When choosing a card from an
ordinary deck, the probability of
getting a five or a red card:
P(5 ) + P(red) – P(5 and red) =
4 26 2 28 7




52 52 52 52 13
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When choosing a card from an
ordinary deck, the probability of
getting a five or a six:
P(5 ) + P(6) – P(5 and 6) =
4
4
0
8
2




52 52 52 52 13
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Mutually Exclusive Events
• Events that are disjoint, cannot happen
together.
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Addition Rule for Mutually Exclusive Events
For any mutually exclusive events A
and B,
P(A or B) = P(A) + P(B)
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When rolling an ordinary die:
P(4 or 6) =
1 1 2 1
  
6 6 6 3
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Survey results:
Education:
Males
Females Row totals
College
Graduates
54
62
116
Not College 31
Graduates
40
71
Column
totals
102
187(Grand total)
85
P(male and college grad) = ?
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Survey results:
Education:
Males
Females Row totals
College
Graduates
54
62
116
Not College 31
Graduates
40
71
Column
totals
102
187(Grand total)
85
54
P(male and college grad) =
187
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Survey results:
Education:
Males
Females Row totals
College
Graduates
54
62
116
Not College 31
Graduates
40
71
Column
totals
102
187(Grand total)
85
P(male or college grad) = ?
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5 | 43
Survey results:
Education:
Males
Females Row totals
College
Graduates
54
62
116
Not College 31
Graduates
40
71
Column
totals
102
187(Grand total)
85
147
P(male or college grad) =
187
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5 | 44
Survey results:
Education:
Males
Females Row totals
College
Graduates
54
62
116
Not College 31
Graduates
40
71
Column
totals
102
187(Grand total)
85
P(male, given college grad) = ?
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5 | 45
Survey results:
Education:
Males
Females Row totals
College
Graduates
54
62
116
Not College 31
Graduates
40
71
Column
totals
102
187(Grand total)
85
54
P(male, given college grad) =
116
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Counting Techniques
• Tree Diagram
• Multiplication Rule of Counting
• Permutations
• Combinations
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Tree Diagram
• a visual display of the total number of
outcomes of an experiment consisting
of a series of events
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Tree diagram for selecting class schedules
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Find the number of paths without
constructing the tree diagram:
Experiment of rolling two dice,
one after the other and observing
any of the six possible outcomes
each time .
Number of paths = 6 x 6 = 36
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Multiplication Rule of Counting
For any series of events, if there are
n1 possible outcomes for event E1
and n2 possible outcomes for event E2,
etc.
then there are
n1 X n2 X ··· X nm possible outcomes
for the series of events E1 through Em.
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Area Code Example
In the past, an area code consisted of 3 digits,
the first of which could be any digit from 2
through 9.
The second digit could be only a 0 or 1.
The last could be any digit.
How many different such area codes were
possible?
8  2  10 = 160
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Ordered Arrangements
In how many different ways could four
items be arranged in order from first to
last?
4  3  2  1 = 24
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Factorial Notation
• n! is read "n factorial"
• n! is applied only when n is a whole
number.
• n! is a product of n with each positive
counting number less than n
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Calculating Factorials
• 5! = 5 • 4 • 3 • 2 • 1 = 120
• 3! = 3 • 2 • 1 = 6
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Definitions
•
1! = 1
• 0! = 1
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Complete the Factorials:
4! = 24
10! = 3,628,800
6! = 720
15! = 1.3077 x 1012
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Permutations
• A Permutation is an arrangement in a
particular order of a group of items.
• There are to be no repetitions of items
within a permutation.
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Listing Permutations
How many different permutations of the
letters a, b, c are possible?
Solution: There are six different
permutations:
abc, acb, bac, bca, cab, cba.
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Listing Permutations
How many different two-letter permutations of
the letters a, b, c, d are possible?
Solution: There are twelve different
permutations:
ab, ac, ad, ba, ca, da, bc, bd, cb, db, cd, dc.
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Permutation Formula
The number of ways to arrange in
order n distinct objects, taking
them r at a time, is:
Pn , r
n!

n  r  !
where n and r are whole numbers and n > r.
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Another Notation for Permutations
P
n r
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Find P7, 3
P7 , 3
7!
7! 5040

 
 210
( 7  3)! 4!
24
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Applying the Permutation Formula
6
P3, 3 = _______
P6, 2
30
= __________
P15, 2
12
P4, 2 = _______
336
P8, 3 = _______
210
= _______
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Application of Permutations
A teacher has chosen eight possible
questions for an upcoming quiz. In how
many different ways can five of these
questions be chosen and arranged in order
from #1 to #5?
Solution: P8,5 =
8!
3!
= 8• 7 • 6 • 5 • 4 = 6720
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Combinations
A combination is a grouping
in no particular order
of items.
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Combination Formula
The number of combinations of n objects
taken r at a time is:
Cn , r
n!

(n  r ) ! r !
where n and r are whole numbers and n > r.
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Other Notations for Combinations:
C
n
r
n
or  
r 
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Find C9, 3
C9, 3
9!
9!
362880



 84
3!( 9  3)! 3!6! 6( 720 )
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Applying the Combination Formula
C5, 3 = 10
C7, 3 = 35
1
C15, 2 = 105
C3, 3 =
C6, 2 = 15
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Application of Combinations
A teacher has chosen eight possible
questions for an upcoming quiz. In how
many different ways can five of these
questions be chosen if order makes no
difference?
Solution: C8,5
8
!
=
= 56
5!3 !
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Determining the Number of
Outcomes of an Experiment
• If the experiment consists of a series of
stage with various outcomes, use the
multiplication rule or a tree diagram.
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Determining the Number of
Outcomes of an Experiment
If the outcomes consist of ordered
subgroups of r outcomes taken from a
group of n outcomes use the
permutation rule.
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Determining the Number of
Outcomes of an Experiment
If the outcomes consist of non-ordered
subgroups of r items taken from a
group of n items use the combination
rule.
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