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Understandable Statistics
Seventh Edition
By Brase and Brase
Prepared by: Lynn Smith
Gloucester County College
Chapter Four
Elementary Probability Theory
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
1
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.
• If an event has probability equal to one,
the event is certain to occur.
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2
Probability Notation
If A represents an event,
P(A)
represents the probability of A.
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3
Three methods to find
probabilities:
• Intuition
• Relative frequency
• Equally likely outcomes
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4
Intuition method
based upon our level of confidence
in the result
Example: I am 95% sure that I
will attend the party.
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5
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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6
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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7
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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8
Equally likely outcomes
No one result is expected to occur
more frequently than any other.
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9
Probability of an event when
outcomes are equally likely =
number of outcomes favorable to event
total number of outcomes
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10
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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11
Statistical Experiment
activity that results in a definite
outcome
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12
Sample Space
set of all possible outcomes of an
experiment
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13
Sample Space for the rolling of
an ordinary die:
1, 2, 3, 4, 5, 6
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14
For the experiment of
rolling an ordinary die:
• P(even number) =
3 = 1
6
2
• P(result less than four) = 3 = 1
6
2
5
• P(not getting a two) =
6
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15
Complement of Event A
the event not A
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16
Probability of a Complement
P(not A) = 1 – P(A)
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17
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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18
Probabilities of an Event and
its Complement
• Denote the probability of an event by the
letter p.
• Denote the probability of the complement
of the event by the letter q.
• p + q must equal 1
• q=1-p
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19
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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20
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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21
If events A and B are
independent,
P(A and B) = P(A) P(B)
•
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22
Conditional Probability
• If events are dependent, the occurrence of
one event changes the probability of the
other.
• The notation P(A|B) is read “the
probability of A, given B.”
• P(A, given B) equals the probability that
event A occurs, assuming that B has
already occurred.
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23
For Dependent 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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24
The Multiplication Rules:
• For independent events:
P(A and B) = P(A) P(B)
•
• For dependent 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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25
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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26
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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27
“And” versus “or”
• And means both events occur together.
• Or means that at least one of the events
occur.
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28
For any events A and B,
P(A or B) =
P(A) + P(B) – P(A and B)
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29
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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30
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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31
For any mutually exclusive
events A and B,
P(A or B) = P(A) + P(B)
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32
When rolling an ordinary die:
P(4 or 6) =
1 1 2 1
  
6 6 6 3
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33
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) = ?
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
34
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
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
35
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) = ?
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
36
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) =
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
147
187
37
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) = ?
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
38
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) = 54
116
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39
Counting Techniques
• Tree Diagram
• Multiplication Rule
• Permutations
• Combinations
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40
Tree Diagram
a method of listing outcomes of an
experiment consisting of a series of
activities
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41
Tree diagram for the
experiment of tossing two coins
H
H
start
T
H
T
T
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42
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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43
Multiplication of Choices
If there are
n possible outcomes for event E1
and m possible outcomes for event E2,
then there are
n x m or nm possible outcomes
for the series of events E1 followed by E2.
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44
Area Code Example
Until a few years ago a three-digit area code was
designed as follows.
The first 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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
45
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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46
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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47
Calculating Factorials
5! = 5 • 4 • 3 • 2 • 1 =
3! = 3 • 2 • 1 =
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120
6
48
Definitions
1! = 1
0! = 1
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49
Complete the Factorials:
4! = 24
10! = 3,628,800
6! = 720
15! = 1.3077 x 1012
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50
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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51
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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52
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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53
Permutation Formula
The number of ways to arrange in
order n distinct objects, taking them
r at a time, is:
Pn , r
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n!

n  r  !
54
Another notation for
permutations:
P
n r
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55
Find P7, 3
7!
7! 5040
P7 , 3 
 
 210
(7  3)! 4!
24
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56
Applying the Permutation
Formula
6
P3, 3 = _______
30
P6, 2
= __________
P15, 2
= _______
12
P4, 2 = _______
336
P8, 3 = _______
210
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57
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 =
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8!
3!
= 8• 7 • 6 • 5 • 4 = 6720
58
Combinations
A combination is a grouping
in no particular order
of items.
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59
Combination Formula
The number of combinations of n objects
taken r at a time is:
Cn , r
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n!

(n  r ) ! r !
60
Other notations for
combinations:
C
n
r
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n
or  
r
 
61
Find C9, 3
C9 , 3
9!
9! 362880



 84
3!(9  3)! 3!6! 6(720)
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62
Applying the Combination
Formula
10
C5, 3 = ______
35
C7, 3 = ________
1
C3, 3 = ______
105
C15, 2 = ________
15
C6, 2 = ______
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63
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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64