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Statistics for
Business and Economics
8th Global Edition
Chapter 3
Elements of Chance: Probability Methods
Copyright © 2013 Pearson Education
Ch. 3-1
Before Learning
It is important to understand that
 The world in which your future occurs is not
deterministic (Ex. Your wife is easily ruffled)
 If you construct and use probability models, you
will have a greater chance of success (Bennett’s
myth with 2 coins)
 There are future events where a probability
model cannot be developed—“Black Swans”
 Ex. 2001 911;
2008 financial crisis;
2010 oil-drilling rig explode in Gulf Coast
Copyright © 2013 Pearson Education
Ch. 3-2
3.1
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How to Understand Probability?
From Gerlamo Cardano in 16th century, we
should count the number of equally possible
outcomes, the proportion relating to an event.
The spirit of a priori probability:
1. Count the number of all possible outcomes
2. Attach equally likely probability to each
3. Count the number of outcomes for an event
We need to image a Random Experiment which
generates equally likely outcomes
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Ch. 3-3
3.1
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How about Asymmetric Dice?
Asymmetric cases with non-equally likely
outcomes are decomposed into more
elementary equally likely outcomes
Ex. A die with two “1”s
We need a random experiment for two reasons:
1. To give an idea as to the nature of the
stochastic phenomena we have in mind
2. To bring out the essential features of the
phenomena and formalize them in precise
mathematical forms
Copyright © 2013 Pearson Education
Ch. 3-4
3.1

Random Experiment
A Random Experiment is defined as a chance
process with the following conditions:
1. All possible distinct outcomes are known a
priori
2. In any particular trial the outcome is not
known a priori but there exists a discernible
regularity of occurrence associated with these
outcomes
3. It can be repeated under identical conditions
Copyright © 2013 Pearson Education
Ch. 3-5
3.1
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Important Terms
Basic Outcome – a possible outcome of a
random experiment
Sample Space (S) – the collection of all
possible outcomes of a random experiment
Ex. Safe hit, hit by pitcher, strikeout, groundball,
fly ball, reach base on error
Event (E) – any subset of basic outcomes from
the sample space
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Ch. 3-6
3.1
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How to Represent the Events
We care about events
Event are formed by combining basic outcomes
An event has occurred when any one of its
basic outcomes occurs
Ex. Throw two coins
A-at least one H: A={HH, HT, TH}
B-two of the same: B={HH, TT}
C-at least one T:
C={HT, TH, TT}
Set Theoretic Operations
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Ch. 3-7
Important Terms

Intersection of Events – If A and B are two
events in a sample space S, then the
intersection, A ∩ B, is the set of all outcomes in
S that belong to both A and B (Venn diagram)
S
A
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AB
B
Ch. 3-8
Important Terms


Joint probability of A and B is the probability of
the intersection of A and B
Ex. Both hitting and reaching the base
S
A
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AB
B
Ch. 3-9
Important Terms
(continued)

A and B are Mutually Exclusive Events if they
have no basic outcomes in common

i.e., the set A ∩ B is empty
S
A
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B
Ch. 3-10
Important Terms
(continued)

Union of Events – If A and B are two events in a
sample space S, then the union, A U B, is the
set of all outcomes in S that belong to either
A or B
S
A
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B
The entire shaded
area represents
AUB
Ch. 3-11
Important Terms
(continued)

Events E1, E2, …,Ek are Collectively Exhaustive
events if E1 U E2 U . . . U Ek = S
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i.e., the events completely cover the sample space
The Complement of an event A is the set of all
basic outcomes in the sample space that do not
belong to A. The complement is denoted A
S
A
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A
Ch. 3-12
Important Terms
(continued)

AB
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Ch. 3-13
Examples
Let the Sample Space be the collection of
all possible outcomes of rolling one die:
S = [1, 2, 3, 4, 5, 6]
Let A be the event “Number rolled is even”
Let B be the event “Number rolled is at least 4”
Then
A = [2, 4, 6]
Copyright © 2013 Pearson Education
and
B = [4, 5, 6]
Ch. 3-14
Examples
(continued)
S = [1, 2, 3, 4, 5, 6]
A = [2, 4, 6]
B = [4, 5, 6]
Complements:
A  [1, 3, 5]
B  [1, 2, 3]
Intersections:
A  B  [4, 6]
Unions:
A  B  [5]
A  B  [2, 4, 5, 6]
A  A  [1, 2, 3, 4, 5, 6]  S
Copyright © 2013 Pearson Education
Ch. 3-15
Examples
(continued)
S = [1, 2, 3, 4, 5, 6]

B = [4, 5, 6]
Mutually exclusive:
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A and B are not mutually exclusive
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A = [2, 4, 6]
The outcomes 4 and 6 are common to both
Collectively exhaustive:
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A and B are not collectively exhaustive
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A U B does not contain 1 or 3
Copyright © 2013 Pearson Education
Ch. 3-16
3.2
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Probability and Its Postulates
Probability – the chance that
an uncertain event will occur
(always between 0 and 1)
0 ≤ P(A) ≤ 1 For any event A
1
.5
0
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Certain
Impossible
Ch. 3-17
Assessing Probability

There are three approaches to assessing the
probability of an uncertain event:
1. classical probability
2. relative frequency probability
3. subjective probability
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Ch. 3-18
Classical Probability

Assumes all outcomes in the sample space are
equally likely to occur
Classical probability of event A:
P(A) 

NA
number of outcomes that satisfy the event A

N
total number of outcomes in the sample space
Requires a count of the outcomes in the sample space
Copyright © 2013 Pearson Education
Ch. 3-19
Comprehend All Outcomes

It is easier to comprehend all possible outcomes
(sample space) by virtually sequential tosses
Ch. 3-20
Comprehend All Outcomes
Ch. 3-21
Permutation vs Combination
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Permutation: The order matters
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Combination: The order does not matter
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With repetition: The thing is return back after being
drawn
Without repetition: The thing can be drawn for at
most one time
Copyright © 2013 Pearson Education
Ch. 3-22
Counting the Possible Outcomes

Use the Permutations with repetition to determine
the number of combinations of n items taken k at a
time
n

k
Ex. The number of all possible outcomes
(sample space) in de Mere’s problem are
and
24
4
6
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36
Ch. 3-23
Permutations and Combinations
The number of possible orderings

The total number of possible ways of arranging x
objects in order is
x!  x(x - 1)(x - 2) ...(2)(1)

x! is read as “x factorial”
Copyright © 2013 Pearson Education
Ch. 3-24
Permutations and Combinations
Permutations without repetition: the number
of possible arrangements when x objects are
to be selected from a total of n objects and
arranged in order [with (n – x) objects left over]
P  n(n  1)(n  2) ...(n  x  1)
n
x
n!

(n  x)!
Copyright © 2013 Pearson Education
Ch. 3-25
Counting the Possible Outcomes

Use the Combinations without repetition to
determine the number of combinations of n items
taken k at a time
n!
C 
k! (n  k)!
n
k
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where
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n! = n(n-1)(n-2)…(1)
0! = 1 by definition
Copyright © 2013 Pearson Education
Ch. 3-26
Permutations and Combinations
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Permutation is an ordered Combination
n
x
P
C 
x!
n
k
n!

x! (n  x)!
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Ch. 3-27
Permutations and Combinations
Example
Suppose that two letters are to be selected
from A, B, C, D and arranged in order. How
many permutations are possible?
 Solution The number of permutations, with
n = 4 and x = 2 , is

4!
P 
 12
(4  2)!
4
2
The permutations are
AB AC AD BA BC BD
CA CB CD DA DB DC
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Ch. 3-28
Permutations and Combinations
Example
(continued)
Suppose that two letters are to be selected
from A, B, C, D. How many combinations are
possible (i.e., order is not important)?
 Solution The number of combinations is
4!
C 
6
2! (4  2)!
4
2
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The combinations are
AB (same as BA)
AC (same as CA)
AD (same as DA)
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BC (same as CB)
BD (same as DB)
CD (same as DC)
Ch. 3-29
3.1
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Questions
Combination with Repetition
Transylvania lottery
Birthday problem (weekly version)
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Ch. 3-30
The Defects of Statistics

From the viewpoint of a priori probability, you
should not surprised about something out-of-mind
happens, should you?
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Ch. 3-31
3.1
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Homework
Homework: Exercises in Sections 3.1 and 3.2
Problems 1~34 in Ross for 5 points
Reminder: The first midterm begins from Oct. 3
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Ch. 3-32