CHAPTER 5
Probability,Review of Basic Concepts
to accompany
Introduction to Business Statistics
fourth edition,by Ronald M,Weiers
Presentation by Priscilla Chaffe-Stengel
Donald N,Stengel
? 2002 The Wadsworth Group
Chapter 5 - Learning Objectives
? Construct and interpret a contingency table
– Frequencies,relative frequencies & cumulative
relative frequencies
? Determine the probability of an event.
? Construct and interpret a probability tree
with sequential events.
? Use Bayes’ Theorem to revise a probability.
? Determine the number of combinations or
permutations of n objects r at a time.
? 2002 The Wadsworth Group
Chapter 5 - Key Terms
? Experiment
? Sample space
? Event
? Probability
? Odds
? Contingency table
? Venn diagram
? Union of events
? Intersection of events
? Complement
? Mutually exclusive events
? Exhaustive events
? Marginal probability
? Joint probability
? Conditional probability
? Independent events
? Tree diagram
? Counting
? Permutations
? Combinations
? 2002 The Wadsworth Group
Chapter 5 - Key Concepts
? The probability of a single event falls
between 0 and 1.
? The probability of the complement of event
A,written A’,is
P(A’) = 1 – P(A)
? The law of large numbers,Over a large
number of trials,the relative frequency with
which an event occurs will approach the
probability of its occurrence for a single trial.
? 2002 The Wadsworth Group
Chapter 5 - Key Concepts
? Odds vs,probability
If the probability event A occurs is,then
the odds in favor of event A occurring are a to
b – a.
– Example,If the probability it will rain
tomorrow is 20%,then the odds it will rain
are 20 to (100 – 20),or 20 to 80,or 1 to 4.
– Example,If the odds an event will occur
are 3 to 2,the probability it will occur is
ab
3
3 ? 2 ?
3
5,
? 2002 The Wadsworth Group
Chapter 5 - Key Concepts
? Mutually exclusive events
– Events A and B are
mutually exclusive if both cannot occur at
the same time,that is,if their intersection is
empty,In a Venn diagram,mutually
exclusive events are usually shown as
nonintersecting areas,If intersecting areas
are shown,they are empty.
? 2002 The Wadsworth Group
Intersections versus Unions
? Intersections -,Both/And”
– The intersection of A and B and C is also
written,
– All events or characteristics occur
simultaneously for all elements contained
in an intersection.
? Unions -,Either/Or”
– The union of A or B or C is also written
– At least one of a number of possible
events occur at the same time,
A B C
A B C,
? 2002 The Wadsworth Group
Working with Unions and
Intersections
? The general rule of addition:
P(A or B) = P(A) + P(B) – P(A and B)
is always true,When events A and B
are mutually exclusive,the last term in
the rule,P(A and B),will become zero
by definition.
? 2002 The Wadsworth Group
Three Kinds of Probabilities
? Simple or marginal probability
– The probability that a single given event will
occur,The typical expression is P(A).
? Joint or compound probability
– The probability that two or more events occur,
The typical expression is P(A and B).
? Conditional probability
– The probability that an event,A,occurs given that
another event,B,has already happened,The
typical expression is P(A|B).
? 2002 The Wadsworth Group
The Contingency Table:
An Example
? Problem 5.15,The following table
represents gas well completions during
1986 in North and South America.
D D’
Dry Not Dry Totals
N North America 14,131 31,575 45,706
N’ South America 404 2,563 2,967
Totals 14,535 34,138 48,673
? 2002 The Wadsworth Group
Example,Problem 5.15
D D’
Dry Not Dry Totals
N North America 14,131 31,575 45,706
N’ South America 404 2,563 2,967
Totals 14,535 34,138 48,673
? 1,What is P(N)? - 1,Simple probability,45,706/48,673
? 2,What is P(D’ and N)? - 2,Joint probability,
31,575/48,673
? 3,What is P(D’ or N)? - 3,Equivalent solutions:
– 3a,(34,138 + 45,706 – 31,575)/48,673 OR,..
– 3b,(31,575 + 2,563 + 14,131)/48,673 OR,..
– 3c,(34,138 + 14,131)/48,673 OR,..
– 3d,(48,673 – 2,563)/48,673 ? 2002 The Wadsworth Group
Simple and Joint Probabilities
Share a Denominator
Note that,when probabilities are
calculated from empirical data,both
simple and joint probabilities use the
entire sample as a denominator.
Watch what happens with conditional
probabilities.
? 2002 The Wadsworth Group
Problem 5.15,continued
D D’
Dry Not Dry Totals
N North America 14,131 31,575 45,706
N’ South America 404 2,563 2,967
Totals 14,535 34,138 48,673
? What is P(N|D)? - Conditional probability,
14,131/14,535
? What is P(D|N)? - Conditional probability,
14,131/45,706
? What is P(D’|N)? - Conditional probability,
31,575/45,706
? What is P(N|D’)? - Conditional probability,
31,575/34,138
Note that conditional probabilities are the ONLY
ones whose denominators are NOT the total sample.
? 2002 The Wadsworth Group
Conditional Probability -
A Definition
? Conditional probability of event A,
given that event B has occurred,
where P(B) > 0
? So,from our prior example,
P ( A | B ) ? P ( A an d B )P ( B )
P ( N | D ) ? P ( N an d D )
P ( D )
?
14,131
48,673
14,535
48,673
? 14,131
14,535
? 2002 The Wadsworth Group
Independent Events
? Events are independent when the
occurrence of one event does not change
the probability that another event will
occur.
– If A and B are independent,P(A|B) = P(A)
because the occurrence of event B does not
change the probability that A will occur.
– If A and B are independent,then
P(A and B) = P(A) ? P(B)
? 2002 The Wadsworth Group
When Events Are Dependent
? Events are dependent when the occurrence
of one event does change the probability
that another event will occur.
– If A and B are dependent,P(A|B) ?? P(A)
because the occurrence of event B does change
the probability that A will occur.
– If A and B are dependent,then
P(A and B) = P(A) ? P(B|A)
? 2002 The Wadsworth Group
The Probability Tree:
Problem 5.15
? Location first
N
N’
45,706/48,673
2,967/48,673
D 14,131/45,706
D’ 31,575/45,706
D 404/2,967
D’ 2,563/2,967
14,131/48,673
31,575/48,673
404/48,673
2,563/48,673
? 2002 The Wadsworth Group
The Probability Tree:
Problem 5.15
? Well condition first
D
D’
14,535/48,673
34,138/48,673
N 14,131/14,535
N’ 404/14,535
N 31,575/ 34,138
N’ 2,563/ 34,138
14,131/48,673
404/48,673
31,575 /48,673
2,563/48,673
? 2002 The Wadsworth Group
What’s the Probability of a Dry
Well? It Depends....
? Does knowing where the well was
drilled change your estimate of the
chances it was dry?
P(D) = 14,535/48,673 = 0.2986
P(D|N’) = 404/2,967 = 0.1362
P(D|N) = 14,131/45,706 = 0.3092
Yes,So the probability the well is dry is
dependent upon its location.
? 2002 The Wadsworth Group
Bayes’ Theorem for the
Revision of Probability
? In the 1700s,Thomas Bayes developed a
way to revise the probability that a first
event occurred from information obtained
from a second event.
? Bayes’ Theorem,For two events A and B
P ( A | B ) ? P ( A a n d B )P ( B ) ? P ( A ) ? P ( B | A )[ P ( A ) ? P ( B | A )] ? [ P ( A ' ) ? P ( B | A ' )]
? 2002 The Wadsworth Group
Revising Probability -
Problem 5.15
Can we compute P(N’|D) from P(D|N’)?
? Using Bayes’ Theorem:
P ( N ' | D ) ?
P ( N ' a n d D )
P ( D )
?
P ( N ' ) ?P ( D | N ' )
[ P ( N ' ) ? P ( D | N ' )] ? [ P ( N ) ? P ( D | N )]
?
( 2,9 6 7 / 48,6 7 3 ) ? ( 4 0 4 / 2,9 6 7 )
[( 2,9 6 7 / 48,6 7 3 ) ? ( 4 0 4 / 2,9 6 7 )] ? [ ( 45,7 0 6 / 48,6 7 3 ) ? ( 14,1 3 1 / 45,7 0 6 )]
?
4 0 4 / 48,6 7 3
( 4 0 4 / 48,6 7 3 ) ? ( 14,1 3 1 / 48,6 7 3 )
?
4 0 4
14,5 3 5
? 2002 The Wadsworth Group
Counting
? Multiplication rule of counting,If there are m
ways a first event can occur and n ways a second
event can occur,the total number of ways the two
events can occur is given by m x n.
? Factorial rule of counting,The number of
ways n objects can be arranged in order.
n! = n x (n – 1) x (n – 2) x,.,x 1
Note that 1! = 0! = 1 by definition.
? 2002 The Wadsworth Group
More Counting
? Permutations,The number of different ways
n objects can be arranged taken r at a time,
Order is important.
? Combinations,The number of ways n objects
can be arranged taken r at a time,Order is not
important.
P(n,r) ? n!(n–r)!
C ( n,r ) ? nr??
??
??
??
????
??
??
??
??
????
? n !r !?( n – r )!
? 2002 The Wadsworth Group
Probability,Review of Basic Concepts
to accompany
Introduction to Business Statistics
fourth edition,by Ronald M,Weiers
Presentation by Priscilla Chaffe-Stengel
Donald N,Stengel
? 2002 The Wadsworth Group
Chapter 5 - Learning Objectives
? Construct and interpret a contingency table
– Frequencies,relative frequencies & cumulative
relative frequencies
? Determine the probability of an event.
? Construct and interpret a probability tree
with sequential events.
? Use Bayes’ Theorem to revise a probability.
? Determine the number of combinations or
permutations of n objects r at a time.
? 2002 The Wadsworth Group
Chapter 5 - Key Terms
? Experiment
? Sample space
? Event
? Probability
? Odds
? Contingency table
? Venn diagram
? Union of events
? Intersection of events
? Complement
? Mutually exclusive events
? Exhaustive events
? Marginal probability
? Joint probability
? Conditional probability
? Independent events
? Tree diagram
? Counting
? Permutations
? Combinations
? 2002 The Wadsworth Group
Chapter 5 - Key Concepts
? The probability of a single event falls
between 0 and 1.
? The probability of the complement of event
A,written A’,is
P(A’) = 1 – P(A)
? The law of large numbers,Over a large
number of trials,the relative frequency with
which an event occurs will approach the
probability of its occurrence for a single trial.
? 2002 The Wadsworth Group
Chapter 5 - Key Concepts
? Odds vs,probability
If the probability event A occurs is,then
the odds in favor of event A occurring are a to
b – a.
– Example,If the probability it will rain
tomorrow is 20%,then the odds it will rain
are 20 to (100 – 20),or 20 to 80,or 1 to 4.
– Example,If the odds an event will occur
are 3 to 2,the probability it will occur is
ab
3
3 ? 2 ?
3
5,
? 2002 The Wadsworth Group
Chapter 5 - Key Concepts
? Mutually exclusive events
– Events A and B are
mutually exclusive if both cannot occur at
the same time,that is,if their intersection is
empty,In a Venn diagram,mutually
exclusive events are usually shown as
nonintersecting areas,If intersecting areas
are shown,they are empty.
? 2002 The Wadsworth Group
Intersections versus Unions
? Intersections -,Both/And”
– The intersection of A and B and C is also
written,
– All events or characteristics occur
simultaneously for all elements contained
in an intersection.
? Unions -,Either/Or”
– The union of A or B or C is also written
– At least one of a number of possible
events occur at the same time,
A B C
A B C,
? 2002 The Wadsworth Group
Working with Unions and
Intersections
? The general rule of addition:
P(A or B) = P(A) + P(B) – P(A and B)
is always true,When events A and B
are mutually exclusive,the last term in
the rule,P(A and B),will become zero
by definition.
? 2002 The Wadsworth Group
Three Kinds of Probabilities
? Simple or marginal probability
– The probability that a single given event will
occur,The typical expression is P(A).
? Joint or compound probability
– The probability that two or more events occur,
The typical expression is P(A and B).
? Conditional probability
– The probability that an event,A,occurs given that
another event,B,has already happened,The
typical expression is P(A|B).
? 2002 The Wadsworth Group
The Contingency Table:
An Example
? Problem 5.15,The following table
represents gas well completions during
1986 in North and South America.
D D’
Dry Not Dry Totals
N North America 14,131 31,575 45,706
N’ South America 404 2,563 2,967
Totals 14,535 34,138 48,673
? 2002 The Wadsworth Group
Example,Problem 5.15
D D’
Dry Not Dry Totals
N North America 14,131 31,575 45,706
N’ South America 404 2,563 2,967
Totals 14,535 34,138 48,673
? 1,What is P(N)? - 1,Simple probability,45,706/48,673
? 2,What is P(D’ and N)? - 2,Joint probability,
31,575/48,673
? 3,What is P(D’ or N)? - 3,Equivalent solutions:
– 3a,(34,138 + 45,706 – 31,575)/48,673 OR,..
– 3b,(31,575 + 2,563 + 14,131)/48,673 OR,..
– 3c,(34,138 + 14,131)/48,673 OR,..
– 3d,(48,673 – 2,563)/48,673 ? 2002 The Wadsworth Group
Simple and Joint Probabilities
Share a Denominator
Note that,when probabilities are
calculated from empirical data,both
simple and joint probabilities use the
entire sample as a denominator.
Watch what happens with conditional
probabilities.
? 2002 The Wadsworth Group
Problem 5.15,continued
D D’
Dry Not Dry Totals
N North America 14,131 31,575 45,706
N’ South America 404 2,563 2,967
Totals 14,535 34,138 48,673
? What is P(N|D)? - Conditional probability,
14,131/14,535
? What is P(D|N)? - Conditional probability,
14,131/45,706
? What is P(D’|N)? - Conditional probability,
31,575/45,706
? What is P(N|D’)? - Conditional probability,
31,575/34,138
Note that conditional probabilities are the ONLY
ones whose denominators are NOT the total sample.
? 2002 The Wadsworth Group
Conditional Probability -
A Definition
? Conditional probability of event A,
given that event B has occurred,
where P(B) > 0
? So,from our prior example,
P ( A | B ) ? P ( A an d B )P ( B )
P ( N | D ) ? P ( N an d D )
P ( D )
?
14,131
48,673
14,535
48,673
? 14,131
14,535
? 2002 The Wadsworth Group
Independent Events
? Events are independent when the
occurrence of one event does not change
the probability that another event will
occur.
– If A and B are independent,P(A|B) = P(A)
because the occurrence of event B does not
change the probability that A will occur.
– If A and B are independent,then
P(A and B) = P(A) ? P(B)
? 2002 The Wadsworth Group
When Events Are Dependent
? Events are dependent when the occurrence
of one event does change the probability
that another event will occur.
– If A and B are dependent,P(A|B) ?? P(A)
because the occurrence of event B does change
the probability that A will occur.
– If A and B are dependent,then
P(A and B) = P(A) ? P(B|A)
? 2002 The Wadsworth Group
The Probability Tree:
Problem 5.15
? Location first
N
N’
45,706/48,673
2,967/48,673
D 14,131/45,706
D’ 31,575/45,706
D 404/2,967
D’ 2,563/2,967
14,131/48,673
31,575/48,673
404/48,673
2,563/48,673
? 2002 The Wadsworth Group
The Probability Tree:
Problem 5.15
? Well condition first
D
D’
14,535/48,673
34,138/48,673
N 14,131/14,535
N’ 404/14,535
N 31,575/ 34,138
N’ 2,563/ 34,138
14,131/48,673
404/48,673
31,575 /48,673
2,563/48,673
? 2002 The Wadsworth Group
What’s the Probability of a Dry
Well? It Depends....
? Does knowing where the well was
drilled change your estimate of the
chances it was dry?
P(D) = 14,535/48,673 = 0.2986
P(D|N’) = 404/2,967 = 0.1362
P(D|N) = 14,131/45,706 = 0.3092
Yes,So the probability the well is dry is
dependent upon its location.
? 2002 The Wadsworth Group
Bayes’ Theorem for the
Revision of Probability
? In the 1700s,Thomas Bayes developed a
way to revise the probability that a first
event occurred from information obtained
from a second event.
? Bayes’ Theorem,For two events A and B
P ( A | B ) ? P ( A a n d B )P ( B ) ? P ( A ) ? P ( B | A )[ P ( A ) ? P ( B | A )] ? [ P ( A ' ) ? P ( B | A ' )]
? 2002 The Wadsworth Group
Revising Probability -
Problem 5.15
Can we compute P(N’|D) from P(D|N’)?
? Using Bayes’ Theorem:
P ( N ' | D ) ?
P ( N ' a n d D )
P ( D )
?
P ( N ' ) ?P ( D | N ' )
[ P ( N ' ) ? P ( D | N ' )] ? [ P ( N ) ? P ( D | N )]
?
( 2,9 6 7 / 48,6 7 3 ) ? ( 4 0 4 / 2,9 6 7 )
[( 2,9 6 7 / 48,6 7 3 ) ? ( 4 0 4 / 2,9 6 7 )] ? [ ( 45,7 0 6 / 48,6 7 3 ) ? ( 14,1 3 1 / 45,7 0 6 )]
?
4 0 4 / 48,6 7 3
( 4 0 4 / 48,6 7 3 ) ? ( 14,1 3 1 / 48,6 7 3 )
?
4 0 4
14,5 3 5
? 2002 The Wadsworth Group
Counting
? Multiplication rule of counting,If there are m
ways a first event can occur and n ways a second
event can occur,the total number of ways the two
events can occur is given by m x n.
? Factorial rule of counting,The number of
ways n objects can be arranged in order.
n! = n x (n – 1) x (n – 2) x,.,x 1
Note that 1! = 0! = 1 by definition.
? 2002 The Wadsworth Group
More Counting
? Permutations,The number of different ways
n objects can be arranged taken r at a time,
Order is important.
? Combinations,The number of ways n objects
can be arranged taken r at a time,Order is not
important.
P(n,r) ? n!(n–r)!
C ( n,r ) ? nr??
??
??
??
????
??
??
??
??
????
? n !r !?( n – r )!
? 2002 The Wadsworth Group
