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Copyright 2011 John Wiley & Sons, Inc. 1
Copyright 2011 John Wiley & Sons, Inc.
Applied Business Statistics, 7Applied Business Statistics, 7thth
ed.ed.
by Ken Blackby Ken Black
Chapter 4Chapter 4
ProbabilityProbability
Copyright 2011 John Wiley & Sons, Inc. 2
Learning Objectives
Comprehend the different ways of assigning
probability.
Understand and apply marginal, union, joint, and
conditional probabilities.
Select the appropriate law of probability to use in
solving problems.
Solve problems using the laws of probability
including the laws of addition, multiplication and
conditional probability
Revise probabilities using Bayes’ rule.
Copyright 2011 John Wiley & Sons, Inc. 3
Probability
Probability – probability of occurrences are
assigned to the inferential process under
conditions of uncertainty
Copyright 2011 John Wiley & Sons, Inc. 4
Methods of Assigning Probabilities
Classical method of assigning probability
(rules and laws)
Relative frequency of occurrence
(cumulated historical data)
Subjective Probability (personal intuition
or reasoning)
Copyright 2011 John Wiley & Sons, Inc. 5
Number of outcomes leading to the event divided by
the total number of outcomes possible
P(E) = ne/N
where N = total number of outcomes, and
ne = number of outcomes in event E
Each outcome is equally likely
Applicable to games of chance
Objective -- everyone correctly using the method
assigns an identical probability
Classical Probability
Copyright 2011 John Wiley & Sons, Inc. 6
Subjective Probability - Comes from a person’s
intuition or reasoning
Subjective -- different individuals may (correctly or
incorrectly) assign different numeric probabilities to
the same event
Degree of belief in the results of the event
Useful for unique (single-trial) experiments
New product introduction
Site selection decisions
Sporting events
Subjective Probability
Copyright 2011 John Wiley & Sons, Inc. 7
Experiment – is a process that produces an outcome
Ex: Rolling two six-sided dice and calculating their sum
Event – an outcome of an experiment
Ex: The sum is at least 10
Elementary event – events that cannot be
decomposed or broken down into other events
Ex: The first die is a six
Sample Space – a complete roster/listing of all
elementary events for an experiment
Ex: {(1,1), (1,2), (1,3), …, (2,1), (2,2), …., (6,5), (6,6)}
Trial: one repetition of the process
Ex: Roll the dice…
Structure of Probability
Copyright 2011 John Wiley & Sons, Inc. 8
Mutually Exclusive Events – events such that the
occurrence of one precludes the occurrence of the
other
These events have no intersection
Independent Events – the occurrence or
nonoccurrence of one has no affect on the occurrence
of the others
Collectively Exhaustive Events – listing of all possible
elementary events for an experiment
Complementary Events – two events, one of which
comprises all the elementary events of an experiment
that are not in the other event
Structure of Probability
Copyright 2011 John Wiley & Sons, Inc. 9
The set of all elementary events for an experiment
Methods for describing a sample space
roster or listing
tree diagram
set builder notation
Venn diagram
Sample Space
Copyright 2011 John Wiley & Sons, Inc. 10
Experiment: randomly select, without replacement,
two families from the residents of Tiny Town
Each ordered pair in the sample space is an
elementary event, for example -- (D,C)
Sample Space: Roster Example
Copyright 2011 John Wiley & Sons, Inc. 11
S = {(x,y) | x is the family selected on the first draw,
and y is the family selected on the second draw}
Concise description of large sample spaces
Sample Space: Set Notation for
Random Sample of Two Families
Copyright 2011 John Wiley & Sons, Inc. 12
{ }
{ }
{ }9,7,6,5,4,3,2,1
6,5,4,3,2
9,7,4,1
=∪
=
=
YX
Y
X
{ }
{ }
{ }
C IBM DEC Apple
F Apple Grape Lime
C F IBM DEC Apple Grape Lime
=
=
∪ =
, ,
, ,
, , , ,
YX
The union of two sets contains an instance of each
element of the two sets.
Union of Sets
Copyright 2011 John Wiley & Sons, Inc. 13
{ }
{ }
{ }
X
Y
X Y
=
=
∩ =
1 4 7 9
2 3 4 5 6
4
, , ,
, , , ,
YX
{ }
{ }
{ }
C IBM DEC Apple
F Apple Grape Lime
C F Apple
=
=
∩ =
, ,
, ,
X ∩ Y
The intersection of two sets contains only those
element common to the two sets.
Intersection of Sets
Copyright 2011 John Wiley & Sons, Inc. 14
{ }
{ }
{ }
X
Y
X Y
=
=
∩ =
1 7 9
2 3 4 5 6
, ,
, , , ,
{ }
{ }
{ }=∩
=
=
FC
LimeGrapeF
AppleDECIBMC
,
,,
YX
0)( =∩ YXP
Events with no common outcomes
Occurrence of one event precludes the occurrence
of the other event
Mutually Exclusive Events
Copyright 2011 John Wiley & Sons, Inc. 15
Independent Events
P X Y P X and P Y X P Y( | ) ( ) ( | ) ( )= =
Occurrence of one event does not affect the
occurrence or nonoccurrence of the other event
The conditional probability of X given Y is equal to
the marginal probability of X.
The conditional probability of Y given X is equal to
the marginal probability of Y.
Copyright 2011 John Wiley & Sons, Inc. 16
Collectively Exhaustive Events
E1 E2 E3
Sample Space with three
collectively exhaustive events
Contains all elementary events for an experiment
Copyright 2011 John Wiley & Sons, Inc. 17
Complementary Events
Sample
Space
A
P Sample Space( ) = 1
P A P A( ) ( )′ = −1
′A
All elementary events not in the event ‘A’ are in its
complementary event.
Copyright 2011 John Wiley & Sons, Inc. 18
Counting the Possibilities
mn Rule
Sampling from a Population with Replacement
Combinations: Sampling from a Population without
Replacement
Copyright 2011 John Wiley & Sons, Inc. 19
mn Rule
If an operation can be done m ways and a second
operation can be done n ways, then there are mn
ways for the two operations to occur in order.
A cafeteria offers 5 salads, 4 meats, 8 vegetables, 3
breads, 4 desserts, and 3 drinks. A meal is two
servings of vegetables, which may be identical.
How many meals are available?
5 * 4 * 8 * 3 * 4 * 3 = 5760
Copyright 2011 John Wiley & Sons, Inc. 20
Combinations: Sampling from a
Population without Replacement
This counting method uses combinations
Selecting n items from a population of N
without replacement
Copyright 2011 John Wiley & Sons, Inc. 21
Combinations
Combinations – sampling “n” items from a
population size N without replacement provides the
formula shown below
A tray contains 1,000 individual tax returns. If 3
returns are randomly selected without replacement
from the tray, how many possible samples are there?
0166,167,00
)!31000(!3
!1000
)!(!
!
=
−
=
−
=





=
nNn
N
n
N
Crn
Copyright 2011 John Wiley & Sons, Inc. 22
Combinations: Sampling from a
Population without Replacement
For example, suppose a small law firm has 16
employees and three are to be selected randomly to
represent the company at the annual meeting of the
American Bar Association.
How many different combinations of lawyers could
be sent to the meeting?
Answer: NCn= 16C3= 16!/(3!×13!) = 560.
Copyright 2011 John Wiley & Sons, Inc. 23
Four Types of Probability
Marginal
The probability
of X occurring
Union
The probability
of X or Y
occurring
Joint
The probability
of X and Y
occurring
Conditional
The probability
of X occurring
given that Y
has occurred
YX YX
Y
X
)(XP P X Y( )∪ P X Y( )∩ P X Y( | )
Copyright 2011 John Wiley & Sons, Inc. 24
General Law of Addition
P X Y P X P Y P X Y( ) ( ) ( ) ( )∪ = + − ∩
YX
Copyright 2011 John Wiley & Sons, Inc. 25
General Law of Addition -- Example
)()()()( SNPSPNPSNP ∩−+=∪
81.0
56.67.70.)(
56.)(
67.)(
70.)(
=
−+=∪
=∩
=
=
SNP
SNP
SP
NPSN
.56
.67.70
Copyright 2011 John Wiley & Sons, Inc. 26
81.
56.67.70.
)()()()(
=
−+=
∩−+=∪ SNPSPNPSNP
.11 .19 .30
.56 .14 .70
.67 .33 1.00
Increase
Storage Space
Yes No Total
Yes
No
Total
Noise
Reduction
Office Design Problem Probability Matrix
Copyright 2011 John Wiley & Sons, Inc. 27
If a worker is randomly selected from the company
described in Demonstration Problem 4.1 (below),
what is the probability that the worker is either
technical or clerical? What is the probability that the
worker is either a professional or a clerical?
Demonstration Problem 4.3
Type of Gender
Position Male Female Total
Managerial 8 3 11
Professional 31 13 44
Technical 52 17 69
Clerical 9 22 31
Total 100 55 155
Copyright 2011 John Wiley & Sons, Inc. 28
Examine the raw value matrix of the company’s human
resources data shown in Demonstration Problem 4.1. In
many raw value and probability matrices like this one, the
rows are non-overlapping or mutually exclusive, as are
the columns. In this matrix, a worker can be classified as
being in only one type of position and as either male or
female but not both. Thus, the categories of type of
position are mutually exclusive, as are the categories of
sex, and the special law of addition can be applied to the
human resource data to determine the union
probabilities.
Demonstration Problem 4.3
Copyright 2011 John Wiley & Sons, Inc. 29
Demonstration Problem 4.3
Let T denote technical, C denote clerical, and P denote
professional. The probability that a worker is either
technical or clerical is
P(T U C) = P (T) + P (C) = 69/155 + 31/155 = 100/155 = .645
The probability that a worker is either professional or
clerical is
P (P U C) = P (P) + P (C) = 44/155 + 31/155 = 75/155 = .484
Copyright 2011 John Wiley & Sons, Inc. 30
P T C P T P C( ) ( ) ( )
.
∪ = +
= +
=
69
155
31
155
645
Demonstration Problem 4.3
Type of Gender
Position Male Female Total
Managerial 8 3 11
Professional 31 13 44
Technical 52 17 69
Clerical 9 22 31
Total 100 55 155
Copyright 2011 John Wiley & Sons, Inc. 31
Type of Gender
Position Male Female Total
Managerial 8 3 11
Professional 31 13 44
Technical 52 17 69
Clerical 9 22 31
Total 100 55 155
P P C P P P C( ) ( ) ( )
.
∪ = +
= +
=
44
155
31
155
484
Demonstration Problem 4.3
Copyright 2011 John Wiley & Sons, Inc. 32
)|()()|()()( YXPYPXYPXPYXP ⋅=⋅=∩
P M
P S M
P M S P M P S M
( ) .
( | ) .
( ) ( ) ( | )
( . )( . ) .
= =
=
∩ = ⋅
= =
80
140
0 5714
0 20
0 5714 0 20 0 1143
Law of Multiplication
Demonstration Problem 4.5
Copyright 2011 John Wiley & Sons, Inc. 33
Law of Multiplication
The intersection of two events is called the joint
probability
General law of multiplication is used to find the joint
probability
General law of multiplication gives the probability
that both events x and y will occur at the same time
P(x|y) is a conditional probability that can be stated
as the probability of x given y
Copyright 2011 John Wiley & Sons, Inc. 34
Law of Multiplication
If a probability matrix is constructed for a problem,
the easiest way to solve for the joint probability is to
find the appropriate cell in the matrix and select the
answer
Copyright 2011 John Wiley & Sons, Inc. 35
Total
.7857
Yes No
.4571 .3286
.1143 .1000 .2143
.5714 .4286 1.00
Married
Yes
No
Total
Supervisor
Probability Matrix
of Employees
20.0)|(
5714.0
140
80
)(
2143.0
140
30
)(
=
==
==
MSP
MP
SP
P M S P M P S M( ) ( ) ( | )
( . )( . ) .
∩ = ⋅
= =0 5714 0 20 0 1143
Law of Multiplication
Demonstration Problem 4.5
Copyright 2011 John Wiley & Sons, Inc. 36
Law of Conditional Probability
If X and Y are two events, the conditional probability
of X occurring given that Y is known or has occurred
is expressed as P(X|Y)
The conditional probability of X given Y is the joint
probability of X and Y divided by the marginal
probability of Y.
)(
)()|(
)(
)(
)|(
YP
XPXYP
YP
YXP
YXP
⋅
=
∩
=
Copyright 2011 John Wiley & Sons, Inc. 37
Law of Conditional Probability - Example
70% of respondents believe noise reduction would
improve productivity.
56% of respondents believed both noise reduction
and increased storage space would improve
productivity
A worker is selected randomly and asked about
changes in the office design
What is the probability that a randomly selected
person believes storage space would improve
productivity given that the person believes noise
reduction improves productivity?
Copyright 2011 John Wiley & Sons, Inc. 38
P N
P N S
P S N
P N S
P N
( ) .
( ) .
( | )
( )
( )
.
.
.
=
∩ =
=
∩
=
=
70
56
56
70
80
NS
.56
.70
Law of Conditional Probability
Copyright 2011 John Wiley & Sons, Inc. 39
Independent Events
Recall: Two events are independent when the
occurrence of one does not affect the probability of
occurrence of the other one
When X and Y are independent, the conditional
probability is equal to the marginal probability
Copyright 2011 John Wiley & Sons, Inc. 40
Geographic Location
Northeast
D
Southeast
E
Midwest
F
West
G
Finance A .12 .05 .04 .07 .28
Manufacturing B .15 .03 .11 .06 .35
Communications C .14 .09 .06 .08 .37
.41 .17 .21 .21 1.00
P A G
P A G
P G
P A
P A G P A
( | )
( )
( )
.
.
. ( ) .
( | ) . ( ) .
=
∩
= = =
= ≠ =
0 07
021
033 028
0 33 028
Independent Events
Demonstration Problem 4.10
Copyright 2011 John Wiley & Sons, Inc. 41
Revision of Probabilities: Bayes’ Rule
)()|()()|()()|(
)()|(
)|(
2211 nn
ii
i
XPXYPXPXYPXPXYP
XPXYP
YXP
⋅⋅⋅++
=
Allows for reversing the order of conditioning (i.e.
P(A|B) can be calculated if you know P(B|A) & P(A))
An extension to the conditional law of probabilities
Copyright 2011 John Wiley & Sons, Inc. 42
Bayes’ Rule
Note, the numerator of Bayes’ Rule and the law of
conditional probability are the same
The denominator is a collective exhaustive listing of
mutually exclusive outcomes of Y
The denominator is a weighted average of the conditional
probabilities with the weights being the prior probabilities
of the corresponding event
Copyright 2011 John Wiley & Sons, Inc. 43
Bayes’ Rule: Ribbon Problem
447.0
)35.0)(12.0()65.0)(08.0(
)35.0)(12.0(
)()|()()|(
)()|(
)|(
553.0
)35.0)(12.0()65.0)(08.0(
)65.0)(08.0(
)()|()()|(
)()|(
)|(
12.0)|(
08.0)|(
35.0)(
65.0)(
=
+
=
⋅+⋅
⋅
=
=
+
=
⋅+⋅
⋅
=
=
=
=
=
SJPSJdPAPAdP
SJPSJdP
dSJP
SJPSJdPAPAdP
APAdP
dAP
SJdP
AdP
SJP
AP
Copyright 2011 John Wiley & Sons, Inc. 44
Bayes’ Rule: Ribbon Problem
Alternative Approach using Tree Diagram
Alamo (A)
0.65
South
Jersey (SJ)
0.35
Defective (d)
0.08
Defective (d)
0.12
Acceptable
0.92
Acceptable
0.88
0.052
0.042
+ 0.094=P(d)
447.0
094.0
042.0
)|( ==dSJP

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Probability Laws and Concepts for Business Statistics

  • 1. Copyright 2011 John Wiley & Sons, Inc. 1 Copyright 2011 John Wiley & Sons, Inc. Applied Business Statistics, 7Applied Business Statistics, 7thth ed.ed. by Ken Blackby Ken Black Chapter 4Chapter 4 ProbabilityProbability
  • 2. Copyright 2011 John Wiley & Sons, Inc. 2 Learning Objectives Comprehend the different ways of assigning probability. Understand and apply marginal, union, joint, and conditional probabilities. Select the appropriate law of probability to use in solving problems. Solve problems using the laws of probability including the laws of addition, multiplication and conditional probability Revise probabilities using Bayes’ rule.
  • 3. Copyright 2011 John Wiley & Sons, Inc. 3 Probability Probability – probability of occurrences are assigned to the inferential process under conditions of uncertainty
  • 4. Copyright 2011 John Wiley & Sons, Inc. 4 Methods of Assigning Probabilities Classical method of assigning probability (rules and laws) Relative frequency of occurrence (cumulated historical data) Subjective Probability (personal intuition or reasoning)
  • 5. Copyright 2011 John Wiley & Sons, Inc. 5 Number of outcomes leading to the event divided by the total number of outcomes possible P(E) = ne/N where N = total number of outcomes, and ne = number of outcomes in event E Each outcome is equally likely Applicable to games of chance Objective -- everyone correctly using the method assigns an identical probability Classical Probability
  • 6. Copyright 2011 John Wiley & Sons, Inc. 6 Subjective Probability - Comes from a person’s intuition or reasoning Subjective -- different individuals may (correctly or incorrectly) assign different numeric probabilities to the same event Degree of belief in the results of the event Useful for unique (single-trial) experiments New product introduction Site selection decisions Sporting events Subjective Probability
  • 7. Copyright 2011 John Wiley & Sons, Inc. 7 Experiment – is a process that produces an outcome Ex: Rolling two six-sided dice and calculating their sum Event – an outcome of an experiment Ex: The sum is at least 10 Elementary event – events that cannot be decomposed or broken down into other events Ex: The first die is a six Sample Space – a complete roster/listing of all elementary events for an experiment Ex: {(1,1), (1,2), (1,3), …, (2,1), (2,2), …., (6,5), (6,6)} Trial: one repetition of the process Ex: Roll the dice… Structure of Probability
  • 8. Copyright 2011 John Wiley & Sons, Inc. 8 Mutually Exclusive Events – events such that the occurrence of one precludes the occurrence of the other These events have no intersection Independent Events – the occurrence or nonoccurrence of one has no affect on the occurrence of the others Collectively Exhaustive Events – listing of all possible elementary events for an experiment Complementary Events – two events, one of which comprises all the elementary events of an experiment that are not in the other event Structure of Probability
  • 9. Copyright 2011 John Wiley & Sons, Inc. 9 The set of all elementary events for an experiment Methods for describing a sample space roster or listing tree diagram set builder notation Venn diagram Sample Space
  • 10. Copyright 2011 John Wiley & Sons, Inc. 10 Experiment: randomly select, without replacement, two families from the residents of Tiny Town Each ordered pair in the sample space is an elementary event, for example -- (D,C) Sample Space: Roster Example
  • 11. Copyright 2011 John Wiley & Sons, Inc. 11 S = {(x,y) | x is the family selected on the first draw, and y is the family selected on the second draw} Concise description of large sample spaces Sample Space: Set Notation for Random Sample of Two Families
  • 12. Copyright 2011 John Wiley & Sons, Inc. 12 { } { } { }9,7,6,5,4,3,2,1 6,5,4,3,2 9,7,4,1 =∪ = = YX Y X { } { } { } C IBM DEC Apple F Apple Grape Lime C F IBM DEC Apple Grape Lime = = ∪ = , , , , , , , , YX The union of two sets contains an instance of each element of the two sets. Union of Sets
  • 13. Copyright 2011 John Wiley & Sons, Inc. 13 { } { } { } X Y X Y = = ∩ = 1 4 7 9 2 3 4 5 6 4 , , , , , , , YX { } { } { } C IBM DEC Apple F Apple Grape Lime C F Apple = = ∩ = , , , , X ∩ Y The intersection of two sets contains only those element common to the two sets. Intersection of Sets
  • 14. Copyright 2011 John Wiley & Sons, Inc. 14 { } { } { } X Y X Y = = ∩ = 1 7 9 2 3 4 5 6 , , , , , , { } { } { }=∩ = = FC LimeGrapeF AppleDECIBMC , ,, YX 0)( =∩ YXP Events with no common outcomes Occurrence of one event precludes the occurrence of the other event Mutually Exclusive Events
  • 15. Copyright 2011 John Wiley & Sons, Inc. 15 Independent Events P X Y P X and P Y X P Y( | ) ( ) ( | ) ( )= = Occurrence of one event does not affect the occurrence or nonoccurrence of the other event The conditional probability of X given Y is equal to the marginal probability of X. The conditional probability of Y given X is equal to the marginal probability of Y.
  • 16. Copyright 2011 John Wiley & Sons, Inc. 16 Collectively Exhaustive Events E1 E2 E3 Sample Space with three collectively exhaustive events Contains all elementary events for an experiment
  • 17. Copyright 2011 John Wiley & Sons, Inc. 17 Complementary Events Sample Space A P Sample Space( ) = 1 P A P A( ) ( )′ = −1 ′A All elementary events not in the event ‘A’ are in its complementary event.
  • 18. Copyright 2011 John Wiley & Sons, Inc. 18 Counting the Possibilities mn Rule Sampling from a Population with Replacement Combinations: Sampling from a Population without Replacement
  • 19. Copyright 2011 John Wiley & Sons, Inc. 19 mn Rule If an operation can be done m ways and a second operation can be done n ways, then there are mn ways for the two operations to occur in order. A cafeteria offers 5 salads, 4 meats, 8 vegetables, 3 breads, 4 desserts, and 3 drinks. A meal is two servings of vegetables, which may be identical. How many meals are available? 5 * 4 * 8 * 3 * 4 * 3 = 5760
  • 20. Copyright 2011 John Wiley & Sons, Inc. 20 Combinations: Sampling from a Population without Replacement This counting method uses combinations Selecting n items from a population of N without replacement
  • 21. Copyright 2011 John Wiley & Sons, Inc. 21 Combinations Combinations – sampling “n” items from a population size N without replacement provides the formula shown below A tray contains 1,000 individual tax returns. If 3 returns are randomly selected without replacement from the tray, how many possible samples are there? 0166,167,00 )!31000(!3 !1000 )!(! ! = − = − =      = nNn N n N Crn
  • 22. Copyright 2011 John Wiley & Sons, Inc. 22 Combinations: Sampling from a Population without Replacement For example, suppose a small law firm has 16 employees and three are to be selected randomly to represent the company at the annual meeting of the American Bar Association. How many different combinations of lawyers could be sent to the meeting? Answer: NCn= 16C3= 16!/(3!×13!) = 560.
  • 23. Copyright 2011 John Wiley & Sons, Inc. 23 Four Types of Probability Marginal The probability of X occurring Union The probability of X or Y occurring Joint The probability of X and Y occurring Conditional The probability of X occurring given that Y has occurred YX YX Y X )(XP P X Y( )∪ P X Y( )∩ P X Y( | )
  • 24. Copyright 2011 John Wiley & Sons, Inc. 24 General Law of Addition P X Y P X P Y P X Y( ) ( ) ( ) ( )∪ = + − ∩ YX
  • 25. Copyright 2011 John Wiley & Sons, Inc. 25 General Law of Addition -- Example )()()()( SNPSPNPSNP ∩−+=∪ 81.0 56.67.70.)( 56.)( 67.)( 70.)( = −+=∪ =∩ = = SNP SNP SP NPSN .56 .67.70
  • 26. Copyright 2011 John Wiley & Sons, Inc. 26 81. 56.67.70. )()()()( = −+= ∩−+=∪ SNPSPNPSNP .11 .19 .30 .56 .14 .70 .67 .33 1.00 Increase Storage Space Yes No Total Yes No Total Noise Reduction Office Design Problem Probability Matrix
  • 27. Copyright 2011 John Wiley & Sons, Inc. 27 If a worker is randomly selected from the company described in Demonstration Problem 4.1 (below), what is the probability that the worker is either technical or clerical? What is the probability that the worker is either a professional or a clerical? Demonstration Problem 4.3 Type of Gender Position Male Female Total Managerial 8 3 11 Professional 31 13 44 Technical 52 17 69 Clerical 9 22 31 Total 100 55 155
  • 28. Copyright 2011 John Wiley & Sons, Inc. 28 Examine the raw value matrix of the company’s human resources data shown in Demonstration Problem 4.1. In many raw value and probability matrices like this one, the rows are non-overlapping or mutually exclusive, as are the columns. In this matrix, a worker can be classified as being in only one type of position and as either male or female but not both. Thus, the categories of type of position are mutually exclusive, as are the categories of sex, and the special law of addition can be applied to the human resource data to determine the union probabilities. Demonstration Problem 4.3
  • 29. Copyright 2011 John Wiley & Sons, Inc. 29 Demonstration Problem 4.3 Let T denote technical, C denote clerical, and P denote professional. The probability that a worker is either technical or clerical is P(T U C) = P (T) + P (C) = 69/155 + 31/155 = 100/155 = .645 The probability that a worker is either professional or clerical is P (P U C) = P (P) + P (C) = 44/155 + 31/155 = 75/155 = .484
  • 30. Copyright 2011 John Wiley & Sons, Inc. 30 P T C P T P C( ) ( ) ( ) . ∪ = + = + = 69 155 31 155 645 Demonstration Problem 4.3 Type of Gender Position Male Female Total Managerial 8 3 11 Professional 31 13 44 Technical 52 17 69 Clerical 9 22 31 Total 100 55 155
  • 31. Copyright 2011 John Wiley & Sons, Inc. 31 Type of Gender Position Male Female Total Managerial 8 3 11 Professional 31 13 44 Technical 52 17 69 Clerical 9 22 31 Total 100 55 155 P P C P P P C( ) ( ) ( ) . ∪ = + = + = 44 155 31 155 484 Demonstration Problem 4.3
  • 32. Copyright 2011 John Wiley & Sons, Inc. 32 )|()()|()()( YXPYPXYPXPYXP ⋅=⋅=∩ P M P S M P M S P M P S M ( ) . ( | ) . ( ) ( ) ( | ) ( . )( . ) . = = = ∩ = ⋅ = = 80 140 0 5714 0 20 0 5714 0 20 0 1143 Law of Multiplication Demonstration Problem 4.5
  • 33. Copyright 2011 John Wiley & Sons, Inc. 33 Law of Multiplication The intersection of two events is called the joint probability General law of multiplication is used to find the joint probability General law of multiplication gives the probability that both events x and y will occur at the same time P(x|y) is a conditional probability that can be stated as the probability of x given y
  • 34. Copyright 2011 John Wiley & Sons, Inc. 34 Law of Multiplication If a probability matrix is constructed for a problem, the easiest way to solve for the joint probability is to find the appropriate cell in the matrix and select the answer
  • 35. Copyright 2011 John Wiley & Sons, Inc. 35 Total .7857 Yes No .4571 .3286 .1143 .1000 .2143 .5714 .4286 1.00 Married Yes No Total Supervisor Probability Matrix of Employees 20.0)|( 5714.0 140 80 )( 2143.0 140 30 )( = == == MSP MP SP P M S P M P S M( ) ( ) ( | ) ( . )( . ) . ∩ = ⋅ = =0 5714 0 20 0 1143 Law of Multiplication Demonstration Problem 4.5
  • 36. Copyright 2011 John Wiley & Sons, Inc. 36 Law of Conditional Probability If X and Y are two events, the conditional probability of X occurring given that Y is known or has occurred is expressed as P(X|Y) The conditional probability of X given Y is the joint probability of X and Y divided by the marginal probability of Y. )( )()|( )( )( )|( YP XPXYP YP YXP YXP ⋅ = ∩ =
  • 37. Copyright 2011 John Wiley & Sons, Inc. 37 Law of Conditional Probability - Example 70% of respondents believe noise reduction would improve productivity. 56% of respondents believed both noise reduction and increased storage space would improve productivity A worker is selected randomly and asked about changes in the office design What is the probability that a randomly selected person believes storage space would improve productivity given that the person believes noise reduction improves productivity?
  • 38. Copyright 2011 John Wiley & Sons, Inc. 38 P N P N S P S N P N S P N ( ) . ( ) . ( | ) ( ) ( ) . . . = ∩ = = ∩ = = 70 56 56 70 80 NS .56 .70 Law of Conditional Probability
  • 39. Copyright 2011 John Wiley & Sons, Inc. 39 Independent Events Recall: Two events are independent when the occurrence of one does not affect the probability of occurrence of the other one When X and Y are independent, the conditional probability is equal to the marginal probability
  • 40. Copyright 2011 John Wiley & Sons, Inc. 40 Geographic Location Northeast D Southeast E Midwest F West G Finance A .12 .05 .04 .07 .28 Manufacturing B .15 .03 .11 .06 .35 Communications C .14 .09 .06 .08 .37 .41 .17 .21 .21 1.00 P A G P A G P G P A P A G P A ( | ) ( ) ( ) . . . ( ) . ( | ) . ( ) . = ∩ = = = = ≠ = 0 07 021 033 028 0 33 028 Independent Events Demonstration Problem 4.10
  • 41. Copyright 2011 John Wiley & Sons, Inc. 41 Revision of Probabilities: Bayes’ Rule )()|()()|()()|( )()|( )|( 2211 nn ii i XPXYPXPXYPXPXYP XPXYP YXP ⋅⋅⋅++ = Allows for reversing the order of conditioning (i.e. P(A|B) can be calculated if you know P(B|A) & P(A)) An extension to the conditional law of probabilities
  • 42. Copyright 2011 John Wiley & Sons, Inc. 42 Bayes’ Rule Note, the numerator of Bayes’ Rule and the law of conditional probability are the same The denominator is a collective exhaustive listing of mutually exclusive outcomes of Y The denominator is a weighted average of the conditional probabilities with the weights being the prior probabilities of the corresponding event
  • 43. Copyright 2011 John Wiley & Sons, Inc. 43 Bayes’ Rule: Ribbon Problem 447.0 )35.0)(12.0()65.0)(08.0( )35.0)(12.0( )()|()()|( )()|( )|( 553.0 )35.0)(12.0()65.0)(08.0( )65.0)(08.0( )()|()()|( )()|( )|( 12.0)|( 08.0)|( 35.0)( 65.0)( = + = ⋅+⋅ ⋅ = = + = ⋅+⋅ ⋅ = = = = = SJPSJdPAPAdP SJPSJdP dSJP SJPSJdPAPAdP APAdP dAP SJdP AdP SJP AP
  • 44. Copyright 2011 John Wiley & Sons, Inc. 44 Bayes’ Rule: Ribbon Problem Alternative Approach using Tree Diagram Alamo (A) 0.65 South Jersey (SJ) 0.35 Defective (d) 0.08 Defective (d) 0.12 Acceptable 0.92 Acceptable 0.88 0.052 0.042 + 0.094=P(d) 447.0 094.0 042.0 )|( ==dSJP

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