Ken Black QA ch04

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Business Statistics, 5th ed.

by Ken Black

Chapter 4

Probability

Discrete Distributions

PowerPoint presentations prepared by Lloyd Jaisingh, Morehead State University



Learning Objectives

• Comprehend the different ways of assigningprobability.

• 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.



Methods of Assigning Probabilities

• Classical method of assigning probability(rules and laws)

• Relative frequency of occurrence(cumulated historical data)

• Subjective Probability (personal intuition orreasoning)



Classical Probability

• Number of outcomes leadingto the event divided by thetotal number of outcomespossible

• Each outcome is equally likely• Determined a priori -- before

performing the experiment

• Applicable to games of chance

• Objective -- everyone correctlyusing the method assigns anidentical probability

Einoutcomesof number

outcomesof numbertotal

:

)(

e

n

n

N Where

N E P e



Relative Frequency Probability

• Based on historicaldata

• Computed afterperforming the

experiment• Number of times an

event occurred dividedby the number of trials

• Objective -- everyone

correctly using themethod assigns anidentical probability

Eproducing

outcomesof number

trialsof numbertotal:

)(

e

n

n

N

Where

N E P

e



Subjective Probability

• Comes from a person’s intuition or reasoning

• Subjective -- different individuals may(correctly) assign different numeric

probabilities to the same event• Degree of belief • Useful for unique (single-trial) experiments

– New product introduction

– Initial public offering of common stock – Site selection decisions – Sporting events



Structure of Probability

• Experiment

• Event

• Elementary Events

• Sample Space

• Unions and Intersections

• Mutually Exclusive Events

• Independent Events• Collectively Exhaustive Events

• Complementary Events



Experiment

• Experiment: a process that produces outcomes – More than one possible outcome

– Only one outcome per trial

• Trial: one repetition of the process

• Elementary Event: cannot be decomposed or broken down into other events

• Event: an outcome of an experiment

– May be an elementary event, or

– May be an aggregate of elementary events

– Usually represented by an uppercase letter, e.g.,A, E1



An Example Experiment

Experiment: randomly select, withoutreplacement, two families from the residents of Tiny Town

Elementary Event: the

sample includes familiesA and C

Event: each family inthe sample has children

in the householdEvent: the sample

families own a total of four automobiles

FamilyChildren inHousehold

Number ofAutomobiles

AB

CD

YesYes

NoYes

32

12

Tiny Town Population



Sample Space

• The set of all elementary events for anexperiment

• Methods for describing a sample space

– roster or listing – tree diagram

– set builder notation

– Venn diagram



Sample Space: Roster Example

• Experiment: randomly select, withoutreplacement, two families from the residents of Tiny Town

• Each ordered pair in the sample space is anelementary event, for example -- (D,C)

FamilyChildren inHousehold

Number ofAutomobiles

ABCD

YesYesNo

Yes

3212

Listing of Sample Space

(A,B), (A,C), (A,D),

(B,A), (B,C), (B,D),(C,A), (C,B), (C,D),(D,A), (D,B), (D,C)



Sample Space: Tree Diagram for

Random Sample of Two Families

A

B

C

D

B

C

D

A

C

D

A

B

D

A

B

C



Sample Space: Set Notation for

Random Sample of Two Families

• S = {(x,y) | x is the family selected on the

first draw, and y is the family selected onthe second draw}

• Concise description of large sample spaces



Sample Space

• Useful for discussion of general principlesand concepts

Listing of Sample Space

(A,B), (A,C), (A,D),(B,A), (B,C), (B,D),(C,A), (C,B), (C,D),(D,A), (D,B), (D,C)

Venn Diagram



Union of Sets

• The union of two sets contains an instanceof each element of the two sets.

X

Y

X Y

1 4 7 9

2 3 4 5 6

1 2 3 4 5 6 7 9

, , ,

, , , ,

, , , , , , ,

C IBM DEC AppleF Apple Grape Lime

C F IBM DEC Apple Grape Lime

, ,

, ,

, , , ,

YX

X Y



Intersection of Sets

• The intersection of two sets contains onlythose element common to the two sets.

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



Mutually Exclusive Events

• Events with nocommon outcomes

• Occurrence of oneevent precludes the

occurrence of theother event

X

Y

X Y

1 7 92 3 4 5 6, ,, , , ,

C IBM DEC AppleF Grape Lime

C F

, ,,

YX

P X Y ( ) 0



Independent Events

• Occurrence of one event does not affect theoccurrence or nonoccurrence of the otherevent

• The conditional probability of X given Y isequal to the marginal probability of X.

• The conditional probability of Y given X isequal to the marginal probability of Y.

P X Y P X and P Y X P Y ( | ) ( ) ( | ) ( )



Collectively Exhaustive Events

• Contains all elementary events for anexperiment

E1 E2 E3

Sample Space with three

collectively exhaustive events



Complementary Events

• All elementary events not in the event ‘A’are in its complementary event.

Sample

Space

A

P Sample Space( ) 1

P A P A( ) ( ) 1

A



Counting the Possibilities

• m n Rule

• Sampling from a Population withReplacement

• Combinations: Sampling from a Populationwithout Replacement



m n Rule

• If an operation can be done in m ways and asecond operation can be done in n ways,then there are m n ways for the two

operations to occur in order.• A cafeteria offers 5 salads, 4 meats, 8vegetables, 3 breads, 4 desserts, and 3drinks. A meal consists of one serving of

each of the items. How many meals areavailable?

• ( Ans: 548343 = 5,760 meals.)



Sampling from a Population with

Replacement

• A tray contains 1,000 individual tax returns.If 3 returns are randomly selected withreplacement from the tray, how many

possible samples are there?• (N)n = (1,000)3 = 1,000,000,000



Combinations: Sampling from a

Population without Replacement

• This counting method uses combinations

• Selecting n items from a population of Nwithout replacement



Combinations: Sampling from a

Population without Replacement

• For example, suppose a small law firm has16 employees and three are to be selectedrandomly to represent the company at the

annual meeting of the American BarAssociation.

• How many different combinations of lawyers could be sent to the meeting?

• Answer: NC n = 16C 3 = 16!/(3!13!) = 560.



Four Types of Probability

• Marginal Probability

• Union Probability

• Joint Probability

• Conditional Probability



Four Types of Probability

Marginal

The probabilityof X occurring

Union

The probabilityof X or Y occurring

Joint

The probabilityof X and Y occurring

Conditional

The probabilityof X occurringgiven that Y has occurred

YX YX

Y

X

P X ( ) P X Y ( ) P X Y ( )P X Y ( | )



General Law of Addition

P X Y P X P Y P X Y ( ) ( ) ( ) ( )

YX



General Law of Addition -- Example

P N S P N P S P N S( ) ( ) ( ) ( )

SN

.56.67.70

P N

P S

P N S

P N S

( ) .

( ) .

( ) .

( ) . . ..

70

67

56

70 67 560 81



Office Design Problem

Probability Matrix

.11 .19 .30

.56 .14 .70

.67 .33 1.00

Increase

Storage Space

Yes No Total

Yes

No

Total

Noise

Reduction




Probability Matrix

.11 .19 .30.56 .14 .70

.67 .33 1.00

Increase

Storage Space

Yes No Total

YesNo

Total

NoiseReduction

P N S P N P S P N S( ) ( ) ( ) ( ). . .

.

70 67 56

81




Probability Matrix

.11 .19 .30

.56 .14 .70

.67 .33 1.00

Increase

Storage Space

Yes No Total

YesNo

Total

NoiseReduction

P N S( ) . . .

.

56 14 11

81



Venn Diagram of the X or Y

but not Both Case

YX



The Neither/Nor Region

YX

P X Y P X Y ( ) ( ) 1



The Neither/Nor Region

SN

P N S P N S( ) ( )

.

.

1

1 81

19



Special Law of Addition

If X and Y are mutually exclusive,

P X Y P X P Y ( ) ( ) ( )

X

Y



Demonstration Problem 4.3

Type of GenderPosition Male Female TotalManagerial 8 3 11Professional 31 13 44

Technical 52 17 69Clerical 9 22 31Total 100 55 155

P T C P T P C ( ) ( ) ( )

.

69

155

31

155

645




Type of GenderPosition Male Female TotalManagerial 8 3 11Professional 31 13 44

Technical 52 17 69Clerical 9 22 31Total 100 55 155

P P C P P P C ( ) ( ) ( )

.

44

155

31

155

484



Law of Multiplication


P X Y P X P Y X P Y P X Y ( ) ( ) ( | ) ( ) ( | )

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


Total

.7857

Yes No

.4571 .3286

.1143 .1000 .2143

.5714 .4286 1.00

Married

Yes

No

Total

Supervisor

Probability Matrix

of Employees

20.0)|(

5714.0140

80)(

2143.0140

30)(

M SP

M P

SP

P M S P M P S M ( ) ( ) ( | )

( . )( . ) .

0 5714 0 20 0 1143

P M S P M P M S

P M S P S P M S

P M P M

( ) ( ) ( )

. . .

( ) ( ) ( )

. . .

( ) ( )

. .

0 5714 0 1143 0 4571

0 2143 0 1143 0 1000

1

1 0 5714 0 4286

P S P S

P M S P S P M S

( ) ( )

. .( ) ( ) ( )

. . .

1

1 0 2143 0 7857

0 7857 0 4571 0 3286



Special Law of Multiplication

for Independent Events

• General Law

• Special Law

P X Y P X P Y X P Y P X Y ( ) ( ) ( | ) ( ) ( | )

If events X and Y are independent,

and P X P X Y P Y P Y X

Consequently

P X Y P X P Y

( ) ( | ), ( ) ( | ).

,

( ) ( ) ( )



Law of Conditional Probability

• The conditional probability of X given Y isthe joint probability of X and Y divided bythe marginal probability of Y.

P X Y P X Y

P Y

P Y X P X

P Y

( | )( )

( )

( | ) ( )

( )



Law of Conditional Probability

NS

.56.70

P N

P N S

P S N

P N S

P N

( ) .

( ) .

( | )( )

( )

.

.

.

70

56

56

70

80




164.

67.

11.

)(

)(

)|(

SP

S N P

S N P

.19 .30

.14 .70

.33 1.00

IncreaseStorage Space

Yes No Total

Yes

NoTotal

Noise

Reduction .11

.56

.67

Reduced Sample

Space for “Increase

Storage Space”

= “Yes”



Independent Events

• If X and Y are independent events, theoccurrence of Y does not affect theprobability of X occurring.

• If X and Y are independent events, theoccurrence of X does not affect theprobability of Y occurring.

If X and Y are independent events ,

, and P X Y P X

P Y X P Y

( | ) ( )

( | ) ( ).



Independent Events

Demonstration Problem 4.10Geographic 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 GP A G

P GP A

P A G P A

( | )( )

( )

.

.. ( ) .

( | ) . ( ) .

0 07

0 210 33 0 28

0 33 0 28



Independent Events


D EA 8 12 20

B 20 30 50

C 6 9 15

34 51 85

P A D

P A

P A D P A

( | ) .

( ) .

( | ) ( ) .

8

342353

20

852353

02353



Revision of Probabilities: Bayes’ Rule

• An extension to the conditional law of probabilities

• Enables revision of original probabilitieswith new information

P X Y P Y X P X

P Y X P X P Y X P X P Y X P X i

i i

n n

( | )( | ) ( )

( | ) ( ) ( | ) ( ) ( | ) ( )

1 1 2 2



Revision of Probabilities

with Bayes' Rule: Ribbon Problem P Alamo

P SouthJersey

P d Alamo

P d SouthJersey

P Alamo d P d Alamo P Alamo P d Alamo P Alamo P d SouthJersey P SouthJersey

P SouthJersey d P d SouthJersey P SouthJersey

P d Alamo P Alamo P d SouthJersey P SouthJersey

( ) .

( ) .

( | ) .

( | ) .

( | ) ( | ) ( )( | ) ( ) ( | ) ( )

( . )( . )

( . )( . ) ( . )( . ).

( | )( | ) ( )

( | ) ( ) ( | ) ( )( . )( . )

( .

0 65

0 35

0 08

0 12

0 08 0 65

0 08 0 65 0 12 0 350 553

0 12 0 35

0 08)( . ) ( . )( . ).

0 65 0 12 0 350 447

i i f i i i




with Bayes’ Rule: Ribbon Problem

ConditionalProbability

0.052

0.042

0.094

0.65

0.35

0.08

0.12

0.0520.094

=0.553

0.0420.094

=0.447

Alamo

South Jersey

Event

PriorProbability

P E i( )

JointProbability

P E d i( )

RevisedProbability

P E d i( | )P d E i( | )

R i i f P b bili i




with Bayes' Rule: Ribbon Problem

Alamo

0.65

South

Jersey0.35

Defective

0.08

Defective

0.12

Acceptable0.92

Acceptable

0.88

0.052

0.042

+ 0.094

P b bili f S



Probability for a Sequence

of Independent Trials

• 25 percent of a bank’s customers are commercial(C) and 75 percent are retail (R).

• Experiment: Record the category (C or R) for

each of the next three customers arriving at thebank.

• Sequences with 1 commercial and 2 retailcustomers.

– C1 R2 R3 – R1 C2 R3

– R1 R2 C3

P b bili f S





• Probability of specific sequences containing1 commercial and 2 retail customers,assuming the events C and R areindependent

P C R R P C P R P R

P R C R P R P C P R

P R R C P R P R P C

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1 2 3

1 2 3

1 2 3

1

4

3

4

3

4

9

64

3

4

1

4

3

4

9

64

3

4

3

4

1

4

9

64

P b bilit f S





• Probability of observing a sequencecontaining 1 commercial and 2 retailcustomers, assuming the events C and R areindependent

P C R R R C R R R C

P C R R P R C R P R R C

( ) ( ) ( )

( ) ( ) ( )

1 2 3 1 2 3 1 2 3

1 2 3 1 2 3 1 2 3

964

964

964

2764

P b bilit f S





• Probability of a specific sequence with 1 commercial and2 retail customers, assuming the events C and R areindependent

• Number of sequences containing 1 commercial and 2retail customers

• Probability of a sequence containing 1 commercial and 2retail customers

P C R R P C P R P R ( ) ( ) ( )9

64

n r C

n

r

n

r n r

!

! !

!

! !

3

1 3 13

39

64

27

64

P b bilit f S




of Dependent Trials

• Twenty percent of a batch of 40 tax returnscontain errors.

• Experiment: Randomly select 4 of the 40 taxreturns and record whether each returncontains an error (E) or not (N).

• Outcomes with exactly 2 erroneous tax returnsE1 E2 N3 N4 E1 N2 E3 N4 E1 N2 N3 E4

N1 E2 E3 N4 N1 E2 N3 E4 N1 N2 E3 E4

P b bilit f S




of Dependent Trials

• Probability of specific sequences containing 2erroneous tax returns (three of the sixsequences)

P E E N N P E P E E P N E E P N E E N

P E N E N P E P N E P E E N P N E N E

( ) ( ) ( | ) ( | ) ( | )

,

, ,.

( ) ( ) ( | ) ( | ) ( | )

1 2 3 4 1 2 1 3 1 2 4 1 2 3

1 2 3 4 1 2 1 3 1 2 4 1 2 3

8

50

7

49

32

48

31

47

55552

5 527 2000 01

8

50

32

49

7

48

31

47

55 552

5 527 2000 01

8

50

32

49

31

48

7

47

55 552

5 527 2000 01

1 2 3 4 1 2 1 3 1 2 4 1 2 3

,

, ,.

( ) ( ) ( | ) ( | ) ( | )

,

, ,.

P E N N E P E P N E P N E N P E E N N

P b bilit f S





• Probability of observing a sequence containingexactly 2 erroneous tax returns

P E E N N E N E N E N N E

N E E N N E N E N N E E

P E E N N P E N E N P E N N E

P N E E N P N E N E P N N E E

(( ) ( ) ( )

( ) ( ) ( ))

( ) ( ) ( )

( ) ( ) ( )

,

1 2 3 4 1 2 3 4 1 2 3 4

1 2 3 4 1 2 3 4 1 2 3 4

1 2 3 4 1 2 3 4 1 2 3 4

1 2 3 4 1 2 3 4 1 2 3 4

55 5525

, ,,

, ,,

, ,,

, ,,

, ,,

, ,

.

527 20055 552

5 527 20055 552

5 527 20055 552

5 527 20055 552

5 527 20055 552

5 527 200

0 06

P b bilit f S




of Dependent Trials• Probability of a specific sequence with exactly 2 erroneous tax

returns

• Number of sequences containing exactly 2 erroneous taxreturns

• Probability of a sequence containing exactly 2 erroneous taxreturns

n r

r

nC

n

r

n

r n r C

!

! !

!

! !

4

2 4 26

655552

5527 200006

,

, ,.

P E E N N ( ),

, ,.1 2 3 4

8

50

7

49

32

48

31

47

55 552

5 527 2000 01



Copyright 2008 John Wiley & Sons, Inc. All rights reserved. Reproduction or translation

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Ken Black QA ch04

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