Business Statistics, 6 ed. bybusiness.uni.edu/slides/ECON-1011_Luk/ch04.pdfExamine the raw value...

Business Statistics, 6th ed.by Ken Black

Chapter 4

Probability

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 probabilityRevise probabilities using Bayes’ rule.

Probability

Probability – probability of occurrences areassigned to the inferential process underconditions of uncertainty

Methods of Assigning Probabilities

Classical method of assigning probability(rules and laws)Relative frequency of occurrence(cumulated historical data)Subjective Probability (personal intuitionor reasoning)

Classical Probability

WhereN

total number of outcomes

number of outcomes in Een

Number of outcomes leading to the event divided by the total number of outcomes possibleEach outcome is equally likelyDetermined a priori ‐‐ before performing the experimentApplicable to games of chanceObjective ‐‐ everyone correctly using the method assigns an identical probability

Classical Probability

Relative Frequency of Occurrence method – the probability of an event is equal to the number of times the event has occurred in the past divided by the total number of opportunities for the event to have occurredFrequency of occurrence is based on what has happened in the past

Relative Frequency Probability

E producing

outcomes ofnumber trialsofnumber total

NWhere

Relative Frequency Probability

Based on historical dataComputed after performing the experimentNumber of times an event occurred divided by the number of trialsObjective ‐‐ everyone correctly using the method assigns an identical probability

Subjective Probability ‐ Comes from a person’s intuition or reasoningSubjective ‐‐ different individuals may (correctly or incorrectly) assign different numeric probabilities to the same eventDegree of belief in the results of the eventUseful for unique (single‐trial) experiments

New product introductionSite selection decisionsSporting events

Subjective Probability

Experiment – is a process that produces an outcomeEvent – an outcome of an experimentElementary event – events that cannot be decomposed or broken down into other eventsSample Space – a complete roster/listing of all elementary events for an experimentTrial: one repetition of the process

Structure of Probability

Set Notation – the use of braces to group membersThe UNION of x, y is formed by combining elements from both sets, and is denoted by x U y. Read as “x or y.”An INTERSECTION is denoted x ∩ y. The symbol is read as “and”. “x and y”

Mutually Exclusive Events – events such that the occurrence of one precludes the occurrence of the other

These events have no intersectionIndependent 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 experimentComplementary Events – two events, one of which comprises all the elementary events of an experiment that are not in the other event

The set of all elementary events for an experimentMethods for describing a sample space

roster or listingtree diagramset builder notationVenn diagram

Sample Space

Family Children in Household

Number of Automobiles

YesYesNoYes

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)

Experiment: randomly select, without replacement, two families from the residents of Tiny TownEach ordered pair in the sample space is an elementary event, for example ‐‐ (D,C)Is (A,B) the same as (B,A)?

Sample Space: Roster Example

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 forRandom Sample of Two Families

9,7,6,5,4,3,2,16,5,4,3,2

9,7,4,1

C IBM DEC Apple

F Apple Grape Lime

C F IBM DEC Apple Grape Lime

, , , ,

The union of two sets contains an instance of each element of the two sets.

Union of Sets

1 4 7 9

2 3 4 5 64

, , , ,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

2 3 4 5 6

, , , ,

FCLimeGrapeF

AppleDECIBMC,

0)( YXP

Events with no common outcomesOccurrence of one event precludes the occurrenceof the other event

Mutually Exclusive Events

Independent Events

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

Occurrence of one event does not affect the occurrence or nonoccurrence of the other eventThe 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.

Collectively Exhaustive Events

E1 E2 E3

Sample Space with three collectively exhaustive events

Contains all elementary events for an experiment

Complementary Events

mpleace A

P SampleSpace( ) 1

P A P A( ) ( ) 1A

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

Counting the Possibilities

mn Rule Sampling from a Population with ReplacementCombinations: Sampling from a Population without Replacement

mn Rule

f 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 (Mistake? ) – Does order matter on the vegetable choice?Counting Coin and Die tosses?

Combinations: Sampling from aPopulation without Replacement

This counting method uses combinationsSelecting n items from a population of Nwithout replacement

Combinations

Combinations – sampling “n” items from a population size N without replacement provides the formula shown belowA tray contains 1,000 individual tax returns. If 3 returns are randomly selected without replacement from the tray, how many possible samples are there?Does order matter?

0166,167,00)!31000(!3

!1000)!(!

Combinations: Sampling from aPopulation 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.Suppose the representatives are chosen such that the order is President, Vice‐President, and Alternate?

Four Types of Probability

Marginal

e probability X occurring

The probability of X or Yoccurring

The probability of X and Yoccurring

Conditional

The probability of X occurring given that Yhas occurred

YX YXY

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

General Law of Addition

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

General Law of Addition ‐‐ Example

)()()()( SNPSPNPSNP

81.056.67.70.)(

56.)(67.)(70.)(

SNPSNPSPNPSN

.56 .67.70

81.56.67.70.

)()()()(

SNPSPNPSNP

.11 .19 .30.56 .14 .70

.67 .33 1.00

Increase Storage SpaceYes No Total

YesNoTotal

Noise Reduction

Office Design Problem Probability Matrix

f a worker is randomly selected from the company described in Demonstration Problem 4.1, 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

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

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

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

Demonstration Problem 4.3

Type of GenderPosition Male Female TotalManagerial 8 3 11Professional 31 13 44Technical 52 17 69Clerical 9 22 31Total 100 55 155

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

Demonstration Problem 4.3

)|()()|()() YXPYPXYPXPYX

P S MP M S P M P S M

( | ) .( ) ( ) ( | )

( . )( . ) .

0 5714

0 5714 0 20 0 1143

Law of MultiplicationDemonstration Problem 4.5

Law of Multiplication

The intersection of two events is called the joint probabilityGeneral law of multiplication is used to find the joint probabilityGeneral law of multiplication gives the probability that both events x and y will occur at the same timeP(x|y) is a conditional probability that can be stated as the probability of x given y

Law of Multiplication

f 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

Yes No

.4571 .3286

.1143 .1000 .2143

.5714 .4286 1.00

Married

pervisor

Probability Matrixof Employees

20.0)|(

5714.014080)(

2143.014030)(

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

11 0 5714 0 4286

3286.04571.07857.0)()()(

7857.02143.01)(1)(

SMPSPSMP

Law of MultiplicationDemonstration Problem 4.5

Law of Conditional Probability

Conditional probability are based on the prior knowledge you have on one of the two events being studiedf X and Y are two events, the conditional probabilityof X occurring given that Y is known or has occurred s expressed as P(X|Y)

)()()|(

YPXPXYP

YPYXPYXP

The conditional probability of X given Y is the jointprobability of X and Y divided by the marginalprobability of Y.

70% of respondents believe noise reduction would mprove productivity.56% of respondents believed both noise reduction and increased storage space would improve productivityA worker is selected randomly and asked about changes in the office designWhat is the probability that a randomly selected person believes storage space would improve productivity given that the person believes noise reduction improves productivity?

P NP N S

P S NP N S

( ) .( ) .

( | )( )

7 05 6

5 67 0

.56 .70

Independent Events

When X and Y are independent, the conditional probability is solved as a marginal probability

Geographic Location

NortheastD

SoutheastE

MidwestF

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

( | )( )

. ( ) .

( | ) . ( ) .

007021

033 028

Independent EventsDemonstration Problem 4.10

Revision of Probabilities: Bayes’ Rule

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

2211 nn

XPXYPXPXYPXPXYPXPXYPYX

An extension to the conditional law of probabilitiesEnables revision of original probabilities with new nformation

Bayes’ Rule

Bays’ rule – extends the use of the law of conditional probabilities to all revision of original probabilities with new information

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 YThe denominator is a weighted average of the conditional probabilities with the weights being the prior probabilities of the corresponding event

Revision of Probabilitieswith Bayes’ Rule: Ribbon ProblemP Alamo

P SouthJerseyP d Alamo

P d SouthJersey

P Alamod P d Alamo P AlamoP d Alamo P Alamo P d SouthJerseyP SouthJersey

P SouthJerseyd P d SouthJerseyP SouthJerseyP d Alamo P Alamo P d SouthJerseyP SouthJersey

( ) .( ) .

( | ) .( | ) .

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

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

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

( . )( . )( .

0 650 350 080 12

0 08 0 650 08 0 65 0 12 0 35

0 12 0 350 08)( . ) ( . )( . )

.0 65 0 12 0 35

Revision of Probabilities with Bayes’ Rule: Ribbon Problem

Conditional Probability

0.0520.094

=0.553

0.0420.094

=0.447

outh Jersey

Prior Probability

P Ei( )

Joint Probability

P E di( )

Revised Probability

P E di( | )P d Ei( | )

Revision of Probabilities with Bayes’ Rule: Ribbon Problem

Alamo0.65

SouthJersey0.35

Defective0.08

Defective0.12

Acceptable0.92

Acceptable0.88

+ 0.094

Business Statistics, 6 ed. bybusiness.uni.edu/slides/ECON-1011_Luk/ch04.pdfExamine the raw value...

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