chapter1.1- probability theory

8/8/2019 Chapter1.1- Probability Theory

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OBS 615:

MANAGERIAL DECISION

MAKING PROCESSES ANDTECHNIQUES


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This course is intended to enable

students:

Understand managerial decision-makingprocesses in organizations

Appreciate the use of various quantitativetechniques in making decisions

Apply the processes and techniques in real

managerial problem solving situations


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The following topics will be

covered

Quantitative methods for Decision making

Linear programming Transportation Problems Assignment problems Network analysis

Inventory management Waiting line methods Simulation Models


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Chapter 1.1

Probability


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Learning Objectives

Concepts: Events, Sample Space,Intersection & Union

Tools: Venn Diagram, Tree Diagram,

Contingency TableProbability, Conditional Probability,

Addition Rule & Multiplication Rule

Use Bayes TheoremSummarize the Rules ofPermutation andCombination


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Basic Concepts

Random Experiment is a process leadingto at least two possible outcomes withuncertainty as to which will occur.

yA coin is thrown

yA consumer is asked which of two

products he or she prefersyThe daily change in an index of stock

market prices is observed


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Sample Spaces

Collection ofall possible outcomes

e.g.: All six faces of a die:

e.g.: All 52 cards

a deck ofbridge cards


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Events and Sample Spaces

An event is a set of basic outcomes fromthe sample space, and it is said to occurif the random experiment gives rise to

one ofits constituent basic outcomes.Simple EventOutcome With 1 Characteristic

Joint Event2 Events Occurring Simultaneously

Compound EventOne or Another Event Occurring


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Simple Event

A: Male

B: Over age 20

C: Has 3 credit cards

D: Red card from a deck of bridge cards

E: Ace card from a deck of bridge cards


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Joint Event

D and E, (DE):

Red, ace card from a bridge deck

A and B, (AB):

Male, over age 20

among a group ofsurvey respondents


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Intersection

Let A and B be two events in the samplespaceS. Theirintersection, denoted AB,is the set of all basic outcomes inSthat

belong to both A and B. Hence, the intersection AB occurs if and

only ifboth A and B occur.

If the events A and B have no commonbasic outcomes, their intersection AB issaid to be the empty set.


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Compound Event

D or E, (DE):Ace or Red card from bridge deck


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Union

Let A and B be two events in thesample spaceS. Theirunion, denoted

AB, is the set of all basic outcomesinSthat belong to at least one of thesetwo events.

Hence, the union AB occurs if andonly ifeither A or B or both occurs


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Event Properties

Mutually Exclusive

Two outcomes that cannotoccur at the same time

E.g. flip a coin, resulting inhead and tail

Collectively ExhaustiveOne outcome in sample space

must occur

E.g. Male or Female


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ii

Special Events

Null Event

C

lub & Diamond on1 Card Draw

Complement of Event

For Event A, AllEvents Not In A:

A' or A

Null Event


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What is Probability?1.Numerical measure of

likelihood that the event

will occurSimple Event

Joint Event

Compound

2.Lies between 0 & 1

3.Sum of events is 1

1

.5

0

Certain

Impossible


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Concept of Probability

A Prioriclassical probability, the probability ofsuccess is based on prior knowledge of the processinvolved.

i.e. the chance of picking a black card from a

deck of bridge cardsEmpirical classical probability, the outcomes are

based on observed data, not on prior knowledge of aprocess.

i.e. the chance that individual selected at randomfrom the Kalosha employee survey if satisfiedwith his or her job. (.89)


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Concept of ProbabilitySubjective probability, the chance of

occurrence assigned to an event by a

particular individual, based on his/herexperience, personal opinion andanalysis of a particular situation.

i.e. The chance of a newly designed style ofmobile phone will be successful in market.


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(There are 2 ways to get one 6 and the other 4)

e.g. P( ) = 2/36

Computing Probabilities

The probability of an event E:

Each of the outcomes in the sample space isequally likely to occur

number of event outcomes( )

total number of possible outcomes in the sample space

P E

X

T

!

!


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Concept of Probability

Experi- Number Number Relativementer of Trials of heads Frequency

Demogen 2048 1061 0.

5

181Buffon 4040 2048 0.5069

Pearson 12000 6019 0.5016

Pearson 24000 12012 0.5005

What conclusion can be drawn from the

observations?


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Presenting Probability &

Sample Space

1. Listing

S = {Head, Tail}2. Venn Diagram

3. Tree Diagram

4. Contingency

Table


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S

A

Venn Diagram

Example: Kalosha Employee Survey

Event:A = Satisfied, = Dissatisfied

P(A) = 356/400 = .89, P() = 44/400 = .11


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KaloshaEmployee

Tree Diagram

Satisfied

Not

Satisfied

Advanced

Not

Advanced

Advanced

Not

Advanced

P(A)=.89

P()=.11

.485

.405

.035

.075

Example: Kalosha Employee Survey JointProbability


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Event

Event B1 B2 Total

A1 P(A1B1) P(A1B2) P(A1)

A2 P(A2B1) P(A2B2) P(A2)

Total P(B1) P(B2) 1

Joint Probability

Using Contingency Table

Joint Probability Marginal (Simple)Probability


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Joint ProbabilityUsing Contingency Table

Kalosha Employee Survey

Satisfied NotSatisfied Total

Advanced

NotAdvanced

Total

.485 .035

.405 .075

.52

.48

.89 .11 1.00

Joint Probability

Simple Probability


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Use of Venn Diagram

Fig. 3.1: AB, Intersection of events A & B,

mutually exclusive

Fig.3

.2: A

B, Union of events A & BFig. 3.3: , Complement of event A

Fig. 3.4 and 3.5:

The events AB and B are mutuallyexclusive, and their union is B.

(A B) ( B) = B


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Use ofVenn Diagram

Let E1, E2,, Ekbe Kmutually exclusive and

collective exhaustive events, and let A be

some other event. Then the Kevents E1A,E2A, , EkA are mutually exclusive,

and their union is A.

(E1A) (E2A) (EkA) = A

(See supplement Fig. 3.7)


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Compound Probability

Addition Rule1. Used to Get Compound Probabilities for

Union of Events

2. P(A or B) = P(A B)= P(A) + P(B) P(A B)

3. ForMutually Exclusive Events:

P(A or B) = P(A B) = P(A) + P(B)4. Probability ofComplement

P(A) + P() = 1. So, P() = 1 P(A)


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Addition Rule: Example

A hamburger chain found that 75%

of all customersuse mustard, 80% use ketchup, 65% use both.What is the probability that a particular customerwill use at least one of these?

A = Customers use mustardB = Customers use ketchup

AB = a particular customer will use at least oneof these

Given P(A) = .75, P(B) = .80, and P(AB) = .65,P(AB) = P(A) + P(B) P(AB)

= .75 + .80 .65= .90


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Conditional Probability

1. Event Probability Given that AnotherEvent Occurred

2. Revise Original Sample Space toAccount forNew Information

Eliminates Certain Outcomes

3. P(A | B) = P(A and B) , P(B)>0P(B)


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Example

Recall the previous hamburger chainexample, what is the probability that aketchup user uses mustard?

P(A|B) = P(AB)/P(B)

= .65/.80 = .8125

Please pay attention to the differencefrom the joint event in wording of thequestion.


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S

Black

Ace

Conditional Probability

Black Happens: Eliminates All Other Outcomes andThus Increase the Conditional Probability

Event (Ace and Black)

(S)Black

Draw a card, what is the probability of black ace?What is the probability of black ace when black happens?


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ColorType Red Black Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

Conditional Probability Using

Contingency TableConditional Event: Draw 1 Card. Note BlackAce

RevisedSampleSpace

P(Ace|Black) =P(Ace and Black)

P(Black)= = 2/262/52

26/52


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Game Show

Imagine you have been selected for a gameshow that offers the chance to win anexpensive new car. The car sits behind one ofthree doors, while monkeys reside behind theother two.

You choose a door, and the host opens one ofthe remaining two, revealing a monkey.

The host then offers you a choice: stick with

your initial choice or switch to the other, still-unopened door.

Do you think thatswitch to the other doorwould increase your chance of winning the car?


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Statistical Independence

1. Event Occurrence Does NotAffectProbability of Another Event

e.g. Toss 1C

oin Twice, Throw3

Dice2. Causality Not Implied

3. Tests For Independence

P(A | B) = P(A), or P(B | A) = P(B),or P(A and B) = P(A)P(B)


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Statistical Independence

Kalosha Employee Survey

Satisfied NotSatisfied Total

Advanced

NotAdvanced

Total

.485 .035

.405 .075

.52

.48

.89 .11 1.00

P(A1)

Note: (.52)(.89) = .4628{ .485

P(B1

)P(A1

and B1

)


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Multiplication Rule

1. Used to Get Joint Probabilities forIntersection of Events (Joint Events)

2. P(A and B) = P(A B)P(A B) = P(A)P(B|A)

= P(B)P(A|B)

3. ForIndependent Events:

P(A and B) = P(AB) = P(A)P(B)


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Randomized Response

An approach for solicitation ofhonest answersto sensitive questions in surveys.

A survey was carried out in spring 1997 to

about 150 undergraduate students takingStatistics for Business and Economics atPeking University.

Each student was faced with two questions.

Students were asked first to flip a coin andthen to answer question a) if the result wasthe national emblem and b) otherwise.


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Practice of Randomized Response

Questiona) Is the second last digit of your office phone

number odd?

b) Recall your undergraduate course work, have youever cheated in midterm or final exams?

Respondents, please do as follows:

1. Flip a coin.

2. If the result is the national emblem (

), answerquestion a); otherwise answer question b).Please circleYes or No below as your answer.

Yes No


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Bayes Theorem

1. Permits RevisingOldProbabilities Based on

New Information2. Application ofConditional Probability

3. Mutually ExclusiveEvents

New

Information

Revised

Probability

Apply Bayes'

Theorem

Prior

Probability


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Bayess Theorem Example

Fifty percent of borrowers repaid their loans. Out ofthose who repaid, 40% had a college degree. Tenpercent of those who defaulted had a college degree.What is the probability that a randomly selectedborrower who has a college degree will repay the loan?

B1= repay, B2= default, A=college degree

P(B1) = .5, P(A|B1) = .4, P(A|B2) = .1, P(B1|A) =?

)()|()()|(

)()|()|(

2211

111BPBAPBPBAP

BPBAP

ABP

!

8.25.

2.

)5)(.1(.)5)(.4(.

)5)(.4(.!!

!


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Event PriorProb

Cond.Prob

JointProb

Post.Prob

Bi P(Bi) P(A|Bi) P(Bi A) P(Bi |A)

B1 .5 .4 .20 .20/.25 = .8

B2 .5 .1 .05 .05/.25 = .2

1.0 P(A) = 0.25 1.0

Bayes Theorem Example

Table Solution

DefaultRepay P(College)

X =


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Random Testing for AIDS?

In 1987, the US Secretary of Health proposedto test blood to estimate how many Americans,and which ones, were infected with AIDS.Continuing calls have been made for

mandatory testing to identify persons with thedisease. These calls met with considerableresistance from statisticians, who speak out

against the mindless proposals for mandatoryAIDS testing....

Refer to Random Testing for AIDS? Chance,

vol. 1, no. 1, 1988


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Permutation and Combination

Counting Rule 1

ExampleIn TV Series: Kangxi Empire,Shilang tossed 50 coins, the number of

outcomes is 22 2 250

. What is theprobability of all coins with heads up?

If any one ofn different mutually exclusive

and collectively exhaustive events can occuron each ofrtrials, the number of possible

outcomes is equal tonnn nr


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Application: Lottery Post Card (Post ofChina)Six digits in each group (2000 version)

Winning the first prize: No.035718

Winning the 5th prize: ending with number3

If there are k1

events on the first trial, k2

events

on the second trial, and krevents on the rth

trial, then the number of possible outcomes

is: k1k

2k

r


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Applicatin: If a license plate consists of3 letters

followed by 3 digits, the total number of

outcomes would be? (most states in the US)

Application: China License PlatesHow many licenses can be issued?

Style 1992: one letter or digit plus 4 digits.

Style 2002: 1) three letters + three digits2) three digits + three letters

3) three digits + three digits


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Counting Rule 2

Example: The number of ways that 5 books could

be arranged on a shelf is: (5)(4)(3)(2)(1) = 120

The number of ways that all n objects can bearranged in orderis

=n

(n

-1)(n

-2)~

(2)(1) =n!

Where n! is called factorial and 0! isdefined as 1.

n

n

P


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Counting Rule 3: Permutation

ExampleWhat is the number of ways of

arranging 3books selected from 5books in

order? (5)(4)(3) = 60

The number of ways of arranging robjectsselected from nobjects in order is

)!(!

rn

nP

n

r

!


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Counting Rule 4: CombinationExampleThe number ofcombinations of3 books

selected from 5 books is

(5)(4)(3)/[(3)(2)(1)] = 10

Note: 3! possible arrangements in order are irrelevant

The number of ways that arranging robjects selectedfrom nobjects, irrespective of the order, is equal to

)!(!

!

rnr

n

r

nC

n

r

!

!


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Examples

Applications: NBA Draft (2002)Randomly choose 4 numbers from 14 numbers

Example:1. World Cup 2002: how many games should a

soccer team play in a group of four teams?

2. There are 10 cities, how many different non-stop tickets? How many different prices ofthe tickets?

NBA Lottery.lnk

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