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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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Permutation and Combination
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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Permutation and Combination
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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Permutation and Combination
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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Permutation and Combination
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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Permutation and Combination
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