Chap002 Probability

8/6/2019 Chap002 Probability

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Chapter 2Chapter 2ProbabilityProbability

COMPLETE

BUSINESSSTATISTICS

byby

AMIR D. ACZELAMIR D. ACZEL

&&

JAYAVEL SOUNDERPANDIANJAYAVEL SOUNDERPANDIAN7th edition.7th edition.

Prepared byPrepared by Lloyd Jaisingh, Morehead StateLloyd Jaisingh, Morehead State

UniversityUniversity

McGraw-Hill/Irwin Copyright 2009 by The McGraw-Hill Companies, Inc. All rights reserved.


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Using Statistics

Basic Definitions: Events, Sample Space, and Probabilities

Basic Rules for Probability

Conditional Probability

Independence of Events Combinatorial Concepts

The Law of Total Probability and Bayes Theorem

The Joint Probability Table

Using the Computer

ProbabilityProbability22

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Define probability, sample space, and event.

Distinguish between subjective and objective probability.

Describe the complement of an event, the intersection, and the union of twoevents.

Compute probabilities of various types of events.

Explain the concept of conditional probability and how to compute it.

Describe permutation and combination and their use in certain probability

computations.

Explain Bayes theorem and its applications.

LEARNINGOBJECTIVES

After studying this chapter, you should be able to:After studying this chapter, you should be able to:

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2-1 Probability is:

A quantitative measure ofuncertainty

A measure of the strength of beliefin the occurrence of anuncertain event

A measure of the degree ofchance or likelihood ofoccurrence of an uncertain event

Measured by a number between 0 and 1 (or between 0% and100%)

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Types of Probability

Objective or Classical Probability

based on equally-likely events

based on long-run relative frequency of events

not based on personal beliefs is the same for all observers (objective)

examples: toss a coin, roll a die, pick a card

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Types of Probability (Continued)

Subjective Probability

based on personal beliefs, experiences, prejudices, intuition - personal

judgment

different for all observers (subjective) examples: Super Bowl, elections, new product introduction, snowfall

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Set - a collection of elements or objects of interest

Empty set (denoted by )

a set containing no elements

Universal set (denoted by S) a set containing all possible elements

Complement (Not). The complement ofA is

a set containing all elements of S not in A

A

2-2 Basic Definitions

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Complement of a Set

A

A

S

Venn Diagram illustrating the Complement of an event

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Intersection (And)

a set containing all elements in both A and B Union (Or)

a set containing all elements in A or B or both

A B

A B

Basic Definitions (Continued)

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A B

Sets: A Intersecting with B

AB

S

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Sets: A Union B

A B

AB

S

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Mutually exclusive or disjoint sets

sets having no elements in common, having no

intersection, whose intersection is the empty set

Partitiona collection of mutually exclusive sets which

together include all possible elements, whoseunion is the universal set

Basic Definitions (Continued)

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Mutually Exclusive or Disjoint Sets

AB

S

Sets have nothing in common

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Sets: Partition

A1

A2

A3

A4

A5

S

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Process that leads to one of several possible outcomes *, e.g.: Coin toss

Heads, Tails

Rolling a die

1, 2, 3, 4, 5, 6

Pick a card AH, KH, QH, ...

Introduce a new product

Each trial of an experiment has a single observed outcome.

The precise outcome of a random experiment is unknown before a trial.

* Also called a basic outcome, elementary event, or simple event

Experiment

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Sample Space or Event Set

Set of all possible outcomes (universal set) for a given experiment

E.g.: Roll a regular six-sided die

S = {1,2,3,4,5,6}

Event

Collection of outcomes having a common characteristic

E.g.: Even number

A = {2,4,6}

Event A occurs if an outcome in the set A occurs

Probability of an event

Sum of the probabilities of the outcomes of which it consists P(A) = P(2) + P(4) + P(6)

Events : Definition

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For example: Roll a die

Six possible outcomes {1,2,3,4,5,6}

If each is equally-likely, the probability of each is 1/6 = 0.1667 =

16.67%

Probability of each equally-likely outcome is 1 divided by the number ofpossible outcomes

Event A (even number)

P(A) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 = 1/2 for e in AP A P e

n A

n S

( ) ( )

( )

( )

!

! ! !

3

6

1

2

P en S

( )( )

!1

Equally-likely Probabilities

(Hypothetical or Ideal Experiments)

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Pick a Card: Sample Space

Event AceUnion of

Events Heart

and Ace

Event Heart

The intersection of theevents Heart and Ace

comprises the single point

circled twice: the ace of hearts

P Heart Ace

n Heart Ace

n S

( )

( )

( )

7

7

!

!

!

16

52

4

13

P Heart

n Heart

n S

( )

( )

( )

! ! !13

52

1

4

P Ace

n Ace

n S

( )

( )

( )

! ! !

4

52

1

13

P Heart Ace

n Heart Ace

n S

( )

( )

( )

+

+

! !1

52

Hearts Diamonds Clubs Spades

A A A A

K K K K

Q Q Q Q

J J J J

10 10 10 10

9 9 9 98 8 8 8

7 7 7 7

6 6 6 6

5 5 5 5

4 4 4 4

3 3 3 3

2 2 2 2

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Range of Values for P(A):

Complements - Probability ofnotA

Intersection - Probability of both A andB

Mutually exclusive events (A and C) :

1)(0 ee AP

P A P A( ) ( )! 1

P A Bn A B

n S( )

( )( )

!

P A C( ) ! 0

2-3 Basic Rules for Probability

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Union - Probability ofAorB orboth (rule of unions)

Mutually exclusive events: IfA and B are mutually exclusive, then

P A Bn A B

n SP A P B P A B( )

( )( )

( ) ( ) ( ) !

!

)()()(0)( BPAPBAPsoBAP !!

Basic Rules for Probability

(Continued)

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Sets: P(A Union B)

)( BAP

AB

S

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Rules of conditional probability:

If events A and D are statistically independent:

so

so

P A BP A B

P B( )( )

( )!

P A B P A B P BP B A P A

( ) ( ) ( )( ) ( )

!!

P A D P A

P D A P D

( ) ( )

( ) ( )

!

!

)()()( DPAPDAP !

Conditional Probability (continued)

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AT&T IBM Total

Telecommunication 40 10 50

Computers 20 30 50

Total 60 40 100

Counts

AT&T IBM

Total

Telecommunication 0.40 0.10 0.50

Computers 0.20 0.30 0.50

Total 0.60 0.40 1.00

Probabilities

2.050.0

10.0

)(

)(

)(

!!

! TP

TIBMP

TIBMP

+

Probability that a project

is undertaken by IBMgiven it is a

telecommunications

project:

Contingency Table - Example 2-2

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P A B P A

P B A P B

and

P A B P A P B

( ) ( )

( ) ( )

( ) ( ) ( )

!

!

!+

Conditions for the statistical independence of events A and B:

P A ce HeartP A ce Heart

P Heart

P A ce

( )( )

( )

( )

!

! ! !

+

1

5213

52

1

13

P HeartA ceP Heart A ce

P A ce

P Heart

( )( )

( )

( )

!

! ! !

+

1

524

52

1

4

)()(52

1

52

13*

52

4)( HeartPAcePHeartAceP !!!+

2-5 Independence of Events

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0976.00024.006.004.0

)()()()()

0024.006.0*04.0

)()()()

!!

!

!!

!

BTPBPTPBTPb

BPTPBTPa

+

+

Events Television (T) and Billboard(B) are

assumed to be independent.

Independence of Events

Example 2-5

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The probability of the union of several independent events

is 1 minus the product of probabilities of their complements:

P A A A An P A P A P A P An( ) ( ) ( ) ( ) ( )1 2 31

1 2 3 ! . .

Example 2-7:

6513.03487.011090.01

)10

()3

()2

()1

(1)10321

(

!!!

! QPQPQPQPQQQQP ..

The probability of the intersection of several independent events

is the product of their separate individual probabilities:

P A A A An P A P A P A P An( ) ( ) ( ) ( ) ( )

1 2 3 1 2 3

!. .

Product Rules for Independent Events

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Consider a pair of six-sided dice. There are six possible outcomes

from throwing the first die {1,2,3,4,5,6} and six possible outcomes

from throwing the second die {1,2,3,4,5,6}. Altogether, there are

6*6 = 36 possible outcomes from throwing the two dice.

In general, if there are n events and the event ican happen in

Ni possible ways, then the number of ways in which the

sequence ofn events may occur isN1N2...Nn.

Pick 5 cards from a deck of 52 -

with replacement 52*52*52*52*52=525 380,204,032

different possible outcomes

Pick 5 cards from a deck of 52 -

without replacement 52*51*50*49*48 = 311,875,200

different possible outcomes

2-6 Combinatorial Concepts

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...

. .Order the letters: A, B, and CA

B

C

B

C

A

B

A

C A

C

B

C

B

A

. ....

......

ABC

ACB

BAC

BCA

CAB

CBA

More on Combinatorial Concepts

(Tree Diagram)

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How many ways can you order the 3 lettersA,B, and C?

There are 3 choices for the first letter, 2 for the second, and 1 for

the last, so there are 3*2*1 = 6 possible ways to order the three

letters A, B, and C.

How many ways are there to order the 6 lettersA,B, C,D, E,

and F? (6*5*4*3*2*1 = 720)

Factorial: For any positive integern, we define n factorialas:

n(n-1)(n-2)...(1). We denote n factorial as n!.

The numbern! is the number of ways in which n objects can

be ordered. By definition 1! = 1 and 0! = 1.

Factorial

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Permutations are the possible ordered selections ofrobjects out

of a total ofn objects. The number of permutations ofn objects

taken rat a time is denoted by nPr, where

What if we chose only 3 out of the 6 lettersA,B, C,D, E, and F?

There are 6 ways to choose the first letter, 5 ways to choose the

second letter, and 4 ways to choose the third letter (leaving 3

letters unchosen). That makes 6*5*4=120 possible orderings or

permutations.

1204*5*61*2*3

1*2*3*4*5*6

!3

!6

)!36(

!6

:

36 !!!!

!

!

P

exampleFor

rnn

rP

n )!(!

Permutations (Order is important)

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Combinations are the possible selections ofritems from a group ofn itemsregardless of the orderof selection. The number of combinations is denotedand is read as n choose r. An alternative notation is nCr. We define the numberof combinations of r out of n elements as:

Suppose that when we pick 3 letters out of the 6 letters A, B, C, D, E, and Fwe chose BCD, or BDC, or CBD, or CDB, or DBC, or DCB. (These are the6 (3!) permutations or orderings of the 3 letters B, C, and D.) But these areorderings of the same combination of 3 letters. How many combinations of 6different letters, taking 3 at a time, are there?

206

120

1*2*3

4*5*6

1)*2*1)(3*2*(3

1*2*3*4*5*6

!3!3

!6

)!36(!3

!6

:

36 !!!!!

!!

!!

C

exampleor

r

n

r)!(nr!

n!C

r

nrn

n

r

Combinations (Order is not Important)

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Example: Template for Calculating

Permutations & Combinations

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P A P A B P A B( ) ( ) ( )!

In terms of conditional probabilities:

More generally (where Bi make up a partition):

P A P A B P A B

P A B P B P A B P B

( ) ( ) ( )

( ) ( ) ( ) ( )

!

!

P A P A Bi

P A Bi

P Bi

( ) ( )

( ) ( )

!

!

2-7 The Law ofTotal Probability and

Bayes Theorem

The law of total probability:

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Event U: Stock market will go up in the next year

Event W: Economy will do well in the next year

66.06.60.)20)(.30(.)80)(.75(.

)()()()(

)()()(

2.8.1)(80.)(

30)(75.)(

!!

!!

!

!!!

!!

WPWUPWPWUP

WUPWUPUP

WPWP

WUPWUP

The Law ofTotal Probability-

Example 2-9

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Bayes theorem enables you, knowing just a little more than theprobability ofA given B, to find the probability of B given A.

Based on the definition of conditional probability and the law of totalprobability.

P B AP A B

P A

PA B

P A B P A B

P A B P B

P A B P B P A B P B

( )( )

( )

( )( ) ( )

( ) ( )

( ) ( ) ( ) ( )

!

!

!

+

+

+ +

Applying the law of total

probability to the denominator

Applying the definition of

conditional probability throughout

Bayes Theorem

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A medical test for a rare disease (affecting 0.1% of the population []) is imperfect:

When administered to an ill person, the test will indicate so with probability

0.92 [ ]

The event is a false negative

When administered to a person who is not ill, the test will erroneously give a

positive result (false positive) with probability 0.04 [ ]

The event is a false positive. .

P I( ) .! 0 001

08.)(92.)( !! IZPIZP

)( IZ

)( IZ96.0)(04.0)( !! IZPIZP

Bayes Theorem - Example 2-10

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P I

P I

P Z I

P Z I

( ) .

( ) .

( ) .

( ) .

!

!

!

!

0001

0999

0 92

0 04

P I ZP I Z

P Z

P I Z

P I Z P I Z

P Z I P I

P Z I P I P Z I P I

( )( )

( )

( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

(. )( . )

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

.

. .

.

.

.

!

!

!

!

!

!

!

+

+

+ +

92 0 0 01

9 2 0 00 1 0 0 4 99 9

0 0 0 0 9 2

0 0 00 92 0 03 99 6

0 0 0 0 9 2

0 4 0 8 8

0 2 2 5

Example 2-10 (continued)

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P I( ) .! 0001

P I( ) .! 0999 P Z I( ) .! 004

P Z I( ) .! 096

P ZI( ) .! 008

P ZI( ) .! 092 P Z I( ) ( . )( . ) .+ ! !0 001 0 92 00092

P Z I( ) ( . )( . ) .+ ! !0 001 0 08 00008

P Z I( ) ( . )( . ) .+ ! !0 999 0 04 03996

P Z I( ) ( . )( . ) .+ ! !0 999 0 96 95904

Prior

Probabilities

Conditional

Probabilities

Joint

Probabilities

Example 2-10 (Tree Diagram)

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Given a partition of events B1,B2 ,...,Bn:

P B A

P A B

P A

P A B

P A B

P AB P B

P AB P B

i

i i

( )

( )

( )

( )

( )

( ) ( )

( ) ( )

1

1

1

1 1

!

!

!

Applying the law of total

probability to the denominator

Applying the definition of

conditional probability throughout

Bayes Theorem Extended

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An economist believes that during periods of high economic growth, the U.S.

dollar appreciates with probability 0.70; in periods of moderate economic

growth, the dollar appreciates with probability 0.40; and during periods of

low economic growth, the dollar appreciates with probability 0.20.

During any period of time, the probability of high economic growth is 0.30,the probability of moderate economic growth is 0.50, and the probability of

low economic growth is 0.50.

Suppose the dollar has been appreciating during the present period. What is

the probability we are experiencing a period of high economic growth?

Partition:H - High growth P(H) = 0.30

M - Moderate growth P(M) = 0.50

L - Low growth P(L) = 0.20

Event A Appreciation!!

!

P AHP A MP A L

( ) .( ) .( ) .

0 700 40

0 20

Bayes Theorem Extended -

Example 2-11

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P H AP H A

P A

P H A

P H A P M A P L A

P A H P H

P A H P H P A M P M P A L P L

( )( )

( )( )

( ) ( ) ( )( ) ( )

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

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

. . .

.

..

!

!

!

!

! !

!

+

+

+ + +

0 70 030

0 70 030 0 40 050 0 20 0 200 21

0 21 0 20 0 04

0 21

0 450467

Example 2-11 (continued)

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Prior

Probabilities

Conditional

Probabilities

Joint

Probabilities

P H( ) .! 0 30

P M( ) .! 0 50

P L( ) .! 0 20

P A H( ) .! 0 70

P A H( ) .! 0 30

P A M( ) .! 0 40

P A M( ) .! 0 60

P A L( ) .! 0 20

P A L( ) .! 0 80

P A H( ) ( . )( . ) .+ ! !0 30 0 70 0 21

P A H( ) ( . )( . ) .+ ! !0 30 0 30 0 09

P A M( ) ( . )( . ) .+ ! !0 50 0 40 0 20

P A M( ) ( . )( . ) .+ ! !0 50 0 60 0 30

P A L( ) ( . )( . ) .+ ! !0 20 0 20 0 04

P A L( ) ( . )( . ) .+ ! !0 20 0 80 0 16

Example 2-11 (Tree Diagram)

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2-8 The Joint Probability Table

Ajoint probability table is similar to a contingency table , except that it

has probabilities in place of frequencies.

The joint probability for Example 2-11 is shown below.

The row totals and column totals are called marginal probabilities.

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The Joint Probability Table

Ajoint probability table is similar to a contingency table , except that it

has probabilities in place of frequencies.

The joint probability for Example 2-11 is shown on the next slide.

The row totals and column totals are called marginal probabilities.

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The Joint Probability Table:

Example 2-11

The joint probability table for Example 2-11 is summarized

below.

High Medium Low TotalTotal

$ Appreciates 0.21 0.2 0.04 0.45

$Depreciates 0.09 0.3 0.16 0.55

TotalTotal 0.30 0.5 0.20 1.00

Marginal probabilities are the row totals and the column totals.

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2-8 Using Computer: Template for Calculating

the Probability of at least one success

2 47

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2-8 Using Computer: Template for Calculating

the Probabilities from a Contingency

Table-Example 2-11

2 48

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2-8 Using Computer: Template for Bayesian

Revision of Probabilities-Example 2-11

9

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2-8 Using Computer: Template for Bayesian

Revision of Probabilities-Example 2-11

Continuation of output from previousslide.

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Chap002 Probability