Complete Business Statistics: Probability

COMPLETE BUSINESS STATISTICS by AMIR D. ACZEL & JAYAVEL SOUNDERPANDIAN 7th edition. Prepared by Lloyd Jaisingh, Morehead State University

Chapter 2 Probability

McGraw-Hill/Irwin

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

2-2

2 Probability         

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

2-3

2

LEARNING OBJECTIVES

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



 



Define probability, sample space, and event. Distinguish between subjective and objective probability. Describe the complement of an event, the intersection, and the union of two events. 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.

2-4

2-1 Probability is:  





A quantitative measure of uncertainty A measure of the strength of belief in the occurrence of an uncertain event A measure of the degree of chance or likelihood of occurrence of an uncertain event Measured by a number between 0 and 1 (or between 0% and 100%)

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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

2-6

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

2-7

2-2 Basic Definitions 

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 of A is  a set containing all elements of S not in A

 A

2-8

Complement of a Set

S

A A

Venn Diagram illustrating the Complement of an event

2-9

Basic Definitions (Continued)





Intersection (And)  A  B – a set containing all elements in both A and B Union (Or)    AB    – a set containing all elements in A or B or both

2-10

Sets: A Intersecting with B S

A B

A B

2-11

Sets: A Union B

S

A B

A B

2-12

Basic Definitions (Continued) • Mutually exclusive or disjoint sets –sets having no elements in common, having no •

intersection, whose intersection is the empty set Partition –a collection of mutually exclusive sets which together include all possible elements, whose union is the universal set

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Mutually Exclusive or Disjoint Sets Sets have nothing in common S

A

B

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Experiment • Process that leads to one of several possible outcomes *, e.g.: 

Coin toss •



Rolling a die •



Heads, Tails 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

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Events : Definition 

Sample Space or Event Set 

Set of all possible outcomes (universal set) for a given experiment  E.g.: Roll a regular six-sided die 



Event 

Collection of outcomes having a common characteristic 

E.g.: Even number 

 

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

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)

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Sets: Partition IU STUDENTS= 1000

S EE (50)

IT (50) BA (550)

BME (100) BT

(250)

GIVEN BA=550 EE=50 BT=250 BME=100 IT=50  Total IU=1000

Event: Pick one “BA” Student from 1000 IU students P ( BA )  

 P ( e) n( BA ) 550   0.55 n( IU ) 1000

Event: Pick one “non-BA” Student from 1000 IU students n( non  BA ) n( BT )  n( EE )  n( IT )  n( BME )  n( IU ) n( IU ) 250  50  50  100 450 550 p ( BA )    1  1  0.45 1000 1000 1000 1000 P ( non  BA ) 

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Equally-likely Probabilities (Hypothetical or Ideal Experiments) • 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%

 P ( e)  • 

1 n( S )

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

Event A (even number)

• •

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

n( A) 3 1   n( S ) 6 2

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2-3 Basic Rules for Probability 

Range of Values for P(A):



Complements - Probability of not A

0  P( A) 1

P( A )  1  P( A) 

Intersection - Probability of both A and B

P( A  B)  n( A  B) n( S ) 

Mutually exclusive events (A and C) :

P( A  C )  0

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Pick a Card: Sample Space Hearts

Diamonds

Clubs

A K Q J 10 9 8 7 6 5 4 3 2

A K Q J 10 9 8 7 6 5 4 3 2

A K Q J 10 9 8 7 6 5 4 3 2

Event ‘Heart’ n ( Heart ) P ( Heart ) 

13 

n(S )

1 

52

Spades A K Q J 10 9 8 7 6 5 4 3 2

Event ‘Ace’ n ( Ace ) P ( Ace ) 

4

1

 n(S )

 52

13

The intersection of the events ‘Heart’ and ‘Ace’ comprises the single point circled twice: the ace of hearts

4

n ( Heart  Ace ) P ( Heart  Ace ) 

1 

n(S )

52

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Pick a Card: Sample Space Union of Events ‘Heart’ And/Or ‘Ace’ P ( Heart  Ace )  n ( Heart  Ace )  n(S ) 16

4 

52

13

Hearts

Diamonds

Clubs

A K Q J 10 9 8 7 6 5 4 3 2

A K Q J 10 9 8 7 6 5 4 3 2

A K Q J 10 9 8 7 6 5 4 3 2

Spades A K Q J 10 9 8 7 6 5 4 3 2

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Basic Rules for Probability (Continued) •

Union - Probability of A or B or both (rule of unions)

P( A  B)  n( A  B)  P( A)  P( B)  P( A  B) n( S )  Mutually exclusive events: If A and B are mutually exclusive, then

P( A  B)  0 so P( A  B)  P( A)  P( B)

Example

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Sets: P(A Union B)

S

A B

P( A  B)

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Contingency Table - Example 2-2 Counts AT& T

IBM

Total

Telecommunication

40

10

50

Computers

20

30

50

Total

60

40

100

Probabilities 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

Probability that a project is undertaken by IBM given it is a telecommunications project:

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2-4 Conditional Probability •

Conditional Probability - Probability of A given B

P( A B) 

P( A  B) , where P( B)  0 P( B)

 Independent events:

P( A B)  P( A) P( B A)  P( B)

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Contingency Table - Example 2-2 Counts AT& T

IBM

Total

Telecommunication

40

10

50

Computers

20

30

50

Total

60

40

100

Probabilities 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

Probability that a project is undertaken by IBM given it is a telecommunications project: P ( IBM  T ) P ( IBM T )  P (T ) 0.10   0.2 0.50

Problem A club has 35 play chess, 58 play bridge ; in which 27 play both chess and bridge. If a member of the club is randomly chosen, what is the conditional probability that she

(a)Play chess given that she plays bridge; (b)Plays bridge given that she plays chess

Let : A be the event of playing chess B be the event of playing bridge N be total of members in the club P(a) = 35/n P(b) = 58/n P(a∩b) = 27/n

(a)the conditional probability that she plays chess given that she plays bridge: P(A | B)    (b)The

P( A  B) P( B)

27 / N  27 / 58 58 / N

conditional probability that she plays bridge given that she plays chess: P(B | A)   

P( A  B) P ( A)

27 / N  27 / 35 35 / N

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Conditional Probability (continued) Rules of conditional probability: P( A B)  P( A  B) so P( A  B)  P( A B) P( B) P( B)  P( B A) P( A)

What is the probability that the number 6 will turn up? The answer is 1/6

Now suppose that I told you that I just tossed a coin and it turned up heads

What is now the probability that the die will show the number 6? The answer is unchanged 1/6 Two events of the die and the coin are independent of each other.

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2-5 Independence of Events Conditions for the statistical independence of events A and B: P( A B)  P( A) P( B A)  P( B) and P( A  B)  P( A) P( B) P ( Ace  Heart ) P ( Heart ) 1 1  52   P ( Ace ) 13 13 52

P ( Ace Heart ) 

P( Ace  Heart) 

P ( Heart  Ace ) P ( Ace ) 1 1  52   P ( Heart ) 4 4 52

P ( Heart Ace ) 

4 13 1 *   P( Ace) P( Heart) 52 52 52

Event “Ace” and Event “ Heart” are independent

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Product Rules for Independent Events 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

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Set (A,B,C,D,E,F) The first card taken

P(A)=1/6

The second card taken If we take twice, how many percent both times we take “A”?

1/6 x 1/6

P(A)=1/6

If we take twice, how many percent at least once we take “A”? Ax non.A + non.Ax A+ AxA P(A)xP( A)  P(A)xP(A)  P(A)xP(A)  1/6 x5/6  5/6x1/6  1/6x1/6  11/36

1- non.A x non.A 1 - P(A) x P(A)  1 - 5/6x5/6  11/36

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Product Rules for Independent Events 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

The probability of the union of several independent events is 1 minus the product of probabilities of their complements: P( A  A  A  An )  1 P( A ) P( A ) P( A ) P( An ) 1 2 3 1 2 3

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2-6 Combinatorial Concepts 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 i can happen in Ni possible ways, then the number of ways in which the sequence of n events may occur is N1N2...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

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More on Combinatorial Concepts (Tree Diagram)

. . .. . . . . . .

Order the letters: A, B, and C

C

B

C

B

A C

C

A

B C

A B

A B

A

. .. .. .

ABC ACB

BAC BCA CAB CBA

2-43

Factorial (Bao nhiêu cách sắp xếp) How many ways can you order the 3 letters A, 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 letters A, B, C, D, E, and F? (6*5*4*3*2*1 = 720) Factorial: For any positive integer n, we define n factorial as: n(n-1)(n-2)...(1). We denote n factorial as n!. The number n! is the number of ways in which n objects can be ordered. By definition 1! = 1 and 0! = 1.

Set (A,B,C) . How many ways choosing 2 out 3 letters? The order is concerned AB BA

AC CA BC CB

6 =3x2

=

3x2x1 1

=

3! (3-2)!

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Permutations (Order is important) What if we chose only 3 out of the 6 letters A, 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.

Permutations are the possible ordered selections of r objects out of a total of n objects. The number of permutations of n objects taken r at a time is denoted by nPr, where

n! P  n r (n  r )!

Forexam ple: 6! 6! 6 * 5 * 4 * 3 * 2 *1    6 * 5 * 4  120 6 P3  (6  3)! 3! 3 * 2 *1

Set (A,B,C) . How many ways choosing 2 out 3 letters? The order is NOT concerned AB BA

(A,B)

AC CA

(A,C)

BC CB

(C,B)

3 =

6 = 2

3x2x1 2X1

=

3! 2! (3-2)!

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Combinations (Order is not Important) Suppose that when we pick 3 letters out of the 6 letters A, B, C, D, E, and F we chose BCD, or BDC, or CBD, or CDB, or DBC, or DCB. (These are the 6 (3!) permutations or orderings of the 3 letters B, C, and D.) But these are orderings of the same combination of 3 letters. How many combinations of 6 different letters, taking 3 at a time, are there?

Combinations are the possible selections of r items from a group of n items  n regardless of the order of selection. The number of combinations is denoted  r and is read as n choose r. An alternative notation is nCr. We define the number of combinations of r out of n elements as:  n n!   n C r  r! (n  r)! r Forexam ple:  n 6! 6! 6 * 5 * 4 * 3 * 2 *1 6 * 5 * 4 120  6 C3       20 r 3 ! ( 6  3 )! 3 ! 3 ! (3 * 2 * 1)(3 * 2 * 1) 3 * 2 * 1 6  

2-51

Example: Template for Calculating Permutations & Combinations

2-53

2-7 The Law of Total Probability and Bayes’ Theorem The law of total probability: P( A)  P( A  B)  P( A  B )

In terms of conditional probabilities: P( A)  P( A  B)  P( A  B )  P( A B) P( B )  P( A B ) P( B )

More generally (where Bi make up a partition): P( A)   P( A  B ) i   P( AB ) P( B ) i i

2-55

Solution Event U: Stock market will go up in the next year Event W: Economy will do well in the next year P(U W ) .75 P(U W )  30 P(W ) .80  P(W )  1.8 .2 P(U )  P(U W )  P(U W )  P(U W ) P(W )  P(U W ) P(W )  (.75)(.80)  (.30)(.20) .60.06 .66

2-56

Bayes’ Theorem • Bayes’ theorem enables you, knowing just a little more than the

•

probability of A given B, to find the probability of B given A. Based on the definition of conditional probability and the law of total probability.

P ( A  B) P ( A) P ( A  B)  P ( A  B)  P ( A  B ) P ( A B) P ( B)  P ( A B) P ( B)  P ( A B ) P ( B )

P ( B A) 

Applying the law of total probability to the denominator

Applying the definition of conditional probability throughout

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

P ( I )  0.001

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

0.92 [ ] P(Z I ) .92  P(Z I ) .08  The event ( Z I ) 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 ( Z I ) is a false positive.

.

P(Z I )  0.04  P(Z I )  0.96

2-59

solution P ( I )  0.001 P ( I )  0.999 P ( Z I )  0.92 P ( Z I )  0.04

P( 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 )

P( I Z ) 

(.92)( 0.001) (.92)( 0.001)  ( 0.04)(.999) 0.00092 0.00092   0.00092  0.03996 .04088 .0225



2-60

Example 2-10 (Tree Diagram) Prior Probabilities

Conditional Probabilities P( Z I )  0.92

P( I )  0001 .

P( I )  0999 .

P( Z I )  008 .

P( Z I )  004 .

Joint Probabilities 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 )  096 . P( Z  I )  (0.999)(0.96) .95904

2-63

Bayes’ Theorem Extended •

Given a partition of events B1,B2 ,...,Bn: P( A  B ) P ( B A)  P( A) P( A  B )   P( A  B ) 1

1

Applying the law of total probability to the denominator

1

i

P( A B ) P( B )   P( A B ) P( B ) 1

1

i

i

Applying the definition of conditional probability throughout

Bayes’ Theorem Extended Example 2-11 





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( A H )  0.70 P( A M )  0.40 P( A L)  0.20

2-64

2-65

Example 2-11 (continued) P( H  A) P( 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)( 0.30)  ( 0.70)( 0.30)  ( 0.40)( 0.50)  ( 0.20)( 0.20) 0.21 0.21   0.21 0.20  0.04 0.45  0.467

2-66

Example 2-11 (Tree Diagram) Prior Probabilities

Conditional Probabilities P ( A H )  0.70

P ( H )  0.30

P ( A H )  0.30

P ( A M )  0.40

Joint Probabilities 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 ( M )  0.50

P ( A M )  0.60 P ( A  M )  ( 0.50)( 0.60)  0.30 P ( L)  0.20

P ( A L)  0.20

P ( A L)  0.80

P ( A  L)  ( 0.20)( 0.20)  0.04

P ( A  L)  ( 0.20)( 0.80)  0.16

2-67

2-8 The Joint Probability Table 

 

A joint 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.

2-68

The Joint Probability Table 

 

A joint 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.

Monty Hall Experiment

7 0

Monty Hall Experiment

7 D A 1 Y 3 . D r . L i s a B j ö r k . E x p e r i m e n t a l E c o n o m i c s : g a

Monty Hall Experiment

Remember

After 10 rounds 

Please take the picture of the result on screen after playing 10 rounds

2-73

The Joint Probability Table: Example 2-11 

The joint probability table for Example 2-11 is summarized below. High

Medium

Low

Total

$ Appreciates

0.21

0.2

0.04

0.45

$Depreciates

0.09

0.3

0.16

0.55

Total

0.30

0.5

0.20

1.00

Marginal probabilities are the row totals and the column totals.

2-74

2-8 Using Computer: Template for Calculating the Probability of at least one success

2-75

2-8 Using Computer: Template for Calculating the Probabilities from a Contingency Table-Example 2-11

2-76

2-8 Using Computer: Template for Bayesian Revision of Probabilities-Example 2-11

2-77

2-8 Using Computer: Template for Bayesian Revision of Probabilities-Example 2-11

Continuation of output from previous slide.