Chapter 2: a Review of Probability Theory

Quantitative Analysis – MGMT 322

EASTERN MEDITERRANEAN UNIVERSITY

MGMT 322

QUANTITATIVE ANALYSIS

Lecture Notes

Prof. Dr. Serhan ÇİFTÇİOĞLU

QUANTITATIVE ANALYSIS

Lecture Notes,

Past Quiz, Exam, & Home Work

Questions

Prof. Dr. Serhan ÇİFTÇİOĞLU

Gazimağusa 2007

1

Quantitative Analysis – MGMT 322

Foreword

These “Lecture Notes” have been prepared to serve as a study guide for the students of “Quantitative Analysis” course (MGMT 322), that I offer for business students. They are designed to outline the critical topics that are covered by the course but also attempt to give an example about the application of each quantitative and statistical tool covered by the course. At the end of these “Lecture Notes” I also added examples of past quizzes, exams, and homework questions, to help interested students both to improve their analytical capacity regarding the application of the quantitative tools, that they have learn, and at the same time better prepare for exams.

I would like to express my gratitude for the excellent work that our teaching assistants and former students of MGMT 322, Meltem İKİNCİ, Erhan ERDOĞAN and Süleyman EFE, and Gift, have done in organizing and typing my in-class lectures.

Assoc. Prof. Dr. Serhan ÇİFTÇİOĞLU

Department of Business Administration

Eastern Mediterranean University

INDEX

Chapter 2: A Review of Probability Theory

A. Probability

B. Event

Example

C. Experiment

D. Sample Space

Example

E. Mutually Exclusive Events

F. NON- MUTUALLY EXCLUSIVE EVENTS…………………….……………………….- 5 -

Example

Alternative Approaches in Probability Theory

1. Classical Approach

Example

2. Relative Frequency Approach

Example

3. Subjective Approach

Example

Probability Rules

A. Marginal (Unconditional) Probability:

B. The Addition Rule:

1. The Addition Rule for Mutually Exclusive Events

Example

2. The Addition Rule for Non-Mutually Exclusive Events

Example

Example

Types of Events

1. Statistically Independent Events

2. Statistically Dependent Events

Types of Probabilities under Independence

1. Marginal (Unconditional) Probability

2. Joint Probability

3. Conditional Probability

Types of Probabilities under Dependence

1. Marginal (Unconditional) Probability

2. Joint Probability

3. Conditional Probability

Bayes’ Theorem

Example

Example

Example

Probability Distribution

Discrete VS Continuous Random Variables...... 17

Discrete Probability Distribution...... 17

Continuous Probability Distribution...... 17

probability distribution of a random variable………...... 18 EXAMPLE...... 19 EXPECTED VALUE OF A RANDOM VARIABLE...... 20

Binomial Probability Distribution

Characteristics of a Bernoulli Type Process

Binomial Formula:

Example

Example

Normal Probability Distribution

Characteristics

Two Parameters: Mean (μ) & Standard Deviation (σ)

Example:

3 Mathematical Facts about Normal Distribution

Example:

Example:

Example:

CHAPTER 3: FORECASTING...... -32-

The Methodology of Forecasting

technical analysis

FUNDAMENTAL ANALYSIS

BASIC STEPS IN FORECASTING

Forecasting Models

For Non-Seasonal Products with Linear Trend Type of Pattern

Naïve Model

moving average model……………………………………………………………………. . -41-

Simple Exponential Model

Important Steps

Time Series Regression Model

Important Points

Smoothing Linear Trend Model

Important Steps

chapter 4: Alternative Decision Making Environments

basic steps in decision making

alternative decision making environments

Alternative Criteria for Decision Making under Uncertainty

The Maxi-Max Criterion

The Maxi-Min Criterion

The Criterion of Realism

The Mini-Max Regret Criterion

Alternative Criteria for Decision Making under Risk

Expected Value Criterion

Example

Criterion of Rationality

The Criterion of Maximum Likelihood

Supplying the Numbers: How to Obtain Mean and Standard Deviation when Data is Missing or Incomplete

Example:

Example:

Chapter 5: Cost – Volume – Profit Analysis

general definition and use of cost- volume- profit analysis

its application

Combining Unit Monetary Values and Probability Distribution

3 – Step Procedure to Obtain Expected Profits

3 – Step Procedure to Obtain Expected Loss

Home Works………………………………………………………………….-72-

Past Quiz Questions………………………………………………………..-78-

Past Exam Questions...... -87-

Chapter 2: A Review of Probability Theory

A. Probability

Probability is the chance or likelihood of something occurring (happening).

We express probabilities as either fractions or decimals.

E.g.: P (H) = 9/10 (as a fraction)

or

P (H) = 0.9 (as a decimal)

Note: The probabilities should neither be < 0, nor >1.

B. Event

An event is defined as either one or more than one of possible outcomes of doing something or an activity.

Example

ACTIVITY: Inflationary process in Turkey in the year 2006:

Events: A: If annual inflation for 2006 is exactly 10%

B: If annual inflation for 2006 is less than 8%

C: If annual inflation for 2006 is either 5% or 7%

C. Experiment

The experiment is the activity that generates events.

Example: Tossing a coin twice.

D. Sample Space

The sample space of an experiment is the list of all possible outcomes of that experiment.

Example

Experiment: Tossing a coin.

Sample Space: S = {HH, TT, HT, TH}

Events: A: Exactly one outcome, that is, getting 2heads = HH

B: More than one outcome, that is, getting at least one tail = (TT, HT, TH)

C: Exactly three outcomes, that is, getting 2tails or getting one head and one tail

= (TT, HT, TH)

E. Mutually Exclusive Events

Mutually exclusive events are those events which if only one of them can take place at a time, and not together at the same time.

F. Non – Mutually Excusive Events

Are those events which can happen together at the same time.

Example 1

Experiment: Tossing a coin twice

Events: X: Getting two Heads: P (HH)

Y: Getting two Heads or two Tails: P (HH or TT)

Z: Getting Head-Tail or Tail-Head: P (HT or TH)

X and Y are non - mutually exclusive but X and Z are mutually exclusive.

Example 2

Experiment: Randomly selecting a student in this class.

Events: A: Selected student is 21 years old.

B: Selected student is male

C: Selected student is female

D: Selected student is 22 years old.

Therefore: Events AB are non-mutually exclusive because they can happen together at

the same time.

Events AD are mutually exclusive because they can not happen together at

the same time.

Events AC are non-mutually exclusive because they can happen together at

the same time.

Events BC are mutually exclusive because they can not happen together at

the same time.

The crucial question we should ask in determining whether events are mutually exclusive is: Can two or more of these events occur at one time? If we answer yes, the events are not mutually exclusive.

Alternative Approaches in Probability Theory

1. Classical Approach

Example 1

Experiment: Randomly selecting a student in a class of 40 students.

Events: A: Selecting a female student, given that the total number of females equal to 10.

B: Selecting a male student, given that the total number of males is equal to 30.

The total number of students is 40

P (A) = 10/40

P (B) = 30/4

2. Relative Frequency Approach

The probability of the event found by using the given past data.

Example

Experiment: Selecting a student randomly.

Events: A: Selecting a student who gets “A” from MGMT 322.

Past Data: 20% of all students who have taken MGMT 322 in the past, got “A”.

P (A) = 0.20

B: Galatasaray wins the championship in 2005, using the past data obtain how

frequently the similar event occurred in the past.

Past Data: 40% of the last 9 years, Galatasaray was the champion.

P (B) = 0.40

3. Subjective Approach

The probability of the event is found by using the personal feeling or experience.

Example

Event: A: Galatasaray wins.

P (A) = 0.90

As a supporter or a football watcher you may have the feeling that Galatasaray

will win 90%.

Probability Rules

A. Marginal (Unconditional) Probability:

Marginal probability is the probability of the occurrence of a single event. Only one event can take place. P (A)

B. The Addition Rule:

1. The Addition Rule for Mutually Exclusive Events

If Events A, B, and C are mutually exclusive (meaning they can not happen together at the same time.)

Then: P(A or B) = P(A) + P(B)

P( A or B or C) = P(A) + P(B) + P(C)

Questions that can be asked and answered by using Addition Rule:

Q1: What is the probability of A or B or C taking place?

Q2: What is the probability of A or B taking place?

Q3: What is the probability of B or C taking place?

Example

Suppose that the number of students in each age group in the class is as follows:

Events: A: Selecting a random student who is 20 years old.

B: Selecting a random student who is 21 years old.

C: Selecting a random student who is 22 years old.

D: Selecting a random student who is 23 years old.

P(A or B) = P(A) + P(B) = 2/23 + 4/23 = 6/23

P(A or C) = P(A) + P(C) = 2/23 + 9/23 = 11/23

P(A or B or C or D) = P(A) + P(B) + P(C) + P(D) = 2/23 + 4/23 + 9/23 + 4/23 = 19/23

2. The Addition Rule for Non-Mutually Exclusive Events

If A and B are non-mutually exclusive events,

Then: P(A or B) = P(A) + P(B) – P(A and B)

Example 1

Suppose that the sex and the eye color of students in a class of 40 people are distributed as follows:

Experiment: Selecting a random student.

Events:

A: Selecting a male student.

B: Selecting a female student.

C: Selecting a black eye student.

D: Selecting a green eye student.

P(A or C) = P(A) + P(C) – P(A and C) = (22/40 + 21/40) – 13/40 = 30/40

P(B or C) = P(B) + P(C) – P(B and C) = (18/40 + 21/40) – 8/40 = 31/40

Example 2

Suppose that Ford Motor Corporation produces 3 different models for Turkish consumers. Annual volume of production and sales of all 3 model combined is 700 units. However, some of the autos sold turned out to be defective. Based on the past data, the production manager estimated the following distribution of defective and non-defective autos for each model manufactured.

Models

1. What is the probability that a randomly selected consumer will buy either model A or C?
2. What is the probability that the consumer will buy either model B or a Defective auto?
3. What is the probability that the consumer will buy either model A or a Non-defective auto?

Solution:

1. P(A or C) = P(A) + P(C) = 150/700 + 350/700 = 500/700
2. P(B or D) = P(B) + P(D) – P(B and D) = (200/700 + 100/700) – 20/700 = 280/700
3. P(A or N) = P(A)+P(N) – P(A and N) = (150/700 + 600/700) – 140/700= 610/700

Example 3

There are 500 companies whose stocks are traded on Istanbul stock exchange. These companies operate in these basic sectors of the economy; Industry, Service, and Mining. Each one of these companies can report their positive profits and losses for each year. Based on the past data, financial analysts estimated the following distribution of the number of companies in each sector that are likely to report positive profits and losses in the year 2007:

Sectors

1. What is the probability of a randomly selected company (out of these companies listed on Istanbul Stock Exchange) to report positive profits in year 2007?
2. What is the probability of a randomly selected company to report either loss in year 2005 or to be from Mining sector?
3. What is the probability of a randomly selected company to be either from Industry or from Service sector?

1. P( Positive profits) = 420/500 = 0.84

1. P( Losses or Mining) = P( Losses) + P( Mining) – P( Losses and Mining)

= 80/50 + 50/500 – 20/500 = 110/500 = 0.22

1. P( Industry or Service) = P(Industry) + P(Service)

= 200/500 + 250/500 = 450/500 = 0.9

Types of Events

1. Statistically Independent Events

These are those events whereby the probability of occurrence of an event is not affected by (or dependent upon) the occurrence of the other event.

Example

Events: A: Izlem goes to Istanbul next Tuesday

B: Istanbul stock market decreases by more than 10% next Friday.

Therefore: Event A and B are Independent events because the occurrence of event “A” does not have an impact on the probability of the occurrence of event “B”.

2. Statistically Dependent Events

These are those events whereby the probability of the occurrence of an event is affected by (or dependent upon) the occurrence of the other event.

Example 1

Events: A: Stock Market Index rises tomorrow

B: Interest rates go down today

Therefore: Event A and B are Dependent because the occurrence of event “B” is expected to increase the probability of the event “A”

Example 2

Events: A: The number of automobiles purchased in Turkey rises

B: The quantity of petroleum consumed in Turkey rises

Therefore: Event A and B are Dependent because the probability of the occurrence of event “B” is affected by or dependent upon the occurrence of event “A”, as the more the automobiles Turkish people purchase, the more the petroleum they will buy to run their automobiles.

Types of Probabilities under Independence

1. Marginal (Unconditional) Probability

P(A): In this case, we are interested in the probability of the occurrence of only one event.

2. Joint Probability

P(AB): This is where you are interested in either the probability of event A and event B happening together at the same time or in succession.

P(AB) = P(A) * P(B)

P(BA) = P(B) * P(A)

A and B are independent events.

3. Conditional Probability

P(A/B): Probability of A, given B!

As long as A and B are independent events P(A) is not affected by P(B) so;

P(A/B) = P(A)

P(B/A) = P(B)

Example

Events: A: The weather is rainy in Cyprus today.

B: Exchange rate of dollar/yen increases today

Therefore: P(A/B) = P(A)

PB/A) = P(B)

Types of Probabilities under Dependence

1. Marginal (Unconditional) Probability

P(A) : In this case, we are interested in the probability of the occurrence of only one event.

2. Joint Probability

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

P(BA) = P(B/A) * P(A)

P(AB) = P(BA)

3. Conditional Probability

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

P(B/A) = P(BA) / P(A)

Example

Suppose that the eye color of students in a class of 40 people given as follows;

Events:

A: Selecting a Male student.

B: Selecting a student with Black Eyes.

C: Selecting a Female student.

D: Selecting a student with Green Eyes.

Q1. If the selected student is a Male student, what is the probability that he has got Green Eyes?

P(D/A) = P(DA) / P(A) = (8/40) / (28/40) = 0.074

Q2. If the selected student is a Female student, what is the probability that she has got Green Eyes.

P(D/C) = P(DC) / P(C) = (2/40) / (12/40) = 0.167

Bayes’ Theorem

Bayes’ Theorem is used for revising prior (before) estimates of probabilities using limited new information to obtain posterior (new or after) probabilities.

Event A gives us P (A); after sometime passes event B occurs which is dependent on event A.

We can use Bayes’ Theorem to obtain posterior probability of A:

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

Example 1

Suppose we have 100 dices in a basket, and we have two types; half of it is Type1 and the other half is Type2. Given that the probability of getting Ace (1) when you roll a Type1 die is 0.3 and that of rolling Ace (1) of Type2 is 0.6. You are also given the initial probability of Type1 as 0.5 and that of Type2 as 0.5.

Assume that you have randomly selected a die and roll it and you found out that Ace (1) has come up, what is the probability of this die to be a Type1 die?

NB: These two events (Ace and Type1 or 2) are dependent because the probability of getting Ace is affected by the occurrence of Type1 or 2.

P( Type1/Ace) = P(Type1 Ace) / P(Ace)

Note: When you are not given the values for the formula like in the example above, use the following five steps below in order to get the Answer:

Step 1: Prepare the following table:

Step 2: Obtain the MARGINAL probability of the secondary event.

P(secondary event) = summation of the individual joint probabilities of the secondary event with each one of the elementary events.

P(Ace) = P(Type1, Ace) + P( Type2, Ace)

Step 3: Compute the joint probability of secondary events with each of the elementary events:

P(Type 1, Ace) = P(Ace/Type1)* P(Type 1)

= 0.3 * 0.5 = 0.15

P(Type 2, Ace) = P(Ace/Type2) * P(Type2)

= 0.6 * 0.5 = 0.30

P(Ace) = 0.15 + 0.30 = 0.45

Step 4: Apply Baye’s Theorem by using the joint probabilities you obtained in step3 to get the final answer.

P(Type1/ Ace) = P(Type 1, Ace) / P(Ace)

= 0.15 / 0.45 = 1/3 This is posterior probability

Example 2

Economists believe that the annual inflation rate in Turkey is affected by the changes in the petroleum prices. The probability of inflation rate increasing in 2001 is estimated to be 0.60. The probability of petroleum prices rising is 0.40. The probability of both inflation rate and the petroleum prices rising is 0.35. On the other hand, the probability of inflation rate rising and at the same time, the petroleum prices not rising is estimated to be 0.20. If petroleum prices do not rise in 2001, what is the probability of inflation rate rising?

Events: I: Inflation rate increasing.

P: Petroleum prices rising.

N: Petroleum prices not rising

Given:

P(I) = 0.6

P(P) = 0.4

P(IP) = 0.35

P(IN) = 0.2

Solution:

P(N) = 1 – P(P) = 1- 0.4 = 0.6

P(I / N) = P(IN) / P(N) = 0.2 / 0.6 = 0.333 ≈ 0.3

Example3

Annual profits of construction sector depend on wage rate paid to labor employed by the sector. The past data about the behavior of annual profits suggests that the probability of profits rising in a given year is 0.80 and the probability of profits falling is 0.20. The past data also suggests that in 60% of all the years during which the profits have increased, wage rate has declined, and it has increased in the remaining 40%. However, in 80% of all years during which profits have fallen, wage rate has increased, and the only remaining 20% wage rate has decreased. Due to the economic crisis in Turkey, wage rate has been declining since the beginning of 2005. Given this downward trend in wages, what is our revised estimate for the probability of annual profits of construction sector to rise in the year?

Events: A: Profits rising

B: Profits falling

C: Wage rate declining

D: Wage rate increasing

Given:

P(A) = 0.80

P(B) = 0.20

P(C/A) = 0.60

P(D/A) = 0.40

P(D/B) = 0.80

P(C/B) = 0.20

Solution: Without using the table

P(A/C) = P(AC) / P(C) or P(CA) / P(A)

0.60 = P(CA) / 0.80

P(CA) or P(AC) = 0.60 * 0.80 = 0.48

P(C) = P(CA) + P(CB)

= [P(C/A) * P(A)] + [P(C/B) * P(B)]

= (0.60 * 0.80) + (0.20 * 0.20)

= 0.48 + 0.04 = 0.52

P(A/C) = 0.48 / 0.52 = 0.92

Example 4

A child chosen at random in a community school system comes from low-income family 15% of the time. Children from low-income families in the community graduate from college only 20% of the time. Children not from low-income families have a 40% chance of graduating from college. As an employer of people from this community, you have been reviewing applicants and noted that the first one had a college degree. What is the probability that the person comes from a low-income family?

Events: L: Child coming from a low-income family

G: Graduate from college

H: Child coming from not a low-income family

Given:

P(L) = 0.15

P(H) = 1- P(H) = 1- 0.15 = 0.85

P(G/L) = 0.20

P(G/H) = 0.40

P(L/G) = ?

Solution:

P(L/G) = P(LG) / P(G)

P(G/L) * P(L) / P(G)

P(G) = [P(G/L) * P(L)] + [P(G/H) * P(H)]

= (0.20 * 0.15) + (0.40 * 0.85)

= 0.03 + 0.34 = 0.37

P(L/G) = 0.03 / 0.37 = 0.08 = 8%

Note: To find the probability of a single event, you need to add up all of its joint probabilities.

Example 5

A new diagnostic test to detect gout has been introduced into Lincoln County Hospital. The manufacturer of the test device has determined that the probability of a false- positive reading (a positive test result for a person who is known not to have gout) is 8%. The probability of a false- negative reading (a negative reading for a person who has the disease) is 4%. County health authorities estimate that approximately 15% of the population in the county has gout. If a test is conducted on a patient from that county and the test result is positive, what is the probability that gout is present?

Solution:

Events:

H: Person who is healthy.

I: Person who is ill.

P: A positive test

N: A negative test.

P(P/H)= 0.08

P(N/I)=0.04

P(I)=0.15

P(H)=1- P(I)=1- 0.15= 0.85 P(H)=0.85

P(I/P)=?

Step 1:

P(N/I)= 0.04

P(P/I)= 1- P(N/I)

P(P/I)= 1- 0.04= 0.96