DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION SYSTEMS Introduction to Business Statistics QM 120...

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DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION SYSTEMS Introduction to Business Statistics QM 120 Chapter 4

Transcript of DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION SYSTEMS Introduction to Business Statistics QM 120...

Page 1: DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION SYSTEMS Introduction to Business Statistics QM 120 Chapter 4.

DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION SYSTEMS

Introduction to Business StatisticsQM 120

Chapter 4

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Experiment, outcomes, and sample space

Probability and statistics are related in an important way. It is used to allow us to evaluate the reliability of our conclusions about the population when we have only sample information.

Data are obtained by observing either uncontrolled events in nature or controlled situation in laboratory. We use the term experiment to describe either method of data collection.

The observation or measurement generated by an experiment may or may not produce a numerical value. Here are some examples: Recording a test grade

Interviewing a householder to obtain his or her opinion in certain issue.

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Experiment, outcomes, and sample space

Table 1: Examples of experiments, outcomes, and sample spaces

Venn diagram is a picture that depicts all possible

outcomes for an experiment while in tree diagram,

each outcome is represented a branch of a tree.

Experiment Outcomes Sample Space

Toss a coin once Head, Tail S = {Head, Tail}

Roll a die once 1, 2, 3, 4, 5, 6 S = {1, 2, 3, 4, 5, 6}

Play a lottery Win, Lose S = {Win, Lose}

Select a student M, F S = {M,F}

Toss a coin twice HH, HT, TH, TT S = {HH, HT, TH, TT}

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● H ● TVenn

Diagram

H

T

Tree Diagram

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Experiment, outcomes, and sample space

Example: Draw the Venn and tree diagrams for the experiment of tossing a coin twice.

Solution:

Example: Draw a tree diagram for three tosses of a coin. List all outcomes for this experiment in a sample space S.Solution

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● TH● HH

● TT● HT

Venn Diagram

TT

TH

HT

HH

H

T

Tree Diagram

H

T

T

H

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Experiment, outcomes, and sample space

A simple event is the outcome that is observed on a single repetition of the experiment. It is often denoted by E with a subscript.

Example: Toss a die and observe the number that appears on the upper face. List the simple event in the experiment.

Solution: When the die is tossed once, there are six possible outcomes. There are the simple events, listed below.

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Event E1: Observe a 1 Event E2: Observe a 2 Event E3: Observe a 3

Event E4: Observe a 4 Event E5: Observe a 5 Event E6: Observe a 6

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Experiment, outcomes, and sample space

We can now define an event (or compound event) as a collection of simple events, often denoted a capital letter.

Example: Tossing a die (continued)We can define the events A and B as follow,A: Observe an odd numberB: Observe a number less than 4

Solution:

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Experiment, outcomes, and sample space

Example: Suppose we randomly select two persons from members of a club and observe whether the person selected is a man or woman. Write all the outcomes from experiment. Draw the Venn and tree diagrams for this experiment.Solution:

Two events are mutually exclusive if, when one event occurs, the other cannot, and vice versa.

The set of all simple events is called the sample space, S.

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Calculating Probability

Probability is a numerical measure of the likelihood that an event will occur.

Two properties of probability

The probability of an even always lies in the range 0 to 1.

0 ≤ P(Ei) ≤ 1

0 ≤ P(A) ≤ 1

The sum of the probabilities of all simple events for an experiment, denoted by ΣP(Ei), is always 1.

∑ P(Ei) = P(E1) + P(E2) + P(E3) + . . . . = 1

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Calculating Probability

Three conceptual approaches to probability

1. Classical probability Two or more events that have the same probability of

occurrence are said to be equally likely events.

The probability of a simple event is equal to 1 divided by the total number of all final outcomes for an equally likely experiment.

Classical probability rule to find probability1

( )Total number of outcomes

Number of outcomes favorable to ( )

Total number of outcomes

iP E

AP A

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Calculating Probability

Example: Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin.

Solution

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Calculating Probability

Example: Find the probability of obtaining an even number in one roll of a die.Solution

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Calculating Probability

Calculating the probability of an event: List all the simple events in the sample space.

Assign an appropriate probability to each simple event.

Determine which simple events result in the event of interest.

Sum the probabilities of the simple events that result in the event of interest.

Always1. Include all simple events in the sample space.2. Assign realistic probabilities to the simple events.

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Calculating Probability

Example: A six years boy has a safe box that contains four banknotes: One-Dinar, Five-Dinar, Ten-Dinar, Twenty-Dinar. His sister which is a three years old girl randomly grabbed three banknotes from the safe box to buy a 30 KD toy. Find the odds (probability) that this girl can buy the toy.Solution

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Calculating Probability

2. Relative frequency concept of probability Suppose we want to know the following probabilities:

The next car coming out of an auto factory is a “lemon”

A randomly selected family owns a home

A randomly selected woman has never smoked

The outcomes above are neither equally likely nor fixed for each sample.

The variation goes to zero as n becomes larger and larger

If an experiment is repeated n times and an event A is observed f times, then

( )f

P An

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Calculating Probability

Example: In a group of 500 women, 80 have played golf at least once. Suppose one of these 500 woman is selected. What is the probability that she played golf at least onceSolution

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Calculating Probability

Example: Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be defective. What is the probability that the next car manufactured at that factory is a defective?Solution:

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Calculating Probability

Example: Lucca Tool Rental would like to assign probabilities to the number of car polishers it rents each day. Office records show the following frequencies of daily rentals for the last 40 days.

Number of Polishers Rented

Number of Days

0 4

1 6

2 18

3 10

4 2

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Calculating Probability

Solution

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Calculating Probability

Law of large numbers: If an experiment is repeated again and again, the probability of an event obtained from the relative frequency approaches the actual probability.

3. Subjective probability Suppose we want to know the following probabilities:

A student who is taking a statistics class will get an A grade.

KSE price index will be higher at the end of the day.

China will dominate the gold medal list in the 2008 Olympics.

Subjective probability is the probability assigned to an event based on subjective judgment, experience, information, and belief.

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Counting Rule

Suppose that an experiment involves a large number N of simple events and you know that all the simple events are equally likely. Then each simple event has probability 1/N and the probability of an event A can be calculated as

The mn rule Consider an experiment that is performed in two stages. If

the first stage can be performed in m ways and for each of these ways, the second stage can be accomplished in n ways, then there mn ways to accomplish the experiment

( )

Where is the number of simple events that result in event A

A

A

nP A

Nn

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Counting Rule

Example: Suppose you want to order a car in one of three styles and in one of four paint colors. To find out how many options are available, you can think of first picking one of the m = 3 styles and then one of n = 4 colors.

Using the mn rule, as shown in the figure below, you have mn= (3)(4) = 12 possible options.

1

1

2

3

4

2

1

2

3

4

3

1

2

3

4

Style Color

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Counting Rule

The extended mn rule If an experiment is performed in k stages, with n1 ways to

accomplish the first stage, n2 to accomplish the second stage…, and nk ways to accomplish the kth stage, the number of ways to accomplish the experiment is

Example: A bus driver can take three routes from city A to city B, four routes from city B to city C, and three routes from city C to city D. For traveling from A to D, the driver must drive from A to B to C to D, how many possible routes from A to D are available

1 2 3... kn n n n

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Counting Rule

Example: A couple is planning their weeding reception. The bird's parents have given thema choice of four reception facilities, three caterers, five DJs, and two limo services. He the couple randomly selects one reception facility, one caterer, one DJ, and one limo service, how many different outcomes are possible?.

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Counting Rule

A counting rule for permutations (orderings) The number of ways we can arrange n distinct objects,

taking them r at a time, is

A counting rule for combination The number of distinct combinations of n distinct objects

that can be formed, taking them r at a time, is

!

( )!

where ! ( 1)( 2).....(3)(2)(1) and 0! 1

nr n r

nP P

n r

n n n n

!( )

!( )!n n

n r r r

nC C

r n r

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Counting Rule

Permutations: Given that position (order) is important, if one has 5 different objects (e.g. A, B, C, D, and E), how many unique ways can they be placed in 3 positions (e.g. ADE, AED, DEA, DAE, EAD, EDA, ABC, ACB, BCA, BAC etc.)

Combinations: If one has 5 different objects (e.g. A, B, C, D, and E), how many ways can they be grouped as 3 objects when position does not matter (e.g. ABC, ABD, ABE, ACD, ACE, ADE are correct but CBA is not ok as is equal to ABC)

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5! (5)(4)(3)(2)(1)60

(5 3)! (2)(1)P

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5! (5)(4)(3)(2)(1)10

3!(5 3)! (3)(2)(1) (2)(1)C

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Marginal & Conditional Probabilities

Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of U.S. companies. The following table gives a two-way classification of their responses.

Marginal probability is the probability of a single event without consideration of any other event

In favor Against Total

15 45 60

4 36 40

Male

Female

Total 19 81 100

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Marginal & Conditional Probabilities

Suppose one employee is selected, he/she maybe classified either on the bases of gender alone or on the bases of opinion.

The probability of each of the following event is called marginal probability P(male) = 60/100

P(female) = 40/100

P(in favor) = 19/100

P(against) = 81/100

Suppose the employee selected is known to be male. What is the probability that he is in favor?

In favor Against Total

15 45 60

4 36 40

Male

Female

Total 19 81 100

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Marginal & Conditional Probabilities

This probability is called conditional probability and is written as “the probability that the employee selected is in favor given that he is a male.”

P( in favor | male )

Conditional probability is the probability that an even will occur given that another event has already occurred. If A and B are two events, then the conditional probability of A given B is written as

P(A | B )

Read as “given”

This event has already occurred

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The event whose probability is to be determined

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Marginal & Conditional Probabilities

Example: Find the conditional probability P(in favor | male) for the data on 100 employeesSolution

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Marginal & Conditional Probabilities

Example: Find the conditional probability P(female|in favor) for the data on 100 employeesSolution

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Marginal & Conditional Probabilities

Example: Consider the experiment of tossing a fair die. Denote by A and B the following events:A={Observing an even number}, B={Observing a number of dots less than or equal to 3}. Find the probability of the event A, given the event B.Solution

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Mutually Exclusive Events

Events that cannot occur together are called mutually exclusive events.

Example: Consider the following events for one roll of a dieA: an even number is observed = {2,4,6}B: an odd number is observed = {1,3,5}C: a number less than 5 is observed = {1,2,3,4}

- A and B are mutually exclusive events but A and C are not.

- How about B and C?- Simple events are mutually exclusive always.

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Independent vs. Dependent events

Two events are said to be independent if the occurrence of one does not affect the probability of the other one.

A and B are said to be independent events if either

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

Example: Refer to the information on 100 employees. Are events “female (F)” and “in favor (A)” independent?

Solution:

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Independent vs. Dependent events

What is the difference between mutually exclusive and independent events? It is common to get confused or not to tell the difference

between these two terminologies.

When two events are mutually exclusive, they cannot both happen. Once the event B has occurred, event A cannot occur, so that P(A|B) = 0, or vice versa. The occurrence of event B certainly affects the probability that event A can occur. Therefore,

Mutually exclusive events must be dependent.

Independent events are never mutually exclusive.

But, dependent events may or may not be mutually exclusive

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Independent vs. Dependent events

Example: A sample of 420 people were asked if they smoke or not and whether they are graduate or not. The following two-way classification table gives their responses

If an person is selected at random from this sample, find the probability that this person is a

College graduate (G).Nonsmoker (NS).Smoker (S) given the person is not a college graduate (NG).College graduate (G) given the person is a nonsmoker (NS).Are the events Smoker and college graduate independent? Why?

College graduate Not a college graduate

Smoker 35 80

Nonsmoker 130 175

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Independent vs. Dependent events

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Independent vs. Dependent events37

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Complementary Events

Two mutually exclusive events that taken together include all the outcome for an experiment (sample space, S) are called complementary events.

Consider the following Venn diagram:

Since two complementary events, taken together, include all the sample space S, the sum of probabilities of all outcomes is 1

P(A) + P(A) = 1 P(A) = 1 - P(A)

The complement of event A, denoted by A and read as “A bar” or “A complement,” is the event that includes all the outcomes for an experiment that are not in AVenn diagram of two complementary evens

AB

AA

B

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Complementary Events

Example: In a lot of five machines, two are defective. If one of machines is randomly selected, what are the complementary events for this experiment and what are their probabilities?Solution: The two complementary events for this experiment are

A = the machine selected is defective

A = the machine selected is not defective

Since there are 2 defective and three non defective machines, the probabilities of each event is

P(A) = 2/5 = 0.4 or simply P(A) = 1 = 0.4 = 0.6P(A) = 3/5 = 0.6

A A

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Intersection of events A and B

Intersection of Events & Multiplication Rule

Intersection of events

Let A and B be two events defined in a sample space. The intersection of A and B represents the collection of all outcomes that are common to both A and B is denoted by any of the followings

A and B, A ∩ B, or AB

Example:Let A = the event that a person owns a PCLet B = the event that aperson owns a mobile

A BA and B

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Intersection of Events & Multiplication Rule

Multiplication rule

Sometimes we may need to find the probability of two events happening together.

The probability of the intersection of two events A and B is called their joint probability. It is written

P(AB) or P(A ∩ B)

It can be obtained by multiplying the marginal probability of one event with the conditional probability of the second one.

Multiplication rule: The probability of the intersection of two events A and B is

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

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Intersection of Events & Multiplication Rule

Example: The following table gives the classification of all employees of a company by gender and college degree.

If one employee is selected at random, what is the probability that the employee is a female and a college graduate?

Solution

College graduate (G)

Not a college graduate

(N) Total

Male (M) 7 20 27

Female (F) 4 9 13

Total 11 29 40

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Intersection of Events & Multiplication Rule

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Intersection of Events & Multiplication Rule

Example: A box contains 20 DVD, 4 of which are defective. If two DVDs are selected at random (without replacement), what is the probability that both are defective?Solution:

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Intersection of Events & Multiplication Rule

If events A and B are independent, their joint probability simplifies from

P(A ∩ B) = P(A)P(B |A) TO P(A ∩ B) = P(A)P(B )

Sometimes we know the joint probability of two events A and B, in this case, the conditional probability of B given A or A given B is

given that P( ) 0 and P( ) 0A B

( )( | )

( )

P A BP A B

P B

( )( | )

( )

P A BP B A

P A

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Intersection of Events & Multiplication Rule

Example: According to a survey, 60% of all homeowners owe money on home mortgages. 36% owe money on both home mortgages and car loans. Find the conditional probability that a homeowner selected at random owes money on a car loan given that he owes money on a home mortgage.Solution:

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Intersection of Events & Multiplication Rule

Example: A computer company has two quality control inspectors, Mr. Smith and Mr. Robertson, who independently inspect each computer before it is shipped to a client. The probability that Mr. Smith fails to detect a defective PC is .02 while it is .01 for Mr. Robertson. Find the probability that both inspectors will fail to detect a defective PC.Solution:

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Intersection of Events & Multiplication Rule

Example: The probability that a patient is allergic to Penicillin is .2. Suppose this drug is administrated to three patients.Finda) The probability that all three of them are allergic to itb) At least one of them is not allergicSolution: Let

A = 1st patient is allergic to penicillin

B = 2nd patient is allergic to penicillin

C = 3rd patient is allergic to penicillin

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Union of events

Let A and B be two events defined in a sample space S. The union of events A and B is the collection of all outcomes that belong either to A and B or to both A and B and is denoted by “ A or B” or “A ∪ B”

A B

S

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Union of Events & Addition Rule

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Example: A company has 1000 employees. Of them, 400 are females and 740 are labor union members. Of the 400 females, 250 are union members. Describe the union of events “female” and “union member”

Solution

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Union of Events & Addition Rule

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Addition rule

The probability of the union of two events A and B is

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Example: A university president has proposed that all students must take a course in ethics as a requirement for graduation. Three hundred faculty members and students from this university were asked about their opinion on this issue and it shown in the following table.

Find the probability that a person selected is a faculty member or in favor.

Opinion

Favor Oppose Neutral Total

Faculty 45 15 10 70

Student 90 110 30 230

Total 135 125 40 300

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Union of Events & Addition Rule

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Solution:

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Union of Events & Addition Rule

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Example: There are a total of 7225 thousand persons with multiple jobs in the US. Of them, 4115 thousand are male, 1742 thousand are single, and 905 thousand are male and single. What is the probability that a selected person is a male or single?Solution:

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Union of Events & Addition Rule

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The probability of the union of two mutually exclusive events A and B is

P(A ∪ B) = P(A) + P(B)

Example: A university president has proposed that all students must take a course in ethics as a requirement for graduation. Three hundred faculty members and students from this university were asked about their opinion on this issue and it shown in the following table.

Find the probability that a person selected is in favor or is neutral

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Union of Events & Addition Rule

Opinion

Favor Oppose Neutral Total

Faculty 45 15 10 70

Student 90 110 30 230

Total 135 125 40 300

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Union of Events & Addition Rule

Solution:

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Example: Eighteen percent of the working lawyers in the United States are female. Two lawyers are selected at random and it is observed whether they are male or female.a) Draw a tree diagram for this experimentb) Find the probability that at least one of the two lawyers is a female.Solution:

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Union of Events & Addition Rule

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Law of Total Probability & Bayes’ Rule

Consider the following Venn diagram

Can we find the area (probability) of A assuming that we know the probability of each Bi & P(A|Bi) ?

B1 B2 Bn-1 Bn

B

A

A1 A2 An-1 An

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Addition Law

Multiplication Law

Conditional Probability

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

( ) ( ) ( | )P A B P B P A B

( ) ( ) ( | )P A B P A P B A

( )( | )

( )

P A BP A B

P B

( )( | )

( )

P A BP B A

P A

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Law of Total Probability & Bayes’ Rule

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1 11

1 1 2 2

( ) ( | )( | )

( ) ( | ) ( ) ( | )

P A P B AP A B

P A P B A P A P B A

Bayes’ theorem (two-event case)

Bayes’ theorem

2 22

1 1 2 2

( ) ( | )( | )

( ) ( | ) ( ) ( | )

P A P B AP A B

P A P B A P A P B A

1 1 2 2

( ) ( | )( | )

( ) ( | ) ( ) ( | ) ... ( ) ( | )i i

in n

P A P B AP A B

P A P B A P A P B A P A P B A

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Law of Total Probability & Bayes’ Rule

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Example: Manufacturing firm that receives shipment of parts from two different suppliers. Currently, 65 percent of the parts purchased by the company are from supplier 1 and the remaining 35 percent are from supplier 2. Historical Data suggest the quality rating of the two supplier are shown in the table:

a) Draw a tree diagram for this experiment with the probability of all outcomesb) Given the information the part is bad, What is the probability the part came from supplier 1?

Good Parts Bad Parts

Supplier 1 98 2

Supplier 2 95 5

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Law of Total Probability & Bayes’ Rule

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Solution:

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Law of Total Probability & Bayes’ Rule

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Example: An insurance company rents 35% of the cars for its customers from Avis and the rest from Hertz. From past records they know that 8% of Avis cars break down and 5% of Hertz cars break down. A customer calls and complains that his rental car broke down. What is the probability that his car was rented from Avis?

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Law of Total Probability & Bayes’ Rule

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Solution:

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Law of Total Probability & Bayes’ Rule

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Example64

A study was recently done in which 500 people were asked to indicate their preferences for one of three products. The following table shows the breakdown of the responses by gender of the respondents.

Product PreferenceGender A B CMale 80 20 10Female 200 70 120

Suppose one person is randomly chosen. Based on this data, what is the probability that the person chose is a female who prefers product C?

A) .31 B) .92 C) .86 D) .24

Page 65: DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION SYSTEMS Introduction to Business Statistics QM 120 Chapter 4.

Example65

The method of probability assessment that relies on an examination of historical data from similar situations is

A) classical assessmentB) relative frequency of occurrence C) subjective assessmentD) historical assessment