Introduction to Business Statistics QM 120 Chapter 4 · Introduction to Business Statistics QM 120 Chapter 4 Spring 2008. Chapter 4: Experiment, outcomes, and sample space 2

DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION SYSTEMS DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION SYSTEMS

Introduction to Business StatisticsQM 120QM 120

Chapter 4

Spring 2008

Chapter 4: Experiment, outcomes, and sample space2

Probability and statistics are related in an important way. Itis used to allow us to evaluate the reliability of ourconclusions about the population when we have onlyconclusions about the population when we have onlysample information.

Data are obtained by observing either uncontrolled eventsi ll d i i i l b W hin nature or controlled situation in laboratory. We use theterm experiment to describe either method of datacollection.

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

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


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

Sample SpaceOutcomesExperiment Sample SpaceOutcomesExperimentS = {Head, Tail}Head, TailToss a coin once

S = {1, 2, 3, 4, 5, 6}1, 2, 3, 4, 5, 6Roll a die onceS = {Win, Lose}Win, LosePlay a lottery

S = {M,F}M, FSelect a studentS = {HH, HT, TH, TT}HH, HT, TH, TTToss a coin twice

Venn diagram is a picture that depicts all possible outcomesf i t hil i t di h t ifor an experiment while in tree diagram, each outcome isrepresented a branch of a tree.


● TH● HH Venn Diagram TH

HT

HHH

Tree Diagram

A simple event is the outcome that is observed on a singletiti f th i t It i ft d t d b E ith

● TT● HTDiagram

TT

THT

Diagram

repetition of the experiment. It is often denoted by E with asubscript.

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


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

E l T i di ( i d)Example: Tossing a die (continued)We can define the events A and B as follow,A: Observe an odd numberA: Observe an odd numberB: Observe a number less than 4

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


Example: Suppose we randomly select two persons frommembers of a club and observe whether the person selected isa man or woman. Write all the outcomes from experiment.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 eventoccurs, the other cannot, and vice versa.

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

Chapter 4: Calculating Probability7

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

T ti f b bilitTwo 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 experimentThe 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


Three conceptual approaches to probability

1. Classical probabilityTwo or more events that have the same probability of occurrence are said to be equally likely eventssaid 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 probability

outcomesofnumberTotal1)(EP i =

outcomes ofnumber Total tofavorable outcomes ofNumber )(

outcomesofnumber TotalAAP =


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


Example: Find the probability of obtaining an even number inone roll of a die.SolutionSolution


Example: A candy dish contains one green and two red candies. You close your eyes, choose two candies one at a time from the dish, and record their colors. What is the probabilityfrom the dish, and record their colors. What is the probability that both candies are red?Solution


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

Always1 Include all simple events in1. Include all simple events in

the sample space.2. Assign realistic probabilities

to the simple events.


Example: A six years boy has a safe box that contains fourbanknotes: One‐Dinar, Five‐Dinar, Ten‐Dinar, Twenty‐Dinar.His sister which is a three years old girl randomly grabbedHis sister which is a three years old girl randomly grabbedthree banknotes from the safe box to buy a 30 KD toy. Find theodds (probability) that this girl can buy the toy.SolutionSolution


2. Relative frequency concept of probabilitySuppose 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 smokedA 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

nfAP =)(


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


Example: Ten of the 500 randomly selected cars manufacturedat a certain auto factory are found to be lemons. What is theprobability that the next car manufactured at that factory is aprobability that the next car manufactured at that factory is alemon?Solution:


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

Number of Polishers Rented Number of Days

0 4

1 61 6

2 18

3 10

4 2


Solution

Number of Polishers Rented Number of Days Probability


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

3. Subjective probability3. Subjective probabilitySuppose 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

The Brazilian team will win world cup 2006

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

Chapter 4: Counting Rule20

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

)( A

NnAP =

The mn ruleevent A inresult that events simple ofnumber the is Where An

N

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

li h th i taccomplish the experiment


Example: Suppose you want toorder a car in one of threestyles and in one of four paint

Style Colorstyles and in one of four paintcolors. To find out how manyoptions are available, you canthink of first picking one of the 1

1

23think of first picking one of the

m = 3 styles and then one of n =4 colors. Using the mn rule, asshown in the figure below you

41

2shown in the figure below, youhave mn= (3)(4) = 12 possibleoptions.

2234

1

3

1

2344


The extended mn ruleIf an experiment is performed in k stages, with n1 ways to accomplish the first stage n2 to accomplish the second stage andaccomplish 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

nnnnExample: A bus driver can take three routes from city A to cityB, four routes from city B to city C, and three routes from cityC i D F li f A D h d i d i

knnnn ...321

C to city D. For traveling from A to D, the driver must drivefrom A to B to C to D, how many possible routes from A to Dare available

Example: A medical technician records a person’s blood typeand Rh factor. Calculate the total outcome of this experimentpusing the mn method and using a tree diagram.


Example: Ahmad has invested in two stocks, ARC Oil andCoal Mining. Ahmad has determined that the possibleoutcomes of these investments three months from now are asoutcomes of these investments three months from now are asfollows.

Investment Gain or Lossin 3 Months (in €000)

A Oil C l Mi iArc Oil Coal Mining105

8-2

0-20


Solution

Ahmad investments can be viewed as a two‐step experiment. It A a i es e s ca be ie e as a o s ep e pe i e Iinvolves two stocks, each with a set of experimental outcomes.

ARC Oil: n1 = 4Coal Mining: n2 = 2gTotal Number of Experimental Outcomes: n1n2 = (4)(2) = 8


Gain 8Gain 8

ARC Oil(Stage 1)

Coal Mining(Stage 2)

ExperimentalOutcomes

Gain 8Gain 8

Gain 10Gain 10Gain 8Gain 8

Lose 2Lose 2

(10, 8) Gain €18,000

(10, -2) Gain €8,000

(5 8) G i €13 000

Gain 5Gain 5Gain 8Gain 8

Gain 8Gain 8

Lose 2Lose 2

(5, 8) Gain €13,000

(5, -2) Gain €3,000

(0 8) Gain €8 000

Gain 8Gain 8Lose 20Lose 20 Lose 2Lose 2

EvenEven(0, 8) Gain €8,000

(0, -2) Lose €2,000

(-20 8) Lose €12 000

Lose 2Lose 2

(-20, 8) Lose €12,000

(-20, -2) Lose €22,000


A counting rule for permutations (orderings)The number of ways we can arrange n distinct objects, taking them r at a time isr at a time, is

)!(!−

==rn

nPP rnn

r

A counting rule for combination

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

=−−= nnnn

gThe number of distinct combinations of n distinct objects that can be formed, taking them r at a time, is

)!(!!)(

rnrnCC n

rnrrn −

===


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

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

Chapter 4: Marginal & Conditional Probabilities28

Suppose all 100 employees of a company were askedwhether they are in favor of or against paying high salariesto CEOs of U S companies The following table gives a two‐to CEOs of U.S. companies. The following table gives a twoway classification of their responses.

TotalAgainstIn favor

604515

40364

Male

Female

Total 1008119

Marginal probability is the probability of a single eventwithout consideration of any other event


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

The probability of each of the following event is calledmarginal probability

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


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

P( in favor | male ) This event has P( in favor | male )

The event whose probability is to be determined Read as “given”

already occurred

Conditional probability is the probability that an even willoccur given that another event has already occurred. If A

is to be determined Read as “given”

occur given that another event has already occurred. If Aand B are two events, then the conditional probability of Agiven B is written as

(A | )P(A | B )


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


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

T t lA i tI f TotalAgainstIn favor604515Male40364Female

Total numberof in favor

1008119TotalFemales who arein favor

favorin are whoemployees ofnumber Totalfavorin are whofemales ofNumber )favorin |female( =P

4 2105.194==


Example: Consider the experiment of tossing a fair die. Denoteby A and B the following events:A={Observing an even number} B={Observing a number ofA {Observing an even number}, B {Observing a number ofdots less than or equal to 3}. Find the probability of the eventA, given the event B.S l tiSolution

Chapter 4: Mutually Exclusive Events34

Events that cannot occur together are called mutuallyexclusive events.

Example: Consider the following events for one roll of a dieA: an even number is observed = {2,4,6}B dd b i b d {1 3 5}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.

Chapter 4: Independent vs. Dependent events35

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

A d B id t b i d d t t if ithA 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. Areevents “female (F)” and “in favor (A)” independent?Solution:


What is the difference between mutually exclusive andindependent events?It i t t f d t t t ll th diff b tIt 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. O e the e e t B ha o u ed e e t A a ot o u o that P(A|B)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 dependentMutually exclusive events must be dependent.

Independent events are never mutually exclusive.

But, dependent events may or may not be mutually exclusive


Example: A sample of 420 people were asked if they smoke or notand whether they are graduate or not. The following two‐wayclassification table gives their responsesg p

Not a college graduateCollege graduate

8035Smoker

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

175130Nonsmoker

p p ,probability that this person is a

College graduate (G).Nonsmoker (NS)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?

Chapter 4: Complementary Events38

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

Consider the following Venn diagram:

The complement of event A, denoted byA and read as “A bar” or “Acomplement,” is the event that includes

AB

AA

B

Since two complementary events, taken together, include all

all the outcomes for an experiment thatare not in AVenn diagram of two complementary evens

Si ce two co p e e ta y eve ts, take toget e , i c ude athe sample space S, the sum of probabilities of all outcomesis 1

P(A) P(A) 1 P(A) 1 P(A)P(A) + P(A) = 1→ P(A) = 1 ‐ P(A)

Chapter 4: Complementary Events39

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

Chapter 4: Intersection of Events & Multiplication Rule40

Intersection of events

Let A and B be two events defined in a sample space. Theintersection of A and B represents the collection of allintersection of A and B represents the collection of alloutcomes that are common to both A and B is denoted byany of the followings

A and B, A ∩ B, or ABExample:Let A = the event that aLet A the event that aperson owns a PCLet B = the event that aperson owns a mobile

A BA and B

Intersection of events A and Bperson owns a mobile


Multiplication rule

Sometimes we may need to find the probability of twoevents happening togetherevents happening together.

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

P(AB) or P(A ∩ B)

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

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

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


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

TotalNot a college graduate

(N)College

graduate (G)

27207Male (M)

If one employee is selected at random what is the probability

1394Female (F)

402911Total

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

SolutionSolution



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


If events A and B are independent, their joint probabilitysimplifies from

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

S ti k th j i t b bilit f t t ASometimes we know the joint probability of two events Aand B, in this case, the conditional probability of B given Aor A given B is

)()( BAPBAP ∩=

)()( BAPABP ∩=

0 )P( and 0 )P( given that ≠≠ BA)(

)(BP )(

)(AP


Example: According to a survey, 60% of all homeowners owemoney on home mortgages. 36% owe money on both homemortgages and car loans Find the conditional probability thatmortgages and car loans. Find the conditional probability thata homeowner selected at random owes money on a car loangiven that he owes money on a home mortgage.S l tiSolution:


Example: A computer company has two quality controlinspectors, Mr. Smith and Mr. Robertson, who independentlyinspect each computer before it is shipped to a client Theinspect each computer before it is shipped to a client. Theprobability that Mr. Smith fails to detect a defective PC is .02while it is .01 for Mr. Robertson. Find the probability that bothinspectors will fail to detect a defective PCinspectors will fail to detect a defective PC.Solution:


Example: The probability that a patient is allergic to Penicillinis .2. Suppose this drug is administrated to three patients.FindFinda) The probability that all three of them are allergic to itb) At least one of them is not allergicS l tiSolution:

49

Chapter 4: Union of Events & Addition Rule

Union of events

Let A and B be two events defined in a sample space S. Theunion of events A and B is the collection of all outcomes thatunion of events A and B is the collection of all outcomes thatbelong either to A and B or to both A and B and is denotedby “ A or B” or “A ∪ B”

S

A B

S

50


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

Solution

51


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 studentsmust take a course in ethics as a requirement for graduation. Threemust take a course in ethics as a requirement for graduation. Threehundred faculty members and students from this university wereasked about their opinion on this issue and it shown in thefollowing tablefollowing table.

Opinion

TotalNeutralOpposeFavor

70101545Fac lt 70101545Faculty

2303011090Student

30040125135Total

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

52


Solution:

Opinion

TotalNeutral (C)Oppose (B)Favor (A)

Faculty (F)

Student (S)

Total

53


Example: There are a total of 7225 thousand persons withmultiple jobs in the US. Of them, 4115 thousand are male, 1742thousand are single and 905 thousand are male and singlethousand are single, and 905 thousand are male and single.What is the probability that a selected person is a male orsingle?S l tiSolution

TotalMarried (B)Single (A) TotalMarried (B)Single (A)

Male (M)

Female (F)

TotalTotal

54


The probability of the union of two mutually exclusiveevents A and B is

P(A B) P(A) P(B)P(A ∪ B) = P(A) + P(B)

Example: A university president has proposed that allp y p p pstudents must take a course in ethics as a requirement forgraduation. Three hundred faculty members and studentsfrom this university were asked about their opinion on thisfrom this university were asked about their opinion on thisissue and it shown in the following table.

Opinion

TotalNeutralOpposeFavor

70101545Faculty

2303011090Student

30040125135T t l

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

30040125135Total

55


Solution:

56


Example: Eighteen percent of the working lawyers in theUnited States are female. Two lawyers are selected at randomand it is observed whether they are male or femaleand 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 afemalefemale.Solution:

57

Chapter 4: Bayes’ Theorem

Addition Law

)()()()( BAPBPAPBAP ∩−+=∪

Multiplication Law

)()()()(

)()()( BAPBPBAP =∩

)()()( ABPAPBAP ∩

Conditional Probability

)()()( ABPAPBAP =∩

y

)()()(

BPBAPBAP ∩

=)(

)()(AP

BAPABP ∩=

58


Example: Manufacturing firm that receives shipment of partsfrom two different suppliers. Currently, 65 percent of theparts purchased by the company are from supplier 1 and theparts purchased by the company are from supplier 1 and theremaining 35 percent are from supplier 2. Historical Datasuggest the quality rating of the two supplier are shown in thetable:table:

Good Parts Bad Parts

Supplier 1 98 2

a) Draw a tree diagram for this experiment with the

Supplier 1 95 5

g pprobability of all outcomesb) Given the information the part is bad, What is theprobability the part came from supplier 1?probability the part came from supplier 1?

59


Solution:

60


)()()( 11 ABPAPBAP

Bayes’ theorem (two-event case)

)()()()()()()(

2211

111 ABPAPABPAP

ABPAPBAP+

=

)()( ABPAP)()()()(

)()()(2211

222 ABPAPABPAP

ABPAPBAP+

=

Bayes’ theorem

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

2211 nn

iii ABPAPABPAPABPAP

ABPAPBAP+++

=)()()()()()( 2211 nn

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Introduction to Business Statistics QM 120 Chapter 4 · Introduction to Business Statistics QM 120 Chapter 4 Spring 2008. Chapter 4: Experiment, outcomes, and sample space 2