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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.
Chapter 4: Experiment, outcomes, and sample space3
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.
Chapter 4: Experiment, outcomes, and sample space4
● 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:
Chapter 4: Experiment, outcomes, and sample space5
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
Chapter 4: Experiment, outcomes, and sample space6
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
Chapter 4: Calculating Probability8
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 =
Chapter 4: Calculating Probability9
Example: Find the probability of obtaining a head and theprobability of obtaining a tail for one toss of a coin.SolutionSo u io
Chapter 4: Calculating Probability10
Example: Find the probability of obtaining an even number inone roll of a die.SolutionSolution
Chapter 4: Calculating Probability11
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
Chapter 4: Calculating Probability12
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.
Chapter 4: Calculating Probability13
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
Chapter 4: Calculating Probability14
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 =)(
Chapter 4: Calculating Probability15
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
Chapter 4: Calculating Probability16
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:
Chapter 4: Calculating Probability17
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
Chapter 4: Calculating Probability18
Solution
Number of Polishers Rented Number of Days Probability
Chapter 4: Calculating Probability19
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
Chapter 4: Counting Rule21
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
Chapter 4: Counting Rule22
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.
Chapter 4: Counting Rule23
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
Chapter 4: Counting Rule24
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
Chapter 4: Counting Rule25
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
Chapter 4: Counting Rule26
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 −
===
Chapter 4: Counting Rule27
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
Chapter 4: Marginal & Conditional Probabilities29
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?
Chapter 4: Marginal & Conditional Probabilities30
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 )
Chapter 4: Marginal & Conditional Probabilities31
Example: Find the conditional probability P(in favor | male) forthe data on 100 employeesSolutionSolution
Chapter 4: Marginal & Conditional Probabilities32
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==
Chapter 4: Marginal & Conditional Probabilities33
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:
Chapter 4: Independent vs. Dependent events36
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
Chapter 4: Independent vs. Dependent events37
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
Chapter 4: Intersection of Events & Multiplication Rule41
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)
Chapter 4: Intersection of Events & Multiplication Rule42
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
Chapter 4: Intersection of Events & Multiplication Rule43
Chapter 4: Intersection of Events & Multiplication Rule44
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:
Chapter 4: Intersection of Events & Multiplication Rule45
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
Chapter 4: Intersection of Events & Multiplication Rule46
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:
Chapter 4: Intersection of Events & Multiplication Rule47
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:
Chapter 4: Intersection of Events & Multiplication Rule48
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
Chapter 4: Union of Events & Addition Rule
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
Chapter 4: Union of Events & Addition Rule
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
Chapter 4: Union of Events & Addition Rule
Solution:
Opinion
TotalNeutral (C)Oppose (B)Favor (A)
Faculty (F)
Student (S)
Total
53
Chapter 4: Union of Events & Addition Rule
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
Chapter 4: Union of Events & Addition Rule
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
Chapter 4: Union of Events & Addition Rule
Solution:
56
Chapter 4: Union of Events & Addition Rule
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
Chapter 4: Bayes’ Theorem
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
Chapter 4: Bayes’ Theorem
Solution:
60
Chapter 4: Bayes’ Theorem
)()()( 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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