CHAPTER 2 PROBABILITYObjectives1. To be able to describe the concept of set, event and probability.2. To be able to apply multiplication rule and addition rule to calculatethe probability.3. To be able to apply Bayes’ Theorem.2.1 Concept of SetDefinition of ExperimentA process that when performed, results in one and only one of manyobservations.Definition of OutcomeThe observations of an experiment.Definition of Sample SpaceThe set of all possible outcomes of a statistical experiment and presentedby the symbol, S.Definition of Sample PointsThe elements of a sample space.Example 2.1Experiment Outcomes Sample space(a) Roll a die 1,2,3,4,5,6 S = {1,2,3,4,5,6}(b) Select a chip Defective, Non-defective S = {Defective, Non-Defective}The sample space for an experiment can be illustrated by:32 Intro to Statistics & Probability1. Venn Diagram: a picture (closed geometric shape such as circle) thatConsists of all possible outcomes for an experiment.2. Tree Diagram: Each outcome is represented by a branch of a tree.Definition of EventEvent: any subset of a sample space.Simple event: an event that includes one and only one of the finaloutcomes for an experiment.Example 2.2Roll a die three times and observe whether you will get “no.6” or not foreach roll.(a) Find the event that you will get “no.6” at most twice.(b) Find the event that you will not get “no.6” at least twice.(c) Find the event that you will not get “no.6” exactly once.SolutionLet G as get ‘no. 6’.G as do not get ‘no. 6’.Then,.H.THTChapter 2 Probability 3 3{ } 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 S = G G G ,G G G ,G G G ,G G G ,G G G ,G G G ,G G G ,G G GLet A be the event that get ‘no. 6’ at most twice.B be the event that will not get ‘no. 6’ at least twice.C be the event that will not get ‘no. 6’ exactly once.Then,(a) { } 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 A = G G G ,G G G ,G G G ,G G G ,G G G ,G G G ,G G G(b) { } 1 2 3 1 2 3 1 2 3 1 2 3 B = G G G ,G G G ,G G G ,G G G(c) { } 1 2 3 1 2 3 1 2 3 C = G G G ,G G G ,G G GDefinition of Complement EventThe complement of an event A with respect to S is the subset of allelements of S that are not in A and is denoted by the symbol A or A' .Definition of Intersection EventThe intersection of two events A and B denoted by the symbol A∩ B , isthe event containing all elements that are common to A and B.Definition of Union of EventThe union of the two events A and B, denoted by the symbol A∪ B , isthe event containing all elements that belong to A or B or both.Definition of Mutually Exclusive EventsTwo events A and B are mutually exclusive or disjoint if A∩ B =φ or{ } that is, if A and B have no element in common.Exercise 2.1If S = {p, q, r, s,t,u, v,w, x, y} and A = {q, s,t,v, x}, B = {p, r,v,w, y},C = {r,u,w, y} and D = {p,t, x}, list the elements of the setscorresponding to the following events:(a) A∪C34 Intro to Statistics & Probability(b) A∩ B(c) C'(d) (A'∪B)∩ D(e) (B ∩C)'(f) (A∩ B'∩D)∪C2.2 Concept of ProbabilityDefinition of ProbabilityThe likelihood of the occurrence of an event that is measured by usingnumerical value.Definition of Equally Likely OutcomesTwo or more outcomes that have the same probability of occurrence.Example 2.3(a) Roll a die once:The probability of getting no. 1 = probability of getting no. 2 =probability of getting no. 3 = probability of getting no. 4 =probability of getting no. 5 = probability of getting no. 6 =61 .(b) Toss a coin once:The probability of getting a head = probability of getting a tail =21 .TheoremIf an experiment can result in any one of N different equally likelyoutcomes, and if exactly n of these outcomes correspond to event A, thenthe probability of event A isNP(A) = nExample 2.4Chapter 2 Probability 3 5A mixture of marbles contains 5 red marbles, 7 green marbles and 8yellow marbles. If a person makes a random selection of one of thesemarbles, find the probability of getting(a) a green marble(b) a yellow marbleSolutionTotal of marbles = 20(a)20P(G) = 7(b)5220P(Y) = 8 =DefinitionThe probability of an event A is the sum of the weights of all samplepoints in A. Therefore,0 ≤ P(A) ≤ 1, and P(S) = 1For an impossible event, M: P(M) = 0For a sure event, C: P(C) = 12.3 Marginal and Conditional ProbabilityDefinition of Marginal ProbabilityThe probability of a single event without consideration of any otherevent.Definition of Conditional ProbabilityThe probability that an event will occur given that another event hasalready occurred.If A and B are two events, then36 Intro to Statistics & ProbabilityMarginal Probability: P(A) or P(B)Conditional Probability:P(A| B) - read as “The probability of A given that B hasalready occurred”.P(B | A) - read as “The probability of B given that A hasalready occurred”.Example 2.5Samples of bottles from three companies are classified for conformanceto proper filling heights. The results from 115 samples are summarized asfollows:ConformsYes (Y) No (N) Total1 (C1) 27 6 332 (C2) 28 9 373 (C3) 35 10 45CompanyTotal 90 25 115Marginal Probability:(a) P(sample of bottles from company 1) = P(C1) =11533(b) P(C2) =11537(c)115P(C3) = 45(d)115P(Y) = 90(e)115P(N) = 25Conditional Probability:(f) P(sample of bottles from company 1 given that it conforms toproper filling heights)Chapter 2 Probability 3 7= P(C1| Y)no.of samples that conform to proper filling heights= no.of samples fromcompany 1and they conform to proper filling heights90= 27(g)90P(C2 | Y) = 28(h)90P(C3 | Y) = 35(i)25P(C1| N) = 6(j)25P(C2 | N) = 9(k)25P(C3 | N) = 10(l)33P(Y | C1) = 27(m)37P(Y | C2) = 28(n)45P(Y | C3) = 35(o)33P(N | C1) = 6(p)37P(N | C2) = 9(q)45P(N | C3) = 102.4 Mutually Exclusive EventsTwo events that cannot occur together are called mutually exclusiveevents.38 Intro to Statistics & ProbabilityExample 2.6A statistical experiment has 15 equally likely outcomes that are denotedby a, b, c, d, e, f, g, h, i, j, k, l, m, n and o. Let event A = {d, f, g, j, k, m, n,o}, B = {c, d, e, f, o}, C = {k, a, c, j} and D = {k, n, m, a, g}.Are events A & B mutually exclusive?Are events A & C mutually exclusive?Are events B & C mutually exclusive?Are events D & B mutually exclusive?SolutionA and B are not mutually exclusive since A∩ B = {d, f ,o}A and C are not mutually exclusive since A∩C = {k, j}B and C are not mutually exclusive since B ∩C = {c}D and B are mutually exclusive since D ∩ B = { } or φ2.5 Independent and Dependent EventsTwo events are said to be independent if the occurrence of one does notaffect the probability of the occurrence of the other.A and B are independent if eitherP(A | B) = P(A) or P(B | A) = P(B)Otherwise two events are said to be dependent.Example 2.7A company had checked causes of defects in 40 computer keyboards lastweek. Of these, 16 have scratches and 15 have spray marks. Of the 16computer keyboards with scratches, 5 have spray marks. Are the events“scratches” and “spray marks” independent? Are they mutuallyexclusive?Chapter 2 Probability 3 9SolutionScratchesS S TotalM 5 10 15M 11 14 25SprayMarksTotal 16 24 40If M and S are independent, then P(M | S) = P(M) or P(S | M) = P(S) .We show only one, that is P(M | S) = P(M) .,16P(M | S) = 5 while8340P(M) = 15 =∴P(M | S) ≠ P(M)Then, M and S are not independent.M and S are not mutually exclusive events since they both can occurtogether.Exercise 2.2A certain national car comes equipped with either an automatic or amanual transmission, and the car is available in one of three colors.Relevant frequency for various combinations of transmission type andcolor are given in the accompanying table.ColorWhite Blue RedTransmissionTypeAutomaticManual425345460278321433Are the events “Automatic” and “Red” mutually exclusive?40 Intro to Statistics & ProbabilityAre the events “Manual” and “Blue” independent?Exercise 2.3A case contains a total of 150 batteries that were manufactured on threemachines. Of them, 40 were manufactured on machine 1. Of the totalbatteries, 100 meet company’s specifications. Of the 40 batteries thatwere manufactured on machine 1, 20 meet company’s specificationswhile of the 80 batteries that were manufactured on machine 3, 60 meetcompany’s specifications. Are the events “do not meet company’sspecifications” and “machine 2” independent?2.6 Complementary EventsThe complement of event A, denoted by A or A' is the event thatincludes all the outcomes for an experiment that are not in A. Twocomplementary events are always mutually exclusive.Take note thatP(A) + P(A) = 1 or P(A) + P(A') = 1andP(A) = 1− P(A) = 1− P(A')2.7 Multiplication RuleThe probability of the intersection of two events is called their jointprobability. It is written as P(A∩ B).The probability will be:( ) ( ) ( | )( ) ( ) ( | ) orP A B P B P A BP A B P A P B A∩ =∩ =In other words, conditional probability will beChapter 2 Probability 4 1( )( | ) ( )( )( | ) ( )P AP B A P A BP BP A B P A B∩=∩=given that P(A) ≠ 0 and P(B) ≠ 0 .Multiplication Rule for Mutually Exclusive EventsIf A and B are two mutually exclusive events, thenP(A∩ B) = 0Example 2.8Disks from 3 suppliers are analyzed for surface finish.Surface FinishExcellent Good BadSupplier 1 140 450 80Supplier 2 60 250 65Supplier 3 200 300 70Suppose one disk is selected at random. Find the following probabilities:(a) P(supplier 1 and good)(b) P(bad and supplier 3)(c) P(supplier 1 and supplier 3)SolutionSurface FinishExcellent(E)Good(G)Bad(B)TotalSupplier 1(S1)140 450 80 67042 Intro to Statistics & ProbabilitySupplier 2(S2)60 250 65 375Supplier 3(S3)200 300 70 570SupplierTotal 400 1000 215 1615(a)323901615450161510001000( 1 ) ( 1| ) ( ) 450 = = ⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝P S ∩G = P S G P G = ⎛(b)323141615P(B ∩ S3) = 70 =(c) P(S1∩ S3) = 0Example 2.9The probability that a sample of fly ash contains sulfate is 0.80; theprobability that this sample contains calcium is 0.78; and the probabilitythat this sample contains both components is 0.75. Find the probabilitythat this sample of fly ash will containa. sulfate given that this sample contains calcium.b. calcium given that this sample contains sulfate.SolutionP(S) = 0.8, P(C) = 0.78, P(S ∩C) = 0.75(a) 0.96150.780.75( )( | ) ( ) = =∩=P CP S C P S C(b) 0.93750.80.75( )( | ) ( ) = =∩=P SP C S P C SExercise 2.4A box contains 30 computer disks, 10 of which are defective. If 3computer disks are selected at random and without replacement from thisbox, what is the probability that exactly 2 non-defective computer disksare chosen?Chapter 2 Probability 4 3Exercise 2.5One bag contains 6 white marbles, 7 black marbles and 7 green marbles.A second bag contains 8 white marbles, 5 black marbles and 4 greenmarbles. Ahmad selects a marble from the first bag and places it in thesecond bag. A marble is now drawn from the second bag by Azli. Whatis the probability that the marble drawn by Azli is green?Multiplication Rule for Independent EventsThe probability of the intersection of two independent events A and BisP(A∩ B) = P(A)P(B)Example 2.10Azila will be having 2 mid-term exams in a week. The probability thatshe will pass an Islamic Civilization subject is 0.92, and the probabilitythat she will pass an IT subject is 0.88. Assume that the events she willpass any of these two subjects are independent. Find the probability that(a) she will pass both subjects.(b) she will pass neither Islamic Civilization nor IT subject.SolutionLet I be the event that Azila will pass an Islamic Civilization subject.T be the event that Azila will pass an IT subject.P(I ) = 0.92, P(T) = 0.88(a) P(I ∩T) = P(I )P(T) = (0.92)(0.88) = 0.8096(b) P(I ∩T) = P(I )P(T ) = (0.08)(0.12) = 0.0096Exercise 2.6If P(C) = 0.65, P(D) = 0.4 and P(C ∩ D) = 0.24 , are the events C andD independent?44 Intro to Statistics & ProbabilityExercise 2.7The probability that a new car needs a service within 6 months ofpurchase is 0.4. Three independent cars are randomly selected.(a) What is the probability that all cars selected will need a servicewithin 6 months of purchase?(b) What is the probability that exactly 2 cars selected will need aservice within 6 months of purchase?(c) What is the probability that two or less cars selected will not needa service within 6 months of purchase?Exercise 2.8A coin is biased so that a tail is 5 times as likely to occur as a head. If thecoin is tossed 3 times, what is the probability of getting 1 tail and 2heads?2.8 Addition RuleThe probability of the union of two events A and B isP(A∪ B) = P(A) + P(B) − P(A∩ B)Addition Rule for Mutually Exclusive EventsThe probability of the union of two mutually exclusive events A and B isP(A∪ B) = P(A) + P(B)Example 2.11All 200 cars were rented by a consulting firm from 3 agencies A, B andC. The cars were checked whether they have bad tyres or not. Based onthe information given, the following two-way classification table wasprepared.Agency Have bad tyres Do not have badtyresA 10 38B 17 80Chapter 2 Probability 4 5C 15 40Suppose one car is selected at random from this firm. Find the followingprobabilities:a. P(a car selected comes from agency A or does not have bad tyres)b. P(a car selected has bad tyres or comes from agency C)c. P(a car selected comes from Agency A or B)SolutionAgency Have bad tyres(T)Do not havebad tyres(T )TotalA 10 38 48B 17 80 97C 15 40 55Total 42 158 200(a)252120038200158200P(A∪T) = P(A) + P(T) − P(A∩T) = 48 + − =(b)100412001520055200P(T ∪C) = P(T) + P(C) − P(T ∩C) = 42 + − =(c)402920097200P(A∪ B) = P(A) + P(B) = 48 + =Exercise 2.9A pair of dice is thrown twice. What is the probability of getting totals of6 and 9?2.9 Bayes’ TheoremIn general, a collection of sets E1, E2 , E3 ,..., Ek such thatE1 ∪ E2 ∪ E3...∪ Ek = S is said to be exhaustive.Total Probability Rule46 Intro to Statistics & ProbabilityAssume E1, E2 , E3 ,..., Ek are k mutually exclusive and exhaustive sets.Then,P(B) = P(B ∩ E1 ) + P(B ∩ E2 ) + ... + P(B ∩ Ek )= P(B | E1 )P(E1 ) + P(B | E2 )P(E2 ) + ... + P(B | Ek )P(Ek )Example 2.12In a certain assembly plant, four machines, A, B, C and D make 25%,20%, 35% and 20% respectively, of the products. It is known from pastexperience that 2%, 4%, 1% and 0.5% of the products made by eachmachine, respectively, are defective. Now, suppose that a finishedproduct is randomly selected. What is the probability that it is defective?SolutionP(A) = 0.25, P(B) = 0.2, P(C) = 0.35, P(D) = 0.2P(DF | A) = 0.02, P(DF | B) = 0.04, P(DF | C) = 0.01,P(DF | D) = 0.005Then,0.0175(0.02)(0.25) (0.04)(0.2) (0.01)(0.35) (0.005)(0.2)( ) ( | ) ( ) ( | ) ( ) ( | ) ( ) ( | ) ( )== + + +P DF = P DF A P A + P DF B P B + P DF C P C + P DF D P DBayes’ TheoremFrom previous, we know that,( )( | ) ( )P BP A B P A B ∩= and( )( | ) ( )P AP B A P B A ∩=then P(A∩ B) = P(B ∩ A) = P(A | B)P(B) = P(B | A)P(A)and we also can writeChapter 2 Probability 4 7( )( | ) ( | ) ( )P BP A B = P B A P AIn general, we obtain the following result, which is known as Bayes’TheoremIf E1, E2, …, Ek are k mutually exclusive and exhaustive events and B isany event, then( | ) ( ) ( | ) ( ) ... ( | ) ( )( | ) ( | ) ( )1 1 2 21 11P B E P E P B E P E P B Ek P EkP E B P B E P E+ + +=Exercise 2.10A firm rents cars from three companies - 20% from company 1, 20%from company 2 and 60% from company 3. If 10% of the cars fromcompany 1, 12% of the cars from company 2 and 4% of the cars fromcompany 3 have bad conditions, what is the probability that(a) the firm will get a car with bad conditions?(b) a car with bad conditions rented by the firm came from company 2?(c) a car without bad conditions rented by the firm came fromcompany 3?Exercise 2.11Marketing managers must estimate the sales of a new model of onetelephone. The records of one major telecommunication companyindicate that 10% of all new telephones sell more than projected, 30%sell close to projected, and 60% sell less than projected. Of those that sellmore than projected, 70% are proposed for additional new features, asare 50% of those that sell close to projected, and 20% of those that sellless than projected.(a) What percentage of telephones produced by this company willbe proposed for additional new features?(b) What percentage of telephones produced by this company thatare proposed for additional features sold less than projected in theirfirst edition?Exercise 2.1248 Intro to Statistics & ProbabilityIt is known from past experience that the probability of selecting an adultover 60 years of age who are smokers is 0.35. Of those adults over 60years of age who are smokers, 55% of them have heart attack. Of thoseadults over 60 years of age who are nonsmokers, 12% of them have heartattack. What is the probability of selecting one of these adults with heartattack is found to be a nonsmoker? What is the probability of selectingone of these adults without heart attack is found to be a smoker?Review Exercises1. List the elements of each of the following sample spaces:(a) the set of integers between 1 and 50 divisible by 3(b) the set S = {x | x3 − x2 − 4x + 4 = 0}(c) the set 9}2{ | 1 3 5 ≤−S = x − ≤ x2. Consider the sample space S = {Toyota, Honda, Mercedes, BMW,Naza, Ferrari, Renault} and the eventsA = {Toyota, Honda, Renault}B = {Honda, Mercedes, BMW}C = {Ferrari}List the elements of the sets corresponding to the following events:(a) B'(b) (A'∩B')∪C(c) A∩ B ∩C'(d) (B ∪C)'(e) (B ∩C)∪ A'(f) (A'∩B')∪(B ∩C')3. When a machine goes down, there is 60% chance that it is due to atechnical problem and a 50% chance that it is due to a human errorproblem. There is a 40% chance that it is due to a technical and ahuman error problem.(a) What is the probability that at least one of these problems are atfault?(b) What is the probability that there is a technical problem but nohuman error problem?Chapter 2 Probability 4 94. Samples of banners produced by a printing company are classified onthe basis of colour intensity and print quality. The results of fivehundred banners are summarized as follows:Print QualityGood BadColour Good 359 57Intensity Bad 77 7Let X denotes the event that a banner has good colour intensity, andlet Y denotes the event that a banner has good print quality. If abanner is selected at random, determine the following probabilities:(a) P(X )(b) P(Y)(c) P(X ')(d) P(X ∩Y)(e) P(X ∪Y)(f) P(X ∪Y')(g) P(X | Y)(h) P(Y | X )(i) If the selected banner has good colour intensity, what is theprobability that the print quality is good?(j) If the selected banner has bad print quality, what is the probabilitythat the colour intensity is good?5. Assume that 2% of all unacceptable cereal cartons taken from aproduction are due to unreadable label and assume that 1% of all theseunacceptable cereal cartons are due to improper package weight. Also,2.5% of this unacceptable cereal cartons will experience at least one ofthese problems. What is the probability that for a randomly selectedunacceptable cereal cartons, improper package weight will beidentified as the cause but there will be no problem due to unreadablelabel?6. Suppose that in an assembly of 500 defective computers, it is foundthat 230 are due to overload, 245 are due to software problem, 208 aredue to hardware problem, 132 are due to overload and softwareproblems, 67 are due to hardware and software problems, 75 are dueto overload and hardware problems and 45 defective computers aredue to all 3 problems. If a defective computer is selected at random50 Intro to Statistics & Probabilityfrom this assembly, find the probability that this computer isdefective due to:(a) overload but not due to software problems.(b) overload and hardware problems but not due to softwareproblems.(c) neither software nor hardware problems.7. If P(A) = 0.6, P(B) = 0.3 and P(A∩ B) = 0.05 , determinethefollowing probabilities:(a) P(A')(b) P(A∪ B)(c) P(A'∩B)(d) P(A∩ B')(e) P[(A∪ B)'](f) P(A'∪B)8. In an inspection of cans of babies’ food, it was found that 8% showedsigns of cracks on top of cans and surface defects on side of cans,30% showed cracks on top of cans, and 40% showed surface defectson side of cans.(a) If a can of baby’s food has surface defects on side of it, what isthe probability that this can will also have cracks on top of it?(b) If a can of baby’s food does not have cracks on top of it, what isthe probability that this can has surface defects on side it?9. Suppose that for a given computer program, there is a 40% chancethat computational errors will occur and a 28% chance thatconfiguration errors will occur. If the computational errors occur,there is an 80% chance that configuration errors will occur.(a) Find the probability that for given computer programs,both computational and configuration errors will not occur.(b) Find the probability that the computational errors will occur giventhat the configuration errors do not occur.10. A batch of 500 soft drink bottles from a bottling company contains10 bottles that are defective. Three bottles are selected, at random,without replacement, from the batch.(a) What is the probability that the third one selected isdefective given that the first was okay and second oneselected was defective?Chapter 2 Probability 5 1(b) What is the probability that the third one selected is okay giventhat the first one selected was defective and the second oneselected was okay?(c) What is the probability that exactly 1 defective bottle is selected?11. Two ordinary dice are thrown. Find the probability that the sum ofthe scores obtained is(a) a multiple of 5.(b) greater than 9.(c) multiple of 5 or is greater than 9.12. One bag contains 8 white and 9 black marbles. A second bagcontains 11 white, 10 black and 12 green marbles. If 1 marble isselected at random from each bag, find the probability that(a) all marbles are black.(b) neither marbles are white.(c) there are 2 marbles with different colours.13. A first year student has final exams for three consecutive days. Theprobability that he will fail Social Ethnic subject is 0.05%, theprobability that he will fail Chemistry is 2% while the probability hewill fail an IT subject is 1%. Assume that the events that he will failany of these three subjects are independent.(a) What is the probability that this student will not fail any of thesesubjects?(b) What is the probability that this student will fail either Chemistryor IT subject?14. A pharmaceutical manufacturing process uses a high speedcompressing machine as part of the process of tablet production.The company will use machine A, B, C and D with probabilities0.23, 0.45, 0.19 and 0.13 respectively. From past experience it isknown that the probability of defective tablet produced by themachines are 0.11, 0.2, 0.07 and 0.05 respectively. Suppose adefective tablet is produced:(a) What is the probability that it is produced by machine D?(b) What is the probability that it is produced by machine B?15. An experiment was conducted to assess the effect of using magnetsat the filler point in the manufacturing of tea filter packs. Usage ofmagnets at the filler point produces 80% of the tea filter packs.From past experience, the technician who controls the filler52 Intro to Statistics & Probabilitypoint discovers that the usage of magnet will affect the weights offilter packs where 5% of the filter packs with weight less than 20gare produced while without using magnets, 3% of filter packs withweight less than 20g are produced. Suppose a filter pack weighingmore than 20g is produced. Which of the two methods (withmagnet and without magnet) is more likely to have supplied thisfilter pack weighing more than 20g?16. Suppose that five workers at a soft drink company are supposedto paste expiration date on each can at the end of the assembly line.A, who pastes 30% of the cans, fails to paste the expiration dateonce in every 300 cans, B, who pastes 15% of the cans, fails topaste the expiration date once in every 250 cans, C, who pastes35% of the cans fails to paste the expiration date once in every 100cans, D, who pastes 7% of the cans, fails to paste the expirationdate once in every 150 cans and E, who pastes 13%, fails to pastethe expiration date once in every 200 cans. If a quality controlinspector finds a can without an expiration date, what is theprobability that C fails to paste it?17. Computer centre had three printers that print at different speeds.Programs are routed to the first available printer.Let A – printer AB – printer BC – printer CJ – printer jamD – destroy the quality of print outAssume:P(A) = 0.6, P(C) = 0.1P(J | A) = 0.2, P(J '| B) = 0.7, P(J | C) = 0.15P(D | J ∩ A) = 0.02, P(D'| J '∩ A) = 0.97, P(D'| J ∩ B) =0.95P(D | J '∩B) = 0.01, P(D | J ∩C) = 0.01, P(D | J '∩C) =0.015(a) Find P(A∩ J '∩D) .(b) Find P(D') .(c) Find P(J ∩ D) .Chapter 2 Probability 5 3(d) Find the probability that a print out came from printer C giventhat the quality of print out is not destroyed but the printer is jam.
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