Solution Manual For Probability, Statistics, and Random Processes for Engineers, 4th Edition

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1Solutions to Chapter 11. The intent of this rather vague problem is to get you to compare the two notions, probabilityas intuition and relative frequency theory.There are many possible answers to how tomake the statement "Ralph is probably guilty of theft" have a numerical value in the relativefrequency theory.First step is to define a repeatable experiment along with its outcomes.The favorable outcome in this case would be ’guilty.’Repeating this experiment a largenumber of times would then give the desired probability in a relative frequency sense.Wethus see that it may entail a lot of work to attach an objective numerical value to such asubjective statement, if in fact it can be done at all.One possible approach would be to look through courthouse statistics for cases similar toRalph’s, similar both in terms of the case itself and the defendant.If we found a sufficientlylarge number of these cases, ten at least, we could then form the probability=, whereis the number of favorable (guilty) verdicts,andis the total number of found cases.Here we effectively assume that the judge and jury are omniscient.Another possibility is tofind a large number of people with personalities and backgroundssimilar to Ralph’s, and to expose them to a very similar situation in which theft is possible.The fraction of these people that then steal in relation to the total number of people, wouldthen give an objective meaning to the phrase "Ralph is probably guilty of theft."2. Note that→3, but36→, i.e.,implies3but not the other way around. Thus if weturn over card 2 andfind a3. So what? It was never stated that a3→. Likewise, withcard 3.On the other hand, if we turn over card 4 andfind a, then the rule is violated.Hence, we must turn over card 4 and card 1, of course.3. First step here is to decide which kind of probability to use.Since no probabilities areexplicitly given, it is reasonable to assume that all numbers are equally likely.Effectivelywe assume that the wheel is “fair."This then allows us to use the classical theory alongwith the axiomatic theory to solve this problem. Now we mustfind the corresponding prob-ability model.We are told in the problem statement that the experiment is “spinning thewheel." We identify the pointed-to numbers as the outcomes.The sample space is thusΩ={123456789}The total number of outcomes is then 9. The probability of eachelemental event{}is then taken as[{}],= 19, as in the classical theory. We are alsotold in the problem statement that the contestant wins if an even number shows. The set ofeven numbers inΩis{2468}We can write this event as a disjoint union of four singleton(atomic) events{2468}={2}∪{4}∪{6}∪{8}Now we can apply axiom 3 of probability to write[{2468}]=[{2}] +[{4}] +[{6}] +[{8}]=19 + 19 + 19 + 19=49We have seen that some ’reasonable’ assumptions are necessary to transform the given wordproblem into something that exactly corresponds to a probability model.It turns out thatthis is a general problem for such word problems, i.e. problems given in natural English.

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24. The experiment involvesflipping a fair coin 3 times. The outcome of each coin toss is eithera head or a tail. Therefore, the sample space of the combined experiment that contains allthe possible outcomes of the 3 tosses, is given byΩ={               }Since all the coins are fair, all the outcomes of the experiment are equally likely. The proba-bility of each singleton event, i.e. an event with a single outcome, is then18. We are interestedinfinding the probability of the event, which is the event of obtaining 2 heads and 1 tail.There are 3 favorable outcomes for this event given by={    }. Therefore,[] =[{}∪{ }∪{ }] =[] +[ ] +[ ] =38. Note thatwe are able to write the probability of the eventas the sum of probability of the singletonevents (from Axiom 3) because the singlteon events of any experiment are mutually exclusive.Why?5. The experiment contains drawing two balls (with replacement) from an urn containing ballsnumbered1,2, and3. The sample space of the experiment is given byΩ={111213212223313233}The event of drawing a ball twice is said to occur when one of the outcomes11,22, or33occurs.Therefore, the event of drawing 2 equal balls, is given by={112233}and[] =[{11}]+[{22}]+[{33}]Since the balls are drawn at random, it can assumed thatdrawing each ball is equally likely. Therefore, the singleton events, or equivalently outcomesof the experiment, are equally likely. Hence,[] = 3(19) =13.6. Let1 2     6represent the six balls. Each outcome will be represented by the two ballsthat were drawn.In thefirst experiment, the balls are drawn without replacement; hence,the two balls drawn cannot have the same index. Then the sample space containing all theoutcomes is given byΩ1={12 13 14 15 1621 23 24 25 2631 32 34 35 3641 42 43 45 4651 52 53 54 5661 62 63 64 65}This can be written compactly asΩ1={()|1≤≤61≤≤6 6=}If thefirst ball is replaced before the second draw, then in addition to the outcomes in theearlier part, there are outcomes where both the two balls drawn are the same. The samplespace for the new experiment is given byΩ2=Ω1∪{11 22 33 44 55 66}This can also be written asΩ2={()|1≤≤61≤≤6}.

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37. Letbe the height of the man andbe the height of the woman. Each outcome of theexperiment can be expressed as a two-tuple( ). Thus(a) The sample spaceΩis the set of all possible pairs of heights for the man and woman.This is given asΩ={( ) :0 0}(b) The event, which is a subset ofΩis given by={( ) :0 0  }8. The word problem describes the physical experiment of drawing numbered balls from an urn.We need tofind a corresponding mathematical model.First we form an appropriate eventspace with meaningful outcomes.Here the physical experiment is ’draw ball from urn,’ sothe outcome in words is ’particular labeled ball drawn,’ which we can identify with its label.So we select asoutcomein our mathematical model, the number on the drawn ball’s face,i.e. the particular label. The outcomes are thus the integers 1,2,3,4,5,6,7,8, 9, and 10. Thesample spaceis thenΩ={12345678910}and is the set of all ten outcomes.Weare told thatis ’the event of drawing a ball numbered no greater than 5.’ Thus we definein our eventfield={12345}The other event specified in the word problem is’theevent of drawing a ball greater than 3 but less than 9.’ In our mathematical eventfield thiscorresponds to={45678}Having constructed our sample space with indicated events,we can use elementary set theory to determine the following answers:={678910}={123910}={45}∪={12345678}={123}={678}∪={123678910}()∪() ={123678}()∪() ={45910}(∪)={910}()={123678910}The last part of the problem asks us to ’express these events in words.’Since we have amathematical model, we should really more precisely ask what each of these eventscorrespondsto in words. We know of course thatcorresponds to ’drawing a ball numbered no greaterthan 5.’We can thus loosely write={0drawing a ball numbered no greater than 5’},although in our mathematical modelis just the set of integers{12345}.So whenwe write={ ’drawing a ball numbered no greater than 5’}, what we really mean is thatthe eventin our mathematical model corresponds to the physical event ’drawing a ballnumbered no greater than 5’ mentioned in the word problem. With this caveat in mind, wecan then write:={0drawing a ball greater than 50}={0drawing a ball not in the range 4-8 inclusive0}={0drawing a ball greater than 3 and no greater than 50}etc.9. The sample space containing four equally likely outcomes is given byΩ={1 2 3 4}. Twoevents={1 2}and={2 3}are given. The required events can be easily obtainedby observation.

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4= set of outcomes inand not in={1}.= set of outcomes inand not in={3}.= set of outcomes inand={2}.∪= set of outcomes inor in={1 2 3}.10.=∪This can be proved using the distributive law on=Ω=(∪) =∪∪= ()∪()∪()Here wefirst write=(∪)and=(∪)Then we can write∪=((∪))∪((∪))=(∪)∪(∪)=∪∪∪=∪∪using the above laws and formulas. Notice that the above two decompositions are into disjointsets. From the third axiom of probability, we know that the probability of union of disjointsets is the sum of the probabilities of the disjoint sets. Therefore, we can add the probabilitiesover the unions.11. In a given random experiment there are fourequally likelyoutcomes1 2 3and4Letthe event,{1 2}[] =[{1 2}] =[{1}] +[{2}] =14+14=12={3 4}[] =[{3 4}] =[{3}] +[{4}] =14+14=12Note that we are told that the four outcomes are equally likely.This means that the foursingleton (atomic) events have equal probability.[] =12= 1−[] = 1−1212. (a) The three axioms of probability are given below(a) [label=()](b) For any event, the probability of the even occuring is always non-negative.[]≥0This ensures that probability is never negative.(c) The probability of occurence of the sample space eventΩis one.[Ω] = 1This ensures that probability of no event exceeds one. Thefirst two axioms ensures thatthe probability is a quantity between0and1, inclusive.(d) For any two events that are disjoint, the probability of the union of the events isthe sum of the probabilities of the two events.[∪] =[] +[]when=This axiom tells us that the probability of any event can be obtained by the sum disjointevents that constitute the event.

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5(b) The event∪can be obtained as the disjoint union of the three sets  .Hence by applying the third axiom of probability, we obtain[∪]=[∪(∪)]=[] +[∪]=[] +[] +[]Now the eventcan be written as the disjoint union ofand(Axiom 3). Therefore[] =[] +[] =⇒[] =[]−[]Similarly[] =[] +[] =⇒[] =[]−[]Therefore[∪] =[] + ([]−[]) + ([]−[]) =[] +[]−[].13. Wefirst form our mathematical model by setting outcomesς= (1 2)where1correspondsto the label on thefirst ball drawn, and2corresponds to the label on the second ball drawn.We can also write the outcomes as stringsς=12The sample spaceΩcan then be identifiedwith the 2-D array11121314152122232425313233343541424344455152535455There are thus 25 outcomes in the sample space.Now the word problem statement usesthe phrase ’at random’ to describe the drawing.This is a technical term that can be read’equally likely.’Thus all the elementary events{12}in our mathematical model must haveequal probability, i.e.[{12}] = 125Armed thusly we can attack the given problem asfollows. Define the event={’sum of labels equalsfive’}, or precisely={41322314}Then we decompose this event into four singleton events as={41}∪{32}∪{23}∪{14}Since different singleton events are disjoint, probability adds, and we have[]=125 + 125 + 125 + 125=425"Dim" ignored that outcomeis different (distinguishable) from outcome."Dense" talkedabout the sums and correctly noted that there were nine of them.However, he incorrectlyassumed that each sum was equally likely.Looking at our sample space above, we cansee that the sum 2 has only one favorable outcome 11, while the sum 6 hasfive favorableoutcomes, just looking at the anti-diagonals of this matrix.14. First we show∩(∪)⊂(∩)∪(∩).Let∈∩(∪).Then∈and∈(∪).∈and∈or∈.

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6Say if∈. Then∈and∈(Step k)Thus∈(∩).And therefore∈(∩)∪(∩). Similar arguments can be made if we consider∈in step k, in which case we will show that∈(∩)and hence∈(∩)∪(∩).Thus we have shown that∩(∪)⊂(∩)∪(∩).Now we show that(∩)∪(∩)⊂∩(∪).Suppose∈(∩)∪(∩). Then∈(∩)or∈(∩).Say∈(∩)Then∈and∈.Or∈and∈(∪). Or in other words,∈∩(∪).Similar arguments can be used to show that if∈(∩), then∈∩(∪).Thus(∩)∪(∩)⊂∩(∪).Thus we have shown that both sets are contained in each other.Hence∩(∪) =(∩)∪(∩).15. We use the set identityΩ=∪Since this union is disjoint, by the additivity of probability(i.e. axiom 3), we get1 =[Ω] =[] +[]which with rearranging becomes the desiredresult.16.(a)∩={12}∩{456}=. Therefore,[∩]=[]=1−[Ω](∵Ω∩Φ=1 =[Ω∪] =[Ω] +[])=1−1 (because[Ω] = 1)=0(b)[∪∪] =[{12}∪{23}∪{456}=[{123456}] =[Ω] = 1.(c) We see that∩=and so[] = 0. Forandto be independent,[] =[][]. Therefore, if either[] = 0or[] = 0or both are zeros,andwill beindependent.17. This problem uses only set theory and just two axioms of probability to get these generalresults.(a) We need to show[] = 0We write the disjoint decompositionΩ=Ω∪and thenuse the additivity of probability (axiom 3) to get[Ω]=[Ω∪]=[Ω] +[]So we must have[] = 0(b) Using set theory, we can write the disjoint decomposition=∪Then by axiom 3, the additivity of probability, we have[]=[∪]=[] +[]

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7or what is the same[] =[]−[](c) Here we simply note∪=Ωis a disjoint decomposition, so that again by axiom 3,[Ω]=[] +[]=1by axiom 2,which is the same as[] = 1−[]18. The outcome is the result of a probabilistic experiment.An event is a collection (set) ofoutcomes.Thefield of events is the complete collection of events that are relevant for thegiven probability problem.19. We start with the mutually exclusive decomposition∪=∪∪yielding[∪] =[] +[] +[]Then consider the two simple disjointdecompositions=∪and=∪which yield[] =[] +[]and[] =[] +[]Putting them alltogether, we have[∪]=[] +[] +[]=([]−[]) +[] + ([]−[])=[] +[]−[]20. From Eq. 1.4-3, we see that⊕= (−)∪(−) =∪. We see thatandare disjoint, i.e.,()∩() =.Therefore, the probability of the union ofandare the sum of the probabilities of the two events. In other words,(⊕) =(∪) =() +()21. We have already (Problem 17) seen that we can write[] =[]−[]and[] =[]−[]. Therefore,(⊕) =() +() =[] +[]−2[].22.(a) For simplicity associate as follows: cat=1, dog=2, goat=3, and pig=4. The outcomesthen become the integers 1,2,3, and 4.The sample spaceΩ={1234}For probabilityinformation we are given:[{12}] = 09 [{34}] = 01 [{4}] = 005and[{2}] = 05Now for every event in ourfield of events, we must be able to specify the probability.This is equivalent to being able to supply the probability for all the singleton events. Tosee if we can do this, we note that singleton events {1} and {3} are missing probabilities,so wefirst write{1}={12}−{2}so that[{1}]=[{12}]−[{2}] = 09−05 = 04

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8Doing the same for the other missing singleton probability[{3}], we write{3}={34}−{4}so that[{3}]=[{34}]−[{4}] = 01−005 = 005Thus we have enough probability information for all the singleton events, and hence all16 = 24subsets ofΩ={1234}The appropriatefieldFof events then consists ofthe following events along with their probabilities:{1}[{1}] = 04{2}[{2}] = 05{3}[{3}] = 005{4}[{4}] = 005{12}[{12}] = 09{13}[{13}] = 045{14}[{14}] = 045{23}[{23}] = 055{24}[{24}] = 055{34}[{34}] = 01{123}[{123}] = 095{124}{124}] = 095{134}[{134}] = 05{234}[{234}] = 06{1234}(=Ω)[{1234}] = 1 =[Ω][] = 0(b) Now the above is not an appropriatefield of events if some of the events do not haveknown probabilities.So if[’pig’={4}] = 005is removed, then we cannot determinethe probabilities of some of the above events.In particular we cannotfind[{3}]Thealternative then is to treat{34}, whose probability is still given, as a singleton andform a smallerfield with just the 8 events formed by unions of {1}, {2}, and {3,4}.Theresultingfield, along with its probabilities is as follows:{1}[{1}] = 04{2}[{2}] = 05{12}[{12}] = 09{34}[{34}] = 01{134}[{134}] = 05{234}[{234}] = 06{1234}(=Ω)[{1234}] = 1 =[Ω][] = 023. First we show that∪(∩)⊂(∪)∩(∪).Suppose∈∩(∪)Then∈Therefore∈(∪), and∈(∪)Hence,∈(∪)∩(∪).Now we show that(∪)∩(∪)⊂∪(∩).Suppose∈(∪)∩(∪)

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9Then∈(∪)and∈(∪)∈and∈If∈, then∈∪(∩)(because⊂(∪(∩)))If∈, then∈and∈.Or in other words,∈(∩)∈∪(∩).Thus we have shown that both the sets are contained in each other. Therefore,∪(∩) =(∪)∩(∪).24. The probability ofis[] =[{1 2}] =[{1}] +[{2}] =14+14=12The event (set)in terms of the outcomes is={3 4}The probabilty ofis[] =[{3 4}] =[{3}] +[{4}] =14+14=12Note that we are told that the four outcomes are equallylikely.This means that the four singleton (atomic) events have equal probability.We verify[] =12= 1−[] = 1−1225. The composition of the urn is:(),(),(),(),(),(),(),().[] = 68,[] = 68,[] == 48is not equal to[][] = 916. Thereforeandare not independent.26. Let = 12represent the outcome of theth toss. Since the tosses are independent:[1 2] =[1][2] = 16·16[1+2= 7|1= 3]=[2= 4|1= 3]=[1= 3 2= 4][1= 3]=[1= 3][2= 4][1= 3](because tosses are independent)=1627. Clearly[] =452and[] = 2652 = 12Then[] =[{pick one of two red aces in 52 cards}] =252Is[] =[][]?Now[] =252=45212=[][]so, yesandare independent events.28. Since it is a fair die, the successive tosses are independent with probability= 16for eachface.From the provided information, we equivalently want the probability of getting a totalof 5 on the two remaining tosses.This can happen in just 4 equally likely outcomes, i.e.(4,1), (3,2), (2,3), and (1,4).The desired probability this then436 = 19.

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1029. We can look at the compound outcomesς= (1 2)as corresponding to the locations in the9×9 array112131415161· · ·911222324252· · ·...13233343· · ·142434· · ·1525· · ·. . .16· · ·.... . ....19· · ·· · ·99with 81 equally likely outcomes. We agree to call the sample space for thefirst experimentΩ1, the sample space for the second experimentΩ2, and the compound sample space simplyΩTo get the sumΣ,1+2= 7we need one of the following outcomes162534435261located on a45◦diagonal in the above table.So there are 6 favorable outcomes for the event{Σ= 7}The event{Σ=odd}contains 40outcomes and the event{Σ=even}contains the remaining81−40 = 41even-sum outcomes.Now the joint event{Σ= 7}∩{Σ=odd}={Σ= 7}since the sum 7 is an odd number.Wecan now calculate the needed probabilities[{Σ=odd}] = 4081and[{Σ= 7}] =681The answer for thefirst question is then[{Σ=7}|{Σ=odd}] =[{Σ= 7}∩{Σ=odd}][{Σ=odd}](by definition)=[{Σ= 7}][{Σ=odd}](by above result)=640The next question is tofind[({17} ×Ω2)∪(Ω1× {27})|{Σ10}]For simplicityof notation, let’s agree to write the compound events{17} ×Ω2andΩ1× {27}assimply{17}and{27}respectively, for the rest of this calculation.So we mustcount the relevant number of outcomes from the above9×9array, where the various sums arefound on45◦diagonals.For the event{Σ10}we count 36 outcomes.For the joint event({17}∪{27})∩{Σ10}, wefind it easier to consider the set of outcomes that makeup the remainder of the event{Σ10}i.e. the event{1≤7}∩{2≤7}∩{Σ10}whichis equal, in words, to the event ’1≤7and2≤7andΣ10’.We could call this thecomplement with respect to{Σ10}of the event({17}∪{27})∩{Σ10}Anyway,wefind from the9×9array that the number of outcomes in{1≤7}∩{2≤7})∩{Σ10}is composed of the following 10 cases:P= 11 = 6 + 5 = 5 + 6 = 7 + 4 = 4 + 7andP= 12 = 5 + 7 = 7 + 5 = 6 + 6andP= 13 = 6 + 7 = 7 + 6andP= 14 = 7 + 7So we subtract these 10 outcomes from the 36 outcomes in the event{Σ10}to obtain26 outcomes in the compound event({17}∪{27})∩{Σ10}The relevantprobabilities are then[{Σ10}] = 3681and[({17}∪{27})∩{Σ10}] = 2681

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11The desired conditional probability is then[({17}∪{27})|{Σ10}] = 26813681 = 2636≈072Finally to compute[{Σ=odd}|{18}]we proceed as follows.For the combinedexperiment, we know there is only one possibility for18and that is1= 9, along withany value for2Thus there are 9 outcomes in the compound event{18}1, so that it’sprobability is981Now the joint event{Σ=odd}∩{18}={1= 9}∩{Σ=odd}={(92)(94)(96)(98)}with four outcomes.Thus since all outcomes are equally likely, wehave[{Σ=odd}∩{18}] =481The desired conditional probability is then[{Σ=odd}|{18}] =[{Σ=odd}∩{18}][{18}]=481981 = 49≈04430. We are given that[] = 0001, whereis the event ’disease is present.’Letdenote theevent ’test is positive,’ so thatis the event ’test is negative.’We are additionally given[|] = 1and[|] = 0005We are asked to compute[|], i.e. the probability that’disease is present given the test is positive.’We use Bayes’ rule and Theorem as follows[|]=[][]=[|][][|][] +[|][]=1×00011×0001 + 0005×0999=11 + 4995≈0167Thus in only about 17% of the cases will a positive test result actually confirm that you sufferfrom the disease.The other 83% of the time you will be needlessly worried!31. Let1denote the set of occupations and let2denote the set of interests and/or hobbies.Then1={’office manager’, ’engineer’, ’doctor’, ’teacher’, ...}2={’nat. defense’, ’books’, ’music’, ’cooking’,...}Letdenote Henrietta’s occupation andher interests.Then[=’office manager’ =’nat. defense’]=[=’office manager’][=’nat. defense’|=’office m≤[=’office manager’]since0≤[=’nat. defense’|=’office manager’]≤1.1Remember, we decided above to write simply{1 }for the compound event{1 } ×Ω2This since, inthis problem, we only compute probabilities for events in the compound experiment.

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1232. Directly from the problem statement[= 3]=3·[= 1][= 2]=2·[= 1]But we also know[= 3] +[= 2] +[= 1] = 1which is always true by axiom 2[Ω] = 1.Therefore[= 1] = 16,[= 2] = 13, and[= 3] = 12.Using Bayes’Theorem, we then compute[= 1|= 1]=[= 1|= 1][= 1]P3=1[= 1|=][=]=(1−)16(1−)16+2 13+2 12=1−1−++32 33. Let,{examinee knows},,{examinee guesses}, and,{getting right answer} .Then[] = [] = 1− [|] = 1and[|] = 1So[|]=[|][][]=1·[|][] +[|][]=+1(1−)=+ (1−)34. There arecontestants and only one most beautiful. Hence[{pick most beautiful}] = 135. Lete,{random drawn chip∈},e,{random drawn chip∈}, ande,{random drawn chip∈}.Also, let,{random drawn chip is defective}Then[]=[|e][e] +[|e][e] +[|e][e]=005×025 + 004×035 + 002×040=00345

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13Hence[e|]=[|e][e][]= 005×02500345= 0363[e|]=[|e][e][]= 004×03500345= 0406[e|]=[|e][e][]= 002×04000345= 023236. From the example[]'logWe set,and construct the following table.[][]0.00.00.50.3460.10.230.60.310.20.320.70.250.30.3610.80.180.40.3670.90.10The peak is quite shallow, therefore the choice ofis not critical near the peak.37.(a) If we associate the 103 villagers with= 103balls and the= 30tents with 30 cells,this becomes a classical occupancy problem.(b) The result is given by Eq.1.8-6, which is repeated here asμ+−1¶=μ30 + 103−1103¶=132!103!29!(c) The result isμ−1−¶=μ103−1103−30¶=102!73!29!To obtain numerical evaluations of these factorial expressions, one might want to useStirling’s formulas:!≈(2)12+12−38. The most natural set of outcomes here are the strings (or vectors) of length, indicatingwhere each ball has landed.There aresuch strings.They are all equally likely.Thenumber of favorable outcomes would be!since there arechoices for thefirst preselectedlocation,−1choices for the second location, etc.The desired probability is then=!Now, since the balls are indistinguishable, we could have considered the so-called distinguish-able outcomes,μ+−1¶in number, however from the description of the experiment inthe problem statement, they would not all be equally likely. So we could not rely on classicaltheory then to give us the probabilities of these outcomes.

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1439. As in problem 1.38, the number of favorable ways is!. However, the total number of waysisnotsince cells can at most hold one ball.For thefirst ball, there arecells; for thesecond ball,−1cells, etc. Thus=(−1)· · ·(−+ 1)=!(−)!Thus=!³!(−)!´=!(−)!!=μ¶−140.(a) Let the tribal leaders be the cells and the rifles be the balls. Then the three tribal leaderscollecting thefive rifles is the analog of puttingfive balls into three cells.(b) These are the distributions shown in non-bold. There arefifteen such distributions.(c) Careful here! If we count only the outcomes in bold we shall get the wrong answer i.e.,6/21=0.286. The reason this answer is wrong is that the outcomes in the columns arenot equally likely. The correct answer is computed using Eq.(1.8-9) i.e.,41.(a) The probability that a specified number appears on the face of a dice is 1/6. Hence theprobability of getting three specified numbers is or 1 in 216. Hence if you win you shouldget $216 for every dollar bet. But the casino payout is only $180:1.(b) The face value of thefirst dice is irrelevant. The probability that the second dice matchesthefirst is 1/6. The probability that the third dice matches thefirst is 1/6. Hence theprobability of getting three unspecified matches is or 1 in 36.(c) Letdenote that dice = 123shows a specified number. Then the probability that(at least) two specified numbers appear is[123] +[132] +[321] +[123]=3×56×16×16 + 16×16×16=00741or about 1 in 14. So per dollar bet you should get $14 but the casino payout is only $10.d.-i. The next six parts can be solved by enumeration i.e., counting.However there is asystematic procedure based on the mathematical operation of convolution that can yieldall of the answers from reading a graph. The details are given in Example 3.3.-5.

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15d. Refer to the table below:We note that there are only three ways of getting a 4: 1+1+2; 1+2+1,2+1+1. Hencetheprobability that the sum equals 4 is3(6×6×6) = 172. Thus the fair payout shouldbe 1:72 instead of 1:60.e. The number of ways of getting a 5 is 6: 3+1+1; 1+3+1; 1+1+3; 2+2+1; 2+1+2; 1+2+2.Hence the probability that the sum equals 5 is6(6×6×6) = 136.A fair payoutwould be 1:36 instead of 1:30.f.-i. follow the same enumeration.j. Let’s think of this a series of throws. The probability that thefirst throw matches oneof the two specified numbers is 2/6.The probability that the next throw matches aspecified number is 1/6.The last throw should not match either of the numbers.Itsprobability is 4/6. In a throw of three dice this can happen in three ways. Hence theprobability is3×26×16×46= 19or 1:9. But the payout is only 1:5.42. Forpackets there are!ways of arranging themselves, but only one way of doing itcorrectly.All the packet arrangements are equally likely.Hence[{correct reassembly}]=1!=(3628800)−1for= 10,≈276×10−743. For three packets, there are six different arrangements (3·2·1 = 6), but only one correct one.Hence on any try,[{success}]and,[{failure}]=16= 1−= 56For afirst correct reassembly on thethtry, there must be−1failures followed by onsuccess on thethtry, thus() = 16μ56¶−1≥1We note in passing that this is a valid PMF, i.e. it sums to one over its support[1∞).Tofind the smallestsuch thatX=116μ56¶−1≥095we note that the complementary event is no successes intrials, with probability1−¡56¢Thus we seek instead the smallestsuch that1−¡56¢≤1−095 = 005Thus'ln(005)ln(56)=165So the answer is= 17.

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