案例datascientist公開課-datascience福利6.interview-wstresearch

ProbabilityformsthebackboneofmanyimportantdatascienceconceptsfrominferentialstatisticstoBayesiannetworks.Itwouldnotbewrongtosaythatthejourneyofmasteringstatisticsbeginswithprobability.Thisskilltestwasconductedtohelpyouidentifyyourskilllevelinprobability.Atotalof1249peopleregisteredforthisskilltest.Thetestwasdesignedtotesttheconceptualknowledgeofprobability.Ifyouareoneofthosewhomissedoutonthisskilltest,herearethequestionsandsolutions.Youmissedontherealtimetest,butcanreadthisarticletofindouthowyoucouldhaveansweredcorrectly.HerearetheleaderboardrankingforalltheBasicsofProbabilityforDataScienceexplainedwithIntroductiontoConditionalProbabilityandBayestheoremfordatascienceLetAandBbeeventsonthesamesamplespace,withP(A)=0.6andP(B)0.7.CanthesetwoeventsbeSolution:ThesetwoeventscannotbedisjointbecauseP(A)+P(B)>1.P(A?B)=P(A)+P(B)-P(A?B).AneventisdisjointifP(A?B)=0.IfAandBaredisjointP(A?B)=0.6+0.7=AndSinceprobabilitycannotbegreaterthan1,thesetwomentionedeventscannotbeAlicehas2kidsandoneofthemisagirl.Whatistheprobabilitythattheotherchildisalsoagirl?YoucanassumethatthereareanequalnumberofmalesandfemalesintheSolution: esfortwokidscanbe{BB,BG,GB,Sinceitismentionedthatoneofthemisagirl,wecanremovetheBBoptionfromthesamplespace.Thereforethesamplespacehas3optionswhileonlyonefitsthesecondcondition.Thereforetheprobabilitythesecondchildwillbeagirltoois1/3.Afairsix-sideddieisrolledtwice.Whatistheprobabilityofgetting2ontherollandnotgetting4onthesecondroll?Solution:Thetwoeventsmentionedareindependent.Therollofthedieisindependentofthesecondroll.Thereforetheprobabilitiescanbedirectlymultiplied. 2)=1/6P(nosecond4)=ThereforeP(getting2andnosecond4)=1/6*5/6=P(A?Cc)willbeonlyP(A).P(onlyA)+P(C)willmakeitP(A?C).P(B?Ac?Cc)isP(onlyB)ThereforeP(A?C)andP(onlyB)willmakeP(A?B?C)Consideratetrahedraldieandrollittwice.Whatistheprobabilitythatthenumberontherollisstrictlyhigherthanthenumberonthesecondroll?Note:Atetrahedraldiehasonlyfoursides(1,2,3andSolution:Thereare6outof16possibilitieswhere rollisstrictlyhigherthanthesecondWhichofthefollowingoptionscannotbetheprobabilityofany-OnlyOnlyOnlyAandBandAandCProbabilityalwaysliewithin0toAnitarandomlypicks4cardsfromadeckof52-cardsandplacesthembackintothedeck(Anysetof4cardsisequallylikely).Then,Babitarandomlychooses8cardsoutofthesamedeck(Anysetof8cardsisequallylikely).Assumethatthechoiceof4cardsbyAnitaandthechoiceof8cardsbyBabitaareindependent.Whatistheprobabilitythatall4cardschosenbyAnitaareinthesetof8cardschosenbyBabita?48C4x48C4x48C8xNoneoftheaboveSolution:(A)Thetotalnumberofpossiblecombinationwouldbe52C4(Forselecting4cardsby*52C8(Forselecting8cardsbySince,the4cardsthatAnitachoosesisamongthe8cardswhichBabitahaschosen,thusthenumberofcombinationspossibleis52C4(Forselectingthe4cardsselectedbyAnita)*48C4(Forselectinganyother4cardsbyBabita,sincethe4cardsselectedbyAnitaarecommon)QuestionContextAplayerisrandomlydealtasequenceof13cardsfromadeckof52-cards.Allsequencesof13cardsareequallylikely.Inanequivalentmodel,thecardsarechosenanddealtoneatatime.Whenchoosingacard,thedealerisequallylikelytopickanyofthecardsthatremaininthedeck.Ifyoudealt13cards,whatistheprobabilitythatthe13thcardisaSolution:Sincewearenottoldanythingaboutthe12cardsthataredealt,theprobabilitythatthe13thcarddealtisaKing,isthesameastheprobabilitythatthecarddealt,orinfactanyparticularcarddealtisaKing,andthisequals:4/52Afairsix-sideddieisrolled6times.Whatistheprobabilityofgettingallesasunique?Solution:Forallthe estobeunique,wehave6choicesfortheturn,5forthesecondturn,4forthethirdturnandsoonThereforetheprobabilityifgettingall eswillbeequaltoAgroupof60studentsisrandomlysplitinto3classesofequalsize.Allpartitionsareequallylikely.JackandJillaretwostudentsbelongingtothatgroup.WhatistheprobabilitythatJackandJillwillendupinthesameclass?Solution:Assignadifferentnumbertoeachstudentfrom1to60.Numbers1to20goingroup1,21to40gotogroup2,41to60gotogroup3.Allpossiblepartitionsareobtainedwithequalprobabilitybyarandomassignmentifthesenumbers,itdoesn’tmatterwithwhichstudentswestart,sowearefreetostartbyassigningarandomnumbertoJackandthenweassignarandomnumbertoJill.AfterJackhasbeenassignedarandomnumberthereare59randomnumbersavailableforJilland19ofthesewillputherinthesamegroupasJack.ThereforetheprobabilityisWehavetwocoins,AandB.ForeachtossofcoinA,theprobabilityofgettingheadis1/2andforeachtossofcoinB,theprobabilityofgettingHeadsis1/3.Alltossesofthesamecoinareindependent.Weselectacoinatrandomandtossittillwegetahead.TheprobabilityofselectingcoinAis?andcoinBis3/4.Whatistheexpectednumberoftossestogettheheads?Solution:IfcoinAisselectedthenthenumberoftimesthecoinwouldbetossedforaguaranteedHeadsis2,similarly,forcoinBitis3.ThusthenumberoftimeswouldbeTosses=2*(1/4)[probabilityofselectingcoinA]+3*(3/4)[probabilityofselectingcoin=Supposealifeinsurancecompanysellsa$240,000oneyeartermlifeinsurancepolicytoa25-yearoldfemalefor$210.Theprobabilitythatthefemalesurvivestheyearis.999592.Findtheexpectedvalueofthispolicyfortheinsurancecompany.Solution:P(companylosesthemoney)=0.99592P(companydoesnotlosethemoney)=0.000408Theamountofmoneycompanylosesifitloses=240,000–210=239790Whilethemoneyitgainsis$210Expectedmoneythecompanywillhavetogive=239790*0.000408=97.8Expectmoneycompanygets=210.Thereforethevalue=210–98=Theabovestatementistrue.YouwouldneedtoknowthatP(A/B)=P(A?B)/P(B)P(Cc?A|A)=P(Cc?A?A)/P(A)=P(Cc?P(B|A?Cc)=P(A?B?Cc)/P(A?Multiplyingthethreewewouldget–P(A?B?Cc),hencetheequationsholdsWhenaneventAindependentofIfandonlyifIfandonlyifIfandonlyifP(A)=0or1Solution:(D)Theeventcanonlybeindependentofitselfwheneitherthereisnochanceofithappeningorwhenitiscertaintohappen.EventAandBisindependentwhenP(A?=P(A)*P(B).NowifB=A,P(A?A)=P(A)whenP(A)=0orSupposeyou’reinthefinalroundof“Let’smakeadeal”gameshowandyouaresupposedtochoosefromthreedoors–1,2&3.Oneofthethreedoorshasacarbehinditandothertwodoorshavegoats.Let’ssayyouchooseDoor1andthehostopensDoor3whichhasagoatbehindit.Toassuretheprobabilityofyourwin,whichofthefollowingoptionswouldyouchoose.SwitchyourRetainyourItdoesn’tmatterprobabilityofwinningorlosingisthesamewithorwithoutrevealingonedoorSolution:Iwould mendreadingthisarticleforadetaileddiscussionoftheMontyHall’sCross-fertilizingaredandawhiteflowerproducesredflowers25%ofthetime.Nowwecross-fertilizefivepairsofredandwhiteflowersandproducefiveoffspring.WhatistheprobabilitythattherearenoredflowerplantsinthefiveSolution:TheprobabilityofoffspringbeingRedis0.25,thustheprobabilityoftheoffspringnotbeingredis0.75.Sinceallthepairsareindependentofeachother,theprobabilitythatalltheoffspringsarenotredwouldbe(0.75)5=0.237.Youcanthinkofthisasabinomialwithallfailures.Aroulettewheelhas38slots–18red,18black,and2green.Youplayfivegamesandalwaysbetonredslots.Howmanygamescanyouexpecttowin?2.3684C)Solution:TheprobabilitythatitwouldbeRedinanyspinis18/38.Now,youareplayingthegame5timesandallthegamesareindependentofeachother.Thus,thenumberofgamesthatyoucanwinwouldbe5*(18/38)=2.3684Aroulettewheelhas38slots,18arered,18areblack,and2aregreen.Youplayfivegamesandalwaysbetonred.Whatistheprobabilitythatyouwinallthe5games?Solution:TheprobabilitythatitwouldbeRedinanyspinis18/38.Now,youareplayingfame5timesandallthegamesareindependentofeachother.Thus,theprobabilitythatyouwinallthegamesis(18/38)5=0.0238Sometestscoresfollowanormaldistributionwithameanof18andastandarddeviationof6.Whatproportionoftesttakershavescoredbetween18and24?NoneoftheaboveSolution:(C)SoherewewouldneedtocalculatetheZscoresforvaluebeing18and24.Wecaneasilyngthatbyputtingsamplemeanas18andpopulationmeanas18withσ=6andcalculatingZ.SimilarlywecancalculateZforsamplemeanas24.Z=(X-Thereforefor26asZ=(18-18)/6=0,lookingattheZtablewefind50%peoplehavescoresbelowFor24asZ=(24-18)/6=1,lookingattheZtablewefind84%peoplehavescoresbelow24.Thereforearound34%peoplehavescoresbetween18and24.Ajarcontains4marbles.3Red&1white.Twomarblesaredrawnwithreplacementaftereachdraw.Whatistheprobabilitythatthesamecolormarbleisdrawntwice?Solution:Ifthemarblesareofthesamecolorthenitwillbe3/4*3/4+1/4*1/4=WhichofthefollowingeventsismostAtleastone6,when6diceareAtleast2sixeswhen12diceareAtleast3sixeswhen18diceareAlltheabovehavesameprobabilitySolution:(A)Probabilityof‘6’turningupinarollofdiceisP(6)=(1/6)&P(6’)=(5/6).Thus,probabilityof∞Case1:(1/6)*(5/6)5=∞Case2:(1/6)2*(5/6)10=∞Case3:(1/6)3*(5/6)15=Thus,thehighestprobabilityisCaseSupposeyouwereinterviewedforatechnicalrole.50%ofthepeoplewhosatfortheinterviewreceivedthecallforsecondinterview.95%ofthepeoplewhogotacallforsecondinterviewfeltgoodabouttheirinterview.75%ofpeoplewhodidnotreceiveasecondcall,alsofeltgoodabouttheirinterview.Ifyoufeltgoodafteryour interview,whatistheprobabilitythatyouwillreceiveasecondinterviewcall?Solution:Let’sassumethereare100peoplethatgavethe roundofinterview.The50peoplegottheinterviewcallforthesecondround.Outofthis95%feltgoodabouttheirinterview,whichis47.5.50peopledidnotgetacallfortheinterview;outofwhich75%feltgoodabout,whichis37.5.Thus,thetotalnumberofpeoplethatfeltgoodaftergivingtheirinterviewis(37.5+47.5)85.Thus,outof85peoplewhofeltgood,only47.5gotthecallfornextround.Hence,theprobabilityofsuccessis(47.5/85)=0.558.AnothermoreacceptedwaytosolvethisproblemistheBaye’stheorem.Ileaveittoyoutocheckforyourself.Acoinofdiameter1-inchesisthrownonatablecoveredwithagridoflineseachtwoinchesapart.Whatistheprobabilitythatthecoinlandsinsideasquarewithouttouchinganyofthelinesofthegrid?Youcanassumethatthethrowinghasnoskillinthrowingthecoinandisthrowingitrandomly.Youcanassumethatthe throwinghasnoskillinthrowingthecoinandisthrowingitrandomly.Solution:Thinkaboutwhereallthecenterofthecoincanbewhenitlandson2inchesgridanditnottouchingthelinesofthegrid.Iftheyellowregionisa1inchsquareandtheoutsidesquareisof2inches.Ifthecenterfallsintheyellowregion,thecoinwillnottouchthegridline.Sincethetotalareais4andtheareaoftheyellowregionis1,theprobabilityis?.Thereareatotalof8bowsof2eachofgreen,yellow,orange&red.Inhowmanywayscanyouselect1bow?A)B)C)D)Solution:Youcanselectonebowoutoffourdifferentbows,soyoucanselectonebowinfourdifferentways.Considerthefollowingprobabilitydensityfunction:WhatistheprobabilityforX≤6i.e.P(x≤6)WhatistheprobabilityforX≤6i.e.Solution:Tocalculatetheareaofaparticularregionofaprobabilitydensityfunction,weneedtointegratethefunctionundertheboundsofthevaluesforwhichweneedtocalculatetheThereforeonintegratingthegivenfunctionfrom0to6,wegetInaclassof30students,approximaywhatistheprobabilitythattwoofthestudentshavetheirbirthdayonthesameday(definedbysamedayandmonth)(assumingit’snotaleapyear)?Forexample–Studentswithbirthday3rdJan1993and3rdJan1994wouldbeafavorableevent.Solution:Thetotalnumberofcombinationspossiblefornotwo stohavethesamebirthdayinaclassof30is30*(30-1)/2=435.Now,thereare365daysinayear(assumingit’snotaleapyear).Thus,theprobabilityofpeoplehavingadifferentbirthdaywouldbe364/365.Nowthereare870combinationspossible.Thus,theprobabilitythatnotwopeoplehavethesamebirthdayis(364/365)^435=0.303.Thus,theprobabilitythattwopeoplewouldhavetheirbirthdaysonthesamedatewouldbe1–0.303=0.696Ahmedisplayingalotterygamewherehemustpick2numbersfrom0to9followedbyanEnglishalphabet(from26-letters).Hemaychoosethesamenumberbothtimes.Ifhisticketmatchesthe2numbersand1letterdrawninorder,hewinsthegrandprizeandreceives$10405.Ifjusthislettermatchesbutoneorbothofthenumbersdonotmatch,hewins$100.Underanyothercircumstance,hewinsnothing.Thegamecostshim$5toplay.Supposehehaschosen04Rtoplay.Whatistheexpectednetprofitfromplayingthis$-$2.81C)$-$-Solution:ExpectedvalueinthisE(X)=P(grandprize)*(10405-5)+P(small)(100-5)+P(losing)*(-5)P(grandprize)=(1/10)*(1/10)*(1/26)P(small)=1/26-1/2600,thereasonweneedtodothisisweneedtoexcludethecasewherehegetstheletterrightandalsothenumbersrights.Hence,weneedtoremovethescenarioofgettingtheletterright.P(losing)=1-1/26-ThereforewecanfitinthevaluestogettheexpectedvalueasAssumeyousellsandwiches.70%peoplechooseegg,andtherestchoosechicken.Whatistheprobabilityofselling2eggsandwichestothenext3Solution:TheprobabilityofsellingEggsandwichis0.7&thatofachickensandwichis0.3.Now,theprobabilitythatnext3customerswouldorder2eggsandwichis0.7*0.7*0.30.147.Theycanordertheminanysequence,theprobabilitieswouldstillbetheQuestioncontext:29–HIVisstillaveryscarydiseasetoevengettestedfor.TheUSmilitarytestsitsrecruitsforHIVwhentheyarerecruited.TheyaretestedonthreeroundsofElisa(anHIVtest)beforetheyaretermedtobepositive.ThepriorprobabilityofanyonehavingHIVis0.00148.ThetruepositiverateforElisais93%andthetruenegativerateis99%.WhatistheprobabilitythatarecruithasHIV,givenhetestedpositiveonElisatest?ThepriorprobabilityofanyonehavingHIVis0.00148.ThetruepositiverateforElisais93%andthetruenegativerateis99%.Solution:

goingthroughtheBayesupdatingsectionof

articleforunderstandingoftheaboveWhatistheprobabilityofhavingHIV,givenhetestedpositiveonElisathesecondtimeaswell.ThepriorprobabilityofanyonehavingHIVis0.00148.ThetruepositiverateforElisais93%andthetruenegativerateis99%.Solution: mendgoingthroughtheBayesupdatingsectionofthisarticlefortheunderstandingoftheabovequestion.Supposeyou’replayingagameinwhichwetossafaircoinmultipletimes.Youhavealreadylostthricewhereyouguessedheadsbutatailsappeared.Whichofthebelowstatementswouldbecorrectinthiscase?YoushouldguessheadsagainsincethetailshasalreadyoccurredthriceanditsmorelikelyforheadstooccurnowYoushouldsaytailsbecauseguessingheadsisnotmakingyouYouhavethesameprobabilityofwinninginguessingeither,hencewhateveryouguessthereisjusta50-50chanceofwinningorlosingNoneoftheseSolution:(C)Thisisaclassicproblemofgambler’sfallacy/montecarlofallacy,wherefalselystartstothinkthattheresultsshouldevenoutinafewturns.Thegamblerstartstobelievethatifwehavereceived3heads,youshouldreceivea3tails.Thisishowevernottrue.Theresultswouldevenoutonlyininfinitenumberoftrials.TheinferenceusingthefrequentistapproachwillalwaysyieldthesameresultastheBayesianapproach.Solution:ThefrequentistApproachishighlydependentonhowwedefinethehypothesiswhileBayesianapproachhelpsusupdateourpriorbeliefs.Thereforethefrequentistapproachmightresultinanoppositeinferenceifwedeclarethehypothesisdifferently.Hencethetwoapproachesmightnotyieldthesameresults.Hospitalrecordsshowthat75%ofpatientssufferingfromadiseasedieduetothatdisease.Whatistheprobabilitythat4outofthe6randomlyselectedpatientsSolution:Thinkofthisasabinomialsincethereareonly2 es,eitherthepatientdiesorheHeren=6,andx=4.p=0.25(probabilityifliving(success))q=0.75(probabilityofP(X)=nCxpxqn-x=6C4(0.25)4(0.75)2=Thestudentsofaparticularclassweregiventwotestsforevaluation.Twenty-fivepercentoftheclassclearedboththetestsandforty-fivepercentofthestudentswereabletoclearthetest.Calculatethepercentageofstudentswhopassedthesecondtestgiventhattheywerealsoabletopassthe Solution:Thisisasimpleproblemofconditionalprobability.LetAbetheeventofpassinginBistheeventofpassinginthesecondtest.P(A?B)ispassinginboththeeventsP(passinginsecondgivenhepassedin one)=P(A?=0.25/0.45whichisaroundWhileitissaidthattheprobabilitiesofhavingaboyoragirlarethesame,let’sassumethattheactualprobabilityofhavingaboyisslightlyhigherat0.51.Supposeacoupleplanstohave3children.Whatistheprobabilitythatexactly2ofthemwillbeboys?Solution:Thinkofthisasabinomialdistributionwheregettingasuccessisaboyandfailureisagirl.Thereforeweneedtocalculatetheprobabilityofgetting2outofthreesuccesses.P(X)=nCxpxqn-x=3C2(0.51)2(0.49)1=Heightsof10year-olds,regardlessofgender,closelyfollowanormaldistributionwithmean55inchesandstandarddeviation6inches.Whichofthefollowingistrue?Wewouldexpectmorenumberof10year-oldstobeshorterthan55inchesthanthenumberofthemwhoaretallerthan55inchesRoughly95%of10year-oldsarebetween37and73inchesA10-year-oldwhois65inchestallwouldbeconsideredmoreunusualthana10-year-oldwhois45inchestallNoneoftheseSolution:(D)NoneoftheabovestatementsareAbout30%ofhumantwinsareidentical,andtherestarefraternal.Identicaltwinsarenecessarilythesamesex,halfaremalesandtheotherhalfarefemales.One-quarteroffraternaltwinsarebothmales,one-quarterbothfemale,andone-halfaremixed:onemale,onefemale.Youhave