Probability On GMAT.docx

Probability On GMAT.docx

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ProbabilityOnGMAT

ProbabilityandGMAT

ThisisabriefcoursetogetyoupreparedtocrackprobabilityquestionsonGMAT.

Introduction

HowprobableisittogetprobabilityquestionsonGMAT?

Probabilityquestionsarebecomingincreasinglycommon.Theytendtobebundledamongthedifficultquestions,sohighscorerswillcommonlyencounter1,2,or3ofthem.Ifyouarealowscorerandarepressedfortime,considerskippingmostofthematerialpast"SimpleProbability."GMATisacomputer-adaptivetest,andlowscorersaren'tlikelytoencounterthemostdifficultprobabilityquestiontypes.

DoIhavetobeageniustosolveprobabilityquestions?

Absolutelynot.BoththisbriefcourseandGMATdonotrequireanymathknowledgebeyondwhatyoulearnedinyourhighschool.Youhavenottobeageniuseither.Justbesuretotrysolvingtheproblemsandgetagripofthesolutiontools,andyou'llcrackit.Totellyouthesecret,themythofthecomplexityoftheprobabilitytheoryissimplyanotherwaytosecurethemathinstructors'wages.

Whatisprobability?

Probabilityisameasureofhowlikelyisaneventtohappen.Itismeasuredinfractionsfrom0to1（0isimpossible,1isunavoidableorcertain）.Sometimesitisdenotedinpercentages,againfrom0%to100%.

Whatareaneventandanoutcome?

Eventisanythingthathappens.Inprobabilitytheorywespeakofeventshavingoutcomesorresults.Acoinflip（anevent）hastwopossibleoutcomes—headsandtails.Adietosshassixpossibleoutcomes.Whenacoinisflipped（aneventistested）,oneoftheoutcomesisobtained.Eitherheadsortails.

Howisprobabilityused?

Aprobabilityiscommonlydenotedasp（SomeEvent）.So,p（Heads）=50%meansthatyouhave1chancein2togetheadsinacoinflip.Thisalsomeansthatifyouflipthecoin100times,you'llgetabout50heads.Butnotexactly50.Youmayget49,or63,orevennoheads.Butyou'remostlikelytogetsuchanumberofheadsthatwillbecloseto50.Thisworksforanyprobability.So,iftheprobabilityofgettingmarriedaftergoingtothecinemais3%,outof1,000moviesyou'llbemarriedabout1,000*3%=30times.Maybe26or34,buttheaverageexpectationis30.That'swhatyouuseprobabilityfor,apartfromcrackingGMAT.

A.SimpleProbability:

TheF/TRule

Ingeneral,theprobabilityofaneventisthenumberoffavorableoutcomesdividedbythetotalnumberofpossibleoutcomes.ThisisknownastheF/TRule,and90%oftheproblemsaresolvedwiththistool.Nokidding.

Probability = （# of favorable outcomes） / （# of possible outcomes） 

Example1

Whatistheprobabilitythatacarddrawnatrandomfromadeckofcardswillbeanace?

Solution

Inthiscasetherearefourfavorableoutcomes:

1.theaceofspades,

2.theaceofhearts,

3.theaceofdiamonds,

4.theaceofclubs.

Sinceeachofthe52cardsinthedeckrepresentsapossibleoutcome,thereare52possibleoutcomes.Therefore,theprobabilityis4/52or1/13.Thesameprinciplecanbeappliedtotheproblemofdeterminingtheprobabilityofobtainingdifferenttotalsfromapairofdice.

Example2

Twofairsix-sideddicearerolled;whatistheprobabilityofhaving5asthesumofthenumbers?

Solution

Thereare36possibleoutcomeswhenapairofdiceisthrown（sixoutcomesforthefirstdietimessixoutcomesforthesecondone）.Sincefouroftheoutcomeshaveatotalof5[（1,4）,（4,1）,（2,3）,（3,2）],theprobabilityofthetwodiceaddingupto5is4/36=1/9.

Example3

Twosix-sideddicearerolled;whatistheprobabilityofhaving12asthesumofthenumbers?

Solution

Wealreadyknowthetotalnumberofpossibleoutcomesis36,andsincethereisonlyoneoutcomethatsumsto12,（6,6—youneedtorolldoublesix）,theprobabilityissimply1/36.

Dinosaurexample

Ablondegirl（G.W.Bush,yourboss,orwhoeveryoulovetooheartily）wasaskedoncewhatistheprobabilityofmeetingadinosaurinthestreet.Theanswerwas:

"50%.Ieithermeetitornot."ThisishowyouDON'TusetheF/Trule.Whencountingtheoutcomes,makesurethat:

∙allofthemareequallylikelytohappen

∙youhavenotleftoutanypossibilitieswhencountingT

∙（quote）FandTareinthesamecurrency,i.e.ifFiscombinationsandTispermutations,you'llgetanerror.

Congratulations!

Nowyou'vecomethroughtheeasypart.Ifyou'refinewithmoderateGMATandamodestschoolinWestVirginiaorNevadadesert,youmayproudlyandhappilyabandonthiscourserighthere.

NOTE:

Thematerialfromhereonthroughtheendofthesectionisdenseandintendedonlyformediumtohighscorers.BecauseGMATisaCAT（computer-adaptivetest）,itisrelativelyunlikelythatlowerscorerswillencounterthesequestions,and,iftheyareshortoftime,theyarebetteroffputtingtheirtimeintoothersections.

B.ProbabilityofMultipleEvents

Forquestionsinvolvingsingleevents,theF/Truleissufficient.Infact,itisoftensufficientforallothercasestoo.But,forquestionsinvolvingmultipleevents,someothertoolsmaybemoreappropriate.EvenwhentheproblemcanbesolvedwithF/T,thesetoolsstillmayprovideamoreelegantsolution.Here'rethetools:

NOTtool

Ifyouknowthattheprobabilityofanevent（oroneoftheoutcomes）isp,theprobabilityofthiseventNOThappening（ortheprobabilityofitNOThavingthisgivenoutcome）,is（1-p）.

p（not A） + p（A） = 1

ANDtool

Iftwo（ormore）independenteventsareoccurring,andyouknowtheprobabilityofeach,theprobabilityofBOTH（orALL）ofthemoccurringtogether（eventAandeventBandeventCetc）isamultiplicationoftheirprobabilities.

p（A and B） = p（A） * p（B）

p（A and B and C ... and Z） = p（A） * p（B） * p（C） * ... * p（Z）

SupposeIwillonlybehappytodayifIgetanemailandwinthelottery.I'vea90%chancetogetanemailand0.1%chancetowinthelottery.Whataremychancesforhappiness?

Sinceemailandlotteryareindependent（gettinganemaildoesn'tchangemylotterychances,andviceversa）,wecanusetheANDtool:

p（emailandlottery）=p（email）*p（lottery）=90%*0.1%=0.09%;SoIhave9chancesin10,000...Notbad.

ORtool

Iftwo（ormore）incompatibleeventsareoccurring,theprobabilityofEITHERofthemoccurring（eventAoreventBoreventCetc）isasumoftheirprobabilities.

p（A or B） = p（A） + p（B）

p（A or B or C ... or Z） = p（A） + p（B） + ... + p（Z）

Incompatiblemeansthattheycan'thappentogether,i.e.p（AandB）=0.Incaseoftwocompatibleevents,theORtoollooksabitmorecomplicated:

p（A or B） = p（A） + p（B） - p（A and B）

IfweknowthatAandBareindependent,wecanapplyANDtooltorewrite:

p（A or B） = p（A） + p（B） - p（A） * p（B）

SupposeIwillnowbehappyinbothcases-eithergettinganemailorwinningthelottery.Whataremychancestohappinessnow?

p（emailorlottery）=p（email）+p（lottery）-p（email）*p（lottery）=90%+0.1%-0.09%=90.01%;Mychancesare9,001in10,000now.I'dratherchoosethisone.

Expressions/Bracketstool

Whenyou'rebeingaskedforsomethingcomplex,tryreducingittoeventsandoutcomes,andwritingaformula.Usebracketstodenotecomplexevents,suchas（AandB）,or（Aand（BorC））,etc.ItiscommontouseANDasifitismultiplicationandORasifitisadditionintheorderpreference,i.e.（AandBorC）=（（AandB）orC）,but（Aand（BorC））<>（AandBorC）.Whenyoufigureouttheformula,it'llbeeasytoreduceittosimplearithmeticoperationsbyusingNOT,AND,andORtools.

Eliminationtricks

Giventhat0<=p（A）<=1,yougetthefollowingrules:

1.p（AandB）<=p（A）

2.p（AorB）>=p（A）

3.p（AandB）<=p（AorB）

Thinkingoftheserulesisoftenanexcellentstrategyforeliminatingcertainanswerchoices.

Example4

Ifafaircoinistossedtwice,whatistheprobabilitythatonthefirsttossthecoinlandsheadsandonthesecondtossthecoinlandstails?

1.1/6

2.1/3

3.1/4

4.1/2

5.1

Solution

SupposefirsttossisA,secondisB.Weknowthatp（A_heads）=50%andthatp（B_tails）=50%.Also,AandBareindependent.So,p（A_headsandB_tails）=p（A_heads）*p（B_tails）=50%*50%=25%=1/4.AnswerisC.

Example5

Ifafaircoinistossedtwicewhatistheprobabilitythatitwilllandeitherheadsbothtimesortailsbothtimes?

1.1/8

2.1/6

3.1/4

4.1/2

5.1

Solution

LetfirsttossbeA,secondB.

p（Ah） = p（At） = p（Bh） = p（Bt） = 1/2

p（Ah and Bh） = p（Ah） * p（Bh） = 1/4

p（At and Bt） = p（At） * p（Bt） = 1/4

p（（Ah and Bh） or （At and Bt）） = p（Ah and Bh） + p（At and Bt） = 1/4 + 1/4 = 1/2

NotethatANDruleworksbecauseAandBareindependent,andORruleworksbecause（AhandBh）and（AtandBt）areincompatible.

Alternatively,youmayuseF/Truletosolvethis.Enumerateoutcomesas（HH,HT,TH,TT）.FavorableareHHandTT.So,p=2/4=1/2.AlthoughinthiscaseF/Truleworksmoregracefully,theAND/ORapproachisstillhelpful-youcanlearnitonsucheasyexamplesasthistoprepareforthemoredifficultones.

Example6

Abowmanhitshistargetin1/2ofhisshots.Whatistheprobabilityofhimmissingthetargetatleastonceinthreeshots?

Solution

Anoptimalwaytosolvethisistothinkthat（missingthetargetatleastonce）=1–（hittingiteverytime）.So,p（hittingiteverytime）=p（shot1_hitandshot2_hitandshot3_hit）=p（shot1_hit）*p（shot2_hit）*p（shot3_hit）=1/2*1/2*1/2=1/8;p（missingatleastonce）=1–p（hittingiteverytime）=1–1/8=7/8.

Alternatively,usetheF/Trule.TheTareHHH,HHM,HMH,HMM,MHH,MHM,MMH,MMM.T=8.TheFareHHM,HMH,H