General Rules of Probability

24 Pages·2003·1.85 MB·English

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CHAPTER
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Inthischapterwecover... General Rules
Independenceandthe
∗
multiplicationrule
of Probability
Applyingthemultiplication
rule
Thegeneraladditionrule
Conditionalprobability
Thegeneralmultiplication
Themathematicsofprobabilitycanprovidemodelstodescribetheflowoftraf-
rule
ficthroughahighwaysystem,atelephoneinterchange,oracomputerproces-
Independence
sor;thegeneticmakeupofpopulations;theenergystatesofsubatomicparticles;
Treediagrams
thespreadofepidemicsorrumors;andtherateofreturnonriskyinvestments.
Althoughweareinterestedinprobabilitybecauseofitsusefulnessinstatistics,
the mathematics of chance is important in many fields of study. This chapter
presentsabitmoreofthetheoryofprobability.
OurstudyofprobabilityinChapter9concentratedonbasicideasandfacts.
Nowwelookatsomedetails.Withmoreprobabilityatourcommand,wecan
modelmorecomplexrandomphenomena.Wehavealreadymetandusedfour
rules.
*Thismoreadvancedchaptergivesmoredetailaboutprobability.Itisnotneededtoreadtherestofthe
book.
280
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Independenceandthemultiplicationrule 281
RULESOFPROBABILITY
Rule1. 0 ≤ P(A) ≤ 1foranyevent A
Rule2. P(S) = 1
Rule3. Foranyevent A,
P(Adoesnotoccur) = 1− P(A)
Rule4. Addition rule:If Aand B aredisjointevents,then
P(Aor B) = P(A)+ P(B)
Independenceandthemultiplicationrule
Rule4,theadditionrulefordisjointevents,describestheprobabilitythatone
ortheotheroftwoevents Aand B occursinthespecialsituationwhen Aand
B cannotoccurtogether.Nowwewilldescribetheprobabilitythatbothevents
Aand B occur,againonlyinaspecialsituation.
Youmayfindithelpfultodrawapicturetodisplayrelationsamongseveral
events.ApicturelikeFigure11.1thatshowsthesamplespaceSasarectangular
areaandeventsasareaswithin S iscalledaVenn diagram.Theevents Aand Venn diagram
B in Figure 11.1 are disjoint because they do not overlap. The Venn diagram
inFigure11.2illustratestwoeventsthatarenotdisjoint.Theevent{Aand B}
appearsastheoverlappingareathatiscommontoboth Aand B.
S
A B
Figure11.1 Venndiagram
showingdisjointevents A
and B.
S
Figure11.2 Venndiagram
showingevents Aand B that A B
A and B
arenotdisjoint.Theevent
{Aand B}consistsof
outcomescommonto A
and B.
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282 CHAPTER11 GeneralRulesofProbability
Supposethatyoutossabalancedcointwice.Youarecountingheads,sotwo
eventsofinterestare
A=firsttossisahead
B =secondtossisahead
Theevents AandBarenotdisjoint.Theyoccurtogetherwheneverbothtosses
give heads. We want to find the probability of the event {Aand B} that both
tossesareheads.
ThecointossingofBuffon,Pearson,andKerrichdescribedatthebeginning
ofChapter9makesuswillingtoassignprobability1/2toaheadwhenwetoss
acoin.So
P(A)=0.5
P(B)=0.5
Whatis P(Aand B)?Ourcommonsensesaysthatitis1/4.Thefirstcoinwill
giveaheadhalfthetimeandthenthesecondwillgiveaheadonhalfofthose
trials,sobothcoinswillgiveheadson1/2×1/2 = 1/4ofalltrialsinthelong
run. This reasoning assumes that the second coin still has probability 1/2 of a
head after the first has given a head. This is true—we can verify it by tossing
twocoinsmanytimesandobservingtheproportionofheadsonthesecondtoss
after the first toss has produced a head. We say that the events “head on the
independence firsttoss”and“headonthesecondtoss”areindependent.Independencemeans
thattheoutcomeofthefirsttosscannotinfluencetheoutcomeofthesecond
toss.
EXAMPLE11.1 Independentornot?
Becauseacoinhasnomemoryandmostcointosserscannotinfluencethefallofthe
coin,itissafetoassumethatsuccessivecointossesareindependent.Forabalanced
cointhismeansthatafterweseetheoutcomeofthefirsttoss,westillassignprobabil-
ity1/2toheadsonthesecondtoss.Ontheotherhand,thecolorsofsuccessivecards
dealtfromthesamedeckarenotindependent.Astandard52-carddeckcontains26
redand26blackcards.Forthefirstcarddealtfromashuffleddeck,theprobability
ofaredcardis26/52=0.50(equallylikelyoutcomes).Onceweseethatthefirst
cardisred,weknowthatthereareonly25redsamongtheremaining51cards.The
probabilitythatthesecondcardisredisthereforeonly25/51=0.49.Knowingthe
outcomeofthefirstdealchangestheprobabilitiesforthesecond.
Ifadoctormeasuresyourbloodpressuretwice,itisreasonabletoassumethatthe
tworesultsareindependentbecausethefirstresultdoesnotinfluencetheinstrument
thatmakesthesecondreading.ButifyoutakeanIQtestorothermentaltesttwice
insuccession,thetwotestscoresarenotindependent.Thelearningthatoccurson
thefirstattemptinfluencesyoursecondattempt.
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Independenceandthemultiplicationrule 283
MULTIPLICATIONRULEFORINDEPENDENTEVENTS
Twoevents Aand B areindependentifknowingthatoneoccursdoes
notchangetheprobabilitythattheotheroccurs.If Aand B are
independent,
P(Aand B) = P(A)P(B)
EXAMPLE11.2 Surviving?
During World War II, the British found that the probability that a bomber is lost
throughenemyactiononamissionoveroccupiedEuropewas0.05.Theprobability
that the bomber returns safely from a mission was therefore 0.95. It is reasonable
to assume that missions are independent. Take A to be the event that a bomber
i
survivesitsithmission.Theprobabilityofsurviving2missionsis
P(A and A )= P(A )P(A )
1 2 1 2
=(0.95)(0.95)=0.9025
Themultiplicationruleappliestomorethantwoevents,providedthatallareinde-
pendent.Sotheprobabilityofsurviving3missionsis
P(A and A and A )= P(A )P(A )P(A )
1 2 3 1 2 3
=(0.95)(0.95)(0.95)=0.8574
Theprobabilityofsurviving20missionsisonly
P(A and A and ... and A )= P(A )P(A )···P(A )
1 2 20 1 2 20
=(0.95)(0.95)···(0.95)
=(0.95)20 =0.3585
Thetourofdutyforanairmanwas30missions.
The multiplication rule P(Aand B) = P(A)P(B) holds if Aand B are in-
dependent but not otherwise. The addition rule P(Aor B) = P(A)+ P(B)
holds if Aand B are disjoint but not otherwise. Resist the temptation to use
these simple rules when the circumstances that justify them are not present.
You must also be certain not to confuse disjointness and independence. If
AandBaredisjoint,thenthefactthat AoccurstellsusthatBcannotoccur—
lookagainatFigure11.1.Sodisjointeventsarenotindependent.Unlikedis-
jointness, we cannot picture independence in a Venn diagram, because it in-
volvestheprobabilitiesoftheeventsratherthanjusttheoutcomesthatmake
uptheevents.
APPLYYOURKNOWLEDGE
11.1 Car colors. Exercise1.3(page7)givesthedistributionofcolorsfor
motorvehiclesmadeinNorthAmerica.Thisdistributionwillbe
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284 CHAPTER11 GeneralRulesofProbability
approximatelytrueforallvehiclessoldinthepastfewyears.Youstand
byahighwayandchoosetworecent-modelvehiclesatrandom.
(a) Itisreasonabletoassumethatthecolorsofthevehiclesyouchoose
areindependent.Explainwhy.
(b) Whatistheprobabilitythatthetwovehiclesyouchooseareboth
black?
(c) Whatistheprobabilitythatthefirstvehicleisblackandthesecond
vehicleiswhite?
11.2 Albinism. Thegeneforalbinisminhumansisrecessive.Thatis,
carriersofthisgenehaveprobability1/2ofpassingittoachild,andthe
childisalbinoonlyifbothparentspassthealbinismgene.Parentspass
theirgenesindependentlyofeachother.Ifbothparentscarrythe
albinismgene,whatistheprobabilitythattheirfirstchildisalbino?If
theyhavetwochildren(whoinheritindependentlyofeachother),
whatistheprobabilitythatbotharealbino?Thatneitherisalbino?
11.3 College-educated laborers? Governmentdatashowthat27%of
employedpeoplehaveatleast4yearsofcollegeandthat14%of
employedpeopleworkaslaborersoroperatorsofmachinesorvehicles.
Nonetheless,wecan’tconcludethatbecause(0.27)(0.14)=0.038
about3.8%ofemployedpeoplearecollege-educatedlaborersor
operators.Whynot?
Applyingthemultiplicationrule
If two events Aand B are independent, the event that Adoes not occur is
also independent of B, and so on. Suppose, for example, that 75% of all reg-
istered voters in a rural district are Republicans. If an opinion poll interviews
twovoterschosenindependently,theprobabilitythatthefirstisaRepublican
Wewantaboy andthesecondisnotaRepublicanis(0.75)(0.25)=0.1875.Themultiplica-
tionrulealsoextendstocollectionsofmorethantwoevents,providedthatall
Misunderstanding
independencecanbe are independent. Independence of events A, B, and C means that no infor-
disastrous.“DearAbby”once mationaboutanyoneoranytwocanchangetheprobabilityoftheremaining
publishedaletterfroma events.Independenceisoftenassumedinsettingupaprobabilitymodelwhen
motherofeightgirls.Sheand
theeventswearedescribingseemtohavenoconnection.Wecanthenusethe
herhusbandhadplanneda
multiplicationrulefreely.Hereisanotherexample.
familyoffourchildren.When
allfourweregirls,theykept
trying.Aftersevengirls,her
EXAMPLE11.3 Underseacables
doctorassuredherthat“the
lawofaverageswasinour
Thefirstsuccessfultransatlantictelegraphcablewaslaidin1866.Thefirsttelephone
favor100to1.”Unfortunately,
cableacrosstheAtlanticdidnotappearuntil1956—thebarrierwasdesigning“re-
havingchildrenisliketossing
peaters,” amplifiers needed to boost the signal, that could operate for years on the
coins.Eightgirlsinarowis
seabottom.Thisfirstcablehad52repeaters.Thelastcoppercable,laidin1983and
highlyunlikely,butonceseven
girlshavebeenborn,itisnot retired in 1994, had 662 repeaters. The first fiber-optic cable was laid in 1988 and
atallunlikelythatthenext has109repeaters.Therearenowmorethan400,000milesofunderseacable,with
childwillbeagirl—anditwas. morebeinglaideveryyeartohandlethefloodofInternettraffic.
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Applyingthemultiplicationrule 285
Repeatersinunderseacablesmustbeveryreliable.Toseewhy,supposethateach
repeaterhasprobability0.999offunctioningwithoutfailurefor25years.Repeaters
failindependentlyofeachother.(Thisassumptionmeansthatthereareno“common
causes”suchasearthquakesthatwouldaffectseveralrepeatersatonce.)Denoteby
A theeventthattheithrepeateroperatessuccessfullyfor25years.
i
Theprobabilitythat2repeatersbothlast25yearsis
P(A and A )= P(A )P(A )
1 2 1 2
=0.999×0.999=0.998
Foracablewith10repeaterstheprobabilityofnofailuresin25yearsis
P(A and A and ... and A )=0.99910 =0.990
1 2 10
Cableswith2or10repeaterswouldbequitereliable.Unfortunately,thelastcopper
transatlanticcablehad662repeaters.Theprobabilitythatall662workfor25years
is
P(A and A and ... and A )=0.999662 =0.516
1 2 662
Thiscablewillfailtoreachits25-yeardesignlifeabouthalfthetimeifeachrepeater
is 99.9% reliable over that period. The multiplication rule for probabilities shows
thatrepeatersmustbemuchmorethan99.9%reliable.
By combining the rules we have learned, we can compute probabilities for
rathercomplexevents.Hereisanexample.
EXAMPLE11.4 FalsepositivesinHIVtesting
ScreeninglargenumbersofbloodsamplesforHIV,thevirusthatcausesAIDS,uses
anenzymeimmunoassay(EIA)testthatdetectsantibodiestothevirus.Appliedto
peoplewhohavenoHIVantibodies,EIAhasprobabilityabout0.006ofproducinga
falsepositive(thatis,apositivetestresultforasamplethathasnoantibodies).Ifthe
140employeesofamedicalclinicaretestedandall140arefreeofHIVantibodies,
whatistheprobabilitythatatleast1falsepositivewilloccur?
Itisreasonabletoassumeaspartoftheprobabilitymodelthatthetestresultsfor
differentindividualsareindependent.Theprobabilitythatthetestispositivefora
singlepersonis0.006,sotheprobabilityofanegativeresultis1−0.006=0.994.
Theprobabilityofatleast1falsepositiveamongthe140peopletestedistherefore
P(atleastonepositive)=1− P(nopositives)
=1− P(140negatives)
=1−0.994140
=1−0.431=0.569
Theprobabilityisgreaterthan1/2thatatleast1ofthe140peoplewilltestpositive
for HIV, even though no one has the virus. That is why samples that test positive
areretestedusingamoreaccurate“Westernblot”test.
APPLYYOURKNOWLEDGE
11.4 Telemarketing. Telephonemarketersandopinionpollsuse
random-digitdialingequipmenttocallresidentialtelephonenumbers
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286 CHAPTER11 GeneralRulesofProbability
atrandom.ThetelephonepollingfirmZogbyInternationalreports
thattheprobabilitythatacallreachesalivepersonis0.2.1 Callsare
independent.
(a) Atelemarketerplaces5calls.Whatistheprobabilitythatnoneof
themreachesaperson?
(b) WhencallsaremadetoNewYorkCity,theprobabilityof
reachingapersonisonly0.08.Whatistheprobabilitythatnone
of5callsmadetoNewYorkCityreachesaperson?
11.5 Bright lights? Astringofholidaylightscontains20lights.Thelights
arewiredinseries,sothatifanylightfails,thewholestringwillgo
dark.Eachlighthasprobability0.02offailingduringtheholiday
season.Thelightsfailindependentlyofeachother.Whatisthe
probabilitythatthestringoflightswillremainbright?
11.6 Playing the slots. Slotmachinesarenowvideogames,withwinning
determinedbyelectronicrandomnumbergenerators.Intheolddays,
slotmachineswerelikethis:youpullthelevertospinthreewheels;
eachwheelhas20symbols,allequallylikelytoshowwhenthewheel
stopsspinning;thethreewheelsareindependentofeachother.
Supposethatthemiddlewheelhas9bellsamongits20symbols,and
theleftandrightwheelshave1belleach.
(a) Youwinthejackpotifallthreewheelsshowbells.Whatisthe
probabilityofwinningthejackpot?
(b) Therearethreewaysthatthethreewheelscanshowtwobellsand
onesymbolotherthanabell.Findtheprobabilityofeachofthese
ways.Whatistheprobabilitythatthewheelsstopwithexactly
twobellsshowingamongthem?
Thegeneraladditionrule
Weknowthatif Aand B aredisjointevents,then P(Aor B) = P(A)+ P(B).
Thisadditionruleextendstomorethantwoeventsthataredisjointinthesense
thatnotwohaveanyoutcomesincommon.TheVenndiagraminFigure11.3
shows three disjoint events A, B, and C. The probability that one of these
eventsoccursis P(A)+ P(B)+ P(C).
S
C
A
Figure11.3 Theaddition
rulefordisjointevents: B
P(Aor B orC)= P(A)+
P(B)+ P(C)whenevents
A, B,andCaredisjoint.
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Thegeneraladditionrule 287
Outcomes here are double- S
counted by P(A) + P(B).
A B
A and B
Figure11.4 Thegeneraladditionrule: P(Aor B)= P(A)+ P(B)− P(Aand B)for
anyevents Aand B.
Ifevents Aand B arenotdisjoint,theycanoccursimultaneously.Theprob-
abilitythatoneortheotheroccursisthenlessthanthesumoftheirprobabili-
ties.AsFigure11.4suggests,theoutcomescommontobotharecountedtwice
when we add probabilities, so we must subtract this probability once. Here is
theadditionruleforanytwoevents,disjointornot.
GENERALADDITIONRULEFORANYTWOEVENTS
Foranytwoevents Aand B,
P(Aor B) = P(A)+ P(B)− P(Aand B)
If Aand B are disjoint, the event {Aand B} that both occur contains no
outcomesandthereforehasprobability0.Sothegeneraladditionruleincludes
Rule4,theadditionrulefordisjointevents.
EXAMPLE11.5 Makingpartner
DeborahandMatthewareanxiouslyawaitingwordonwhethertheyhavebeenmade
partnersoftheirlawfirm.Deborahguessesthatherprobabilityofmakingpartneris
0.7andthatMatthew’sis0.5.(ThesearepersonalprobabilitiesreflectingDeborah’s
assessment of chance.) This assignment of probabilities does not give us enough
information to compute the probability that at least one of the two is promoted.
Inparticular,addingtheindividualprobabilitiesofpromotiongivestheimpossible
result1.2.IfDeborahalsoguessesthattheprobabilitythatbothsheandMattheware
madepartnersis0.3,thenbythegeneraladditionrule
P(atleastoneispromoted)=0.7+0.5−0.3=0.9
Theprobabilitythatneitherispromotedisthen0.1byRule3.
Venn diagrams are a great help in finding probabilities because you can
justthinkofaddingandsubtractingareas.Figure11.5showssomeeventsand
their probabilities for Example 11.5. What is the probability that Deborah
is promoted and Matthew is not? The Venn diagram shows that this is the
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288 CHAPTER11 GeneralRulesofProbability
Figure11.5 Venndiagram
andprobabilitiesfor Neither D nor M
Example11.5. 0.1
M and not D
0.2
D and not M D and M
0.4 0.3
D = Deborah is made partner
M = Matthew is made partner
probabilitythatDeborahispromotedminustheprobabilitythatbotharepro-
moted, 0.7−0.3 = 0.4. Similarly, the probability that Matthew is promoted
and Deborah is not is 0.5−0.3 = 0.2. The four probabilities that appear in
thefigureaddto1becausetheyrefertofourdisjointeventsthatmakeupthe
entiresamplespace.
APPLYYOURKNOWLEDGE
11.7 Tastesinmusic. Musicalstylesotherthanrockandpoparebecoming
morepopular.Asurveyofcollegestudentsfindsthat40%likecountry
music,30%likegospelmusic,and10%likeboth.
(a) MakeaVenndiagramwiththeseresults.
(b) Whatpercentofcollegestudentslikecountrybutnotgospel?
(c) Whatpercentlikeneithercountrynorgospel?
11.8 Prosperity and education. Callahouseholdprosperousifitsincome
exceeds$100,000.Callthehouseholdeducatedifthehouseholder
completedcollege.SelectanAmericanhouseholdatrandom,andlet
Abetheeventthattheselectedhouseholdisprosperousand B the
eventthatitiseducated.AccordingtotheCurrentPopulation
Survey,in2001 P(A) = 0.138, P(B) = 0.261,andtheprobabilitythat
ahouseholdisbothprosperousandeducatedis P(Aand B) = 0.082.2
(MarkC,Burnett/Stock,Boston)
(a) DrawaVenndiagramthatshowstherelationbetweentheevents
Aand B.Whatistheprobability P(Aor B)thatthehousehold
selectediseitherprosperousoreducated?
(b) Inyourdiagram,shadetheeventthatthehouseholdiseducated
butnotprosperous.Whatistheprobabilityofthisevent?
Conditionalprobability
Theprobabilityweassigntoaneventcanchangeifweknowthatsomeother
eventhasoccurred.Thisideaisthekeytomanyapplicationsofprobability.
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Conditionalprobability 289
EXAMPLE11.6 Raceandethnicity
The 2000 census allowed people to choose what race they consider themselves to
be,andseparatelywhetherornottheyconsiderthemselvestobe“Hispanic/Latino.”
HerearetheprobabilitiesforarandomlychosenAmerican:
Hispanic NotHispanic Total
Asian 0.000 0.036 0.036
Black 0.003 0.121 0.124
White 0.060 0.691 0.751
Other 0.062 0.027 0.089
Politicallycorrect
Total 0.125 0.875 In1950,theRussian
mathematicianB.V.
The“Total”rowandcolumnareobtainedfromtheprobabilitiesinthebodyofthe Gnedenko(1912–1995)wrote
TheTheoryofProbability,atext
table by the addition rule. For example, the probability that a randomly chosen
thatwaspopulararoundthe
Americanisblackis
world.Theintroduction
P(black)= P(blackandHispanic)+ P(blackandnotHispanic) containsamystifying
=0.003+0.121=0.124 paragraphthatbegins,“We
notethattheentire
Now we are told that the person chosen is Hispanic. That is, he or she is one of
developmentofprobability
the12.5%inthe“Hispanic”columnofthetable.Theprobabilitythatapersonis theoryshowsevidenceofhow
black,giventheinformationthatheorsheisHispanic,istheproportionofblacksinthe itsconceptsandideaswere
“Hispanic”column: crystallizedinaseverestruggle
0.003 betweenmaterialisticand
P(black|Hispanic)= =0.024 idealisticconceptions.”Itturns
0.125
outthat“materialistic”is
Thisisaconditional probability.Youcanreadthebar|as“giventheinformation jargonfor“Marxist-Leninist.”
that.” Itwasgoodforthehealthof
RussianscientistsintheStalin
eratoaddsuchstatementsto
Although12.4%oftheentirepopulationisblack,only2.4%ofHispanicsare theirbooks.
black.It’scommonsensethatknowingthatoneevent(thepersonisHispanic)
occursoftenchangestheprobabilityofanotherevent(thepersonisblack).The conditional probability
examplealsoshowshowweshoulddefineconditionalprobability.Theideaof
aconditionalprobability P(B | A)ofoneevent B giventhatanotherevent A
occursistheproportionofalloccurrencesof Aforwhich B alsooccurs.
CONDITIONALPROBABILITY
When P(A) > 0,theconditional probabilityof B given Ais
P(Aand B)
P(B | A) =
P(A)
Besuretokeepinmindthedistinctrolesoftheevents AandBin P(B | A).
Event Arepresents the information we are given, and B is the event whose