概率论总结范文.doc

概率论总结范文 北京邮电大学国际学院高欢Introduction tothe Probability1.1the fundamentalknowledge aboutprobability Experiments:It isused inprobability theoryto describevirtually everyprogress whoseoute isnot knownin advancewith certainty.1.repeat again and again2.Many outesExperimentscharacteristic3.Know allthe outes,but donot knowwhich onewill be obtained insome experiment4.Simple events,multiple eventsthe sample space:the collectionof allpossible outesof an experiment.For example:roll asix-sided dice,the sample space containsthe sixnumbers1,2,3,4,5,6.It canbe writtenas S=1,2,3,4,5,61.2the definitionof ProbabilityIn a given experiment,it isnecessary toassign to each event A in the samplespace Sa numberPr(A)which indicatesthe probability that Awill our1.For everyevent A,Pr(A)0Axiom2.Pr(S)=13.For everyinfinite sequence of disjoint events A1,A2,.11PrPr()iiiiAA=?=?1.Pr(?)=02.For everyfinite sequenceof ndisjointevents A1,.,An,Theorems11PrPr()nniiiiAA=?=?3.For everyevent A,Pr(Ac)=1-Pr(A)4.For everyevent A,0Pr()1A5.For everytwo events A and B,Pr(AB)=Pr(A)+Pr(B)-Pr(AB)Corollaries fromthe Axioms:If A?B,then Pr(BAc)=Pr(B)?Pr(A)If A?B,then Pr(A)Pr(B)Example1Consider two events A and Bsuch that Pr(A)=1/3and Pr(B)=1/2.Determine the value of Pr(BAc)for each of the following conditions:(a)A and B aredisjoint;(b)A?B;(c)Pr(AB)=1/8Solution:(a)Pr(BAC)=1/2(b)Pr(BAC)=1/2-1/3=1/6(c)Pr(BAC)=Pr(B)-Pr(AB)=1/2-1/8=3/81.3simple SampleSpaces and Combination MethodThe definitionof simple samplespace:A samplespace Scontaining noutes s1,.,sn iscalled asimple samplespace if the probabilityassigned toeach of the outess1,sn is1/n.If an event Ain thissimplesamplespace containsexactly moutes,then Pr(A)=mn.Example1:Consider anexperiment inwhich afair coinis tossedonce anda balanceddie isrolled once.(a)Describe the samplespacefor thisexperiment.(b)What isthe probability that ahead will be obtained on thecoin andan oddnumber will beobtained on the die?Solution:(a)123456head(head,1)(head,2)(head,3)(head,4)(head,5)(head,6)tail(tail,1)(tail,2)(tail,3)(tail,4)(tail,5)(tail,6)S=(head,1)(head,2)(head,3)(head,4)(head,5)(head,6)(tail,1)(tail,2)(tail,3)(tail,4)(tail,5)(tail,6)(b)3=1/4Pr(A)=26Binomial cofficients!()!nnkk nk?=?For allnumbers x and yand eachpositive integern,(x+y)n=0nkn k?knx yk=?Example2The UnitedStates Senatecontains twosenators from eachof the50states.(a)If amittee ofeight senatorsis selected at random,what isthe probability that itwill contain at least one of the twosenators from a certainspecified state?(b)what isthe probabilitythat agroup of50senators selected at randomwill containone senatorfromeachstate?Solution:(a)Pr(A)=12798226988100C CCCC+(b)Pr(B)=50501002C1.4The probability of aUnion ofEvents Theorem1:Pr(123AAA)=Pr(A1)+Pr(A2)+Pr(A3)-Pr(A1A2)+Pr(A2A3)+Pr(A1A3)+Pr(A1A2A3).Example1:Suppose thatfour guestscheck theirhats when they arriveat arestaurant,and that these hatsare returnedto themin a random orderwhen theyleave.Determine the probabilitythatno guestswill receivethe properhat.Solution:Pr(A)=4433A=38Example2:Among agroup of200students,137students areenrolled in a mathematicsclass,50students areenrolled ina historyclass,and124students areenrolled ina musilass.Furthermore,the number of students enrolled inboth the math andhistory classes is33,the numberof studentsenrolled inboth historyand musilassesis29,the numberof studentsenrolled inboth mathand musicis92.Finally thenumberof studentsenrolled in allthree classesis18.We shalldetermine the probabilitythata studentselectedat random fromthe groupof200students willbe enrolled inatleast oneof thethree classes.Solution:let Adenote theevent that the selectedstudent isenrolled in themathclass.Let Bdenote theevent that the selectedstudent isenrolledin the historyclass.Let Cdenote theevent thatthe selectedstudent isenrolledinthe musilass.Pr(A)=137/200Pr(B)=50/200Pr(C)=124/200Pr(AB)=33/200Pr(BC)=29/200Pr(AC)=92/200Pr(ABC)=18/200We canget Pr(ABC)=7/82conditional probability2.1the definitionof conditionalprobability Pr(A|B)=Pr()Pr()ABB Pr(AB)=Pr(B)Pr(A|B)or Pr(AB)=Pr(A)Pr(B|A)Suppose that A1,A2,.An areevents such that Pr(A1A2.An-1)0.Then Pr(A1A2.An)=Pr(A1)Pr(A2|A1)Pr(A3|A1A2)Pr(An|A1A2An-1)Suppose thatA1,A2,An,B areevents such thatPr(A1A2.An-1|B)0.Then Pr(A1A2An|B)=Pr(A1|B)Pr(A2|A1B)Pr(An|A1A2.An-1B)Example1Selecting four balls.Suppose thatfourballsare selectedone ata time,without replacement,from abox containingr redballs andb blueballs(r=2,b=2).We shalldetermine the probabilityofobtaining thesequenceofoutes red,blue,red,blue.Solution:If welet Rjdenote theevent thata redball isobtainedon the jthdraw and let Bjdenote theevent thata blueball isobtainedonthe jthdraw,then Pr(R1B2R3B4)=Pr(R1)Pr(B2|R1)Pr(R3|R1B2)Pr(B4|R1B2R3)1rbr?+?+?=1123brb rbrbrb?+?2.2Independent EventsDefinition:A andB are independent if and only if Pr(A|B)=Pr(A)and Pr(B|A)=Pr(B).Pr(AB)=Pr(A)Pr(B)=0Theorem IftwoeventsA andB are independent,then the eventsAand BCare alsoindependent Pr(ABC)=Pr(A)Pr(BC)In particular,in orderfor threeeventsA,B,andCto be independent,thefollowingfour relationsmust besatisfied:Pr(AB)=Pr(A)Pr(B)Pr(AC)=Pr(A)Pr(C)Pr(BC)=Pr(B)Pr(C)Pr(ABC)=Pr(A)Pr(B)Pr(C)If AandB are independentevent,then ACand BCare alsoindependent Proof:Pr(ACBC)=1-Pr(AB)=1-Pr(A)-Pr(B)+Pr(A)Pr(B)=1-Pr(B)-Pr(A)1-Pr(B)=1-Pr(B)1-Pr(A)=Pr(AC)Pr(BC)Example1Two studentsAandBareboth registeredfor acertain course.Assume thatstudent Aattends class80percent of the time,student Battends class60percent ofthe time,and theabsences of two studentsare independent.(a)what isthe probabilitythat atleast oneofthetwo studentswillbein class on agiven day?(b)if atleastoneofthetwo studentsis inclassonagivenday,what isthe probabilitythatAis inclass thatday.Solution:(a)Pr(AB)=Pr(A)+Pr(B)-Pr(AB)=0.8+0.6-0.80.6=0.92(b)Pr(A|AB)=Pr()Pr()AABAB=0.8/0.92=0.8696Example2If AandBare independentevents,Pr(A)=0.4,Pr(B)=0.7,what isthevalue ofPr(AC|BC)Solution:Pr(AC|BC)=Pr(AC)=1-Pr(A)=0.62.3Bayestheorem Theorem1suppose thattheeventsB1Bk from a partitionofthe space Sand Pr(Bj)0for j=1k.Then,for everyeventAin S,Pr(A)=1Pr()Pr(|)kjjjBA B=Theorem2Bayestheorem.Let theevent B1.,Bk froma partitionofthespace SsuchthatPr(Bj)0for j=1,.k,andletA beaneventsuchthatPr(A)0.Then,for i=1,k,Pr(Bi|A)=1Pr()Pr(|)Pr()Pr(|)iikjjjBA BBAB=Example1The percentagesof votersclassed asLiberals inthree differentelection districtare dividedas follows:In thefirst district,21percent;inthesecond district,45percent;and inthe thirddistrict,75percent.If adistrict is selectedatrandom anda voterisselectedatrandomfrom thatdistrict,what isthe probabilitythat shewillbea Liberal?Solution:Pr(A)=1110.210.450.75333+=0.47Example2There is a sicktree,its hostleft thehome togo ona journey.He askhis neighborto waterthe tree.Suppose ifthe sicktree isnt bewatered,the probabilityof treegoing dead is0.8,ifthe tree iswatered,the probabilityof goingdeadis0.15.and wealso knowthe probabilitythattheneighbor remembersto waterthe tree is0.9.(a)when thehost eback,what isthe probabilitythatthe treeisstill alive?(b)whenthehost cameback,he foundthetreehad beendead,what isthe probabilitythattheneighbor forgotto waterthetree.Solution:(a)Pr(A)=0.90.85=0.7850.10.2+(b)Pr(C1|B)=0.10.8+0.10.80.90.15=0.3723Random Variablesand Distribution3.1the additionalkonwledege aboutrandom variableDefinition of a random variable Consideranexperimentfor which thesamplespace isdenoted byS.A real-valued functionthat isdefined onthespaceS iscalled a random variable.In otherwords,inaparticular experimenta random variable Xwould besome functionthat assigna real number X(s)toeachpossible outesS.The uniform distribution onInteger/discrete dis