Probability and Stochastic Processes - 公告

1 Probability and Stochastic ProcessesA Friendly Introduction for Electrical and Computer EngineersSecond EditionQuiz SolutionsRoy D. Yates and David J. GoodmanMay 22, 2004 The MATLAB section quizzes at the end of each chapter use programs available fordownload as the This archive has programs of general pur-pose programs for solving Probability problems as well as associatedwith examples or quizzes in the text. Also available is a the general We have made a substantial effort to check the solution to every quiz. Nevertheless,there is a nonzero Probability (in fact, a Probability close to unity) that errors will befound. If you find errors or have suggestions or comments, please send email errors are found, corrected solutions will be posted at the Solutions Chapter 1 Quiz the Venn diagrams for parts (a)-(g) below, the shaded area represents the (1)R=Tc(2)M O(3)M OMOTMOTMOT(4)R M(4)R M(6)Tc MQuiz (1)A1={vvv, vvd,vdv, vdd}(2)B1={dvv,dvd,ddv,ddd}(3)A2={vvv, vvd,dvv,dvd}(4)B2={vdv, vdd,ddv,ddd}(5)A3={vvv,ddd}(6)B3={vdv,dv d}(7)A4={vvv, vvd,vdv,dvv, vdd,dvd,ddv}(8)B4={ddd,ddv,dvd,vdd}Recal l thatAiandBiare collectively exhaustive ifAi Bi=S.

2 Also,AiandBiaremutually exclusive ifAi Bi= . Since we have written down each pairAiandBiabove,we can simply check for these pairA1andB1are mutually exclusive and collectively exhaustive. The pairA2andB2are mutually exclusive and collectively exhaustive. The pairA3andB3are mutuallyexclusive butnotcollectively exhaustive. The pairA4andB4are not mutually exclusivesincedvdbelongs toA4andB4. However,A4andB4are collectively are exactly 50 equally likely outcomes:s51throughs100. Each of these outcomeshas Probability (1)P[{s79}] = (2)P[{s100}] = (3)P[A]=P[{s90,..,s100}] =11 (4)P[F]=P[{s51,..,s59}] =9 (5)P[T 80]=P[{s80,..,s100}] =21 (6)P[T<90]=P[{s51,s52,..,s89}] =39 (7)P[aCgrade or better]=P[{s70,..,s100}] =31 (8)P[student passes]=P[{s60,..,s100}] =41 can describe this experiment by the event space consisting of the four possibleeventsVB,VL,DB, andDL.

3 We represent these events in the ?B??In a roundabout way, the problem statement tells us how to fill in the table. In particular,P[V]= [VL]+P[VB](1)P[L]= [VL]+P[DL](2)SinceP[VL]= , we can conclude thatP[VB]= and thatP[DL]= This allows us to fill in two more table ?The remaining table entry is filled in by observing that the probabilities must sum to impliesP[DB]= and the complete table the various probabilities is now straightforward:3(1)P[DL]= (2)P[D L]=P[VL]+P[DL]+P[DB]= + + (3)P[VB]= (4)P[V L]=P[V]+P[L] P[VL]= + (5)P[V D]=P[S]=1(6)P[LB]=P[LLc]=0 Quiz (1) The Probability of exactly two voice calls isP[NV=2]=P[{vvd,vdv,dvv}]= (1)(2) The Probability of at least one voice call isP[NV 1]=P[{vdd,dvd,ddv, vvd,vdv,dvv, vvv}](2)=6( )+ (3)An easier way to get the same answer is to observe thatP[NV 1]=1 P[NV<1]=1 P[NV=0]=1 P[{ddd}]= (4)(3) The conditional Probability of two voice calls followed by a data call given that therewere two voice calls isP[{vvd}|NV=2]=P[{vvd},NV=2]P[NV=2]=P[{ vvd}]P[NV=2]= (5)(4) The conditional Probability of two data calls followed by a voice call given therewere two voice calls isP[{ddv}|NV=2]=P[{ddv},NV=2]P[NV=2]=0(6 )

4 The joint event of the outcomeddvand exactly two voice calls has Probability zerosince there is only one voice call in the outcomeddv.(5) The conditional Probability of exactly two voice calls given at least one voice call isP[NV=2|Nv 1]=P[NV=2,NV 1]P[NV 1]=P[NV=2]P[NV 1]= (7)(6) The conditional Probability of at least one voice call given there were exactly twovoice calls isP[NV 1|NV=2]=P[NV 1,NV=2]P[NV=2]=P[NV=2]P[NV=2]=1(8)Given that there were two voice calls, there must have been at least one voice this experiment, there are four outcomes with probabilitiesP[{vv}] =( )2= [{vd}] =( )( )= [{dv}] =( )( )= [{dd}] =( )2= checking the independence of any two eventsAandB, it s wise to avoid intuitionand simply check whetherP[AB]=P[A]P[B]. Using the probabilities of the outcomes,we now can test for the independence of events.

5 (1) First, we calculate the Probability of the joint event:P[NV=2,NV 1]=P[NV=2]=P[{vv}]= (1)Next, we observe thatP[NV 1]=P[{vd,dv, vv}]= (2)Finally, we make the comparisonP[NV=2]P[NV 1]=( )( ) =P[NV=2,NV 1](3)which shows the two events are dependent.(2) The Probability of the joint event isP[NV 1,C1=v]=P[{vd,vv}]= (4)From part (a),P[NV 1]= Further,P[C1=v]= so thatP[NV 1]P[C1=v]=( )( )= =P[NV 1,C1=v](5)Hence, the events are dependent.(3) The problem statement that the calls were independent implies that the events thesecond call is a voice call,{C2=v}, and the first call is a data call,{C1=d}areindependent events. Just to be sure, we can do the calculations to check:P[C1=d,C2=v]=P[{dv}]= (6)SinceP[C1=d]P[C2=v]=( )( )= , we confirm that the events areindependent. Note that this shouldn t be surprising since we used the information thatthe calls were independent in the problem statement to determine the probabilities ofthe outcomes.

6 (4) The Probability of the joint event isP[C2=v,NVis even]=P[{vv}]= (7)Also, each event has probabilityP[C2=v]=P[{dv, vv}]= ,P[NVis even]=P[{dd,vv}]= (8)Thus,P[C2=v]P[NVis even]=( )( )= SinceP[C2=v,NVis even] = , the events are the event that that the user is found on pagei. The tree for the experimentis user is found unless all three paging attempts fail. Thus the Probability the user isfound isP[F]=1 P Fc1Fc2Fc3 =1 ( )3= (1)Quiz (1) We can view choosing each bit in the code word as a subexperiment. Each subex-periment has two possible outcomes: 0 and 1. Thus by the fundamental principle ofcounting, there are 2 2 2 2=24=16 possible code words.(2) An experiment that can yield all possible code words with two zeroes is to choosewhich 2 bits (out of 4 bits) will be zero. The other two bits then must be ones.

7 Thereare 42 =6 ways to do this. Hence, there are six code words with exactly two this problem, it is also possible to simply enumerate the six code words:1100,1010,1001,0101,0110,0011.(3) When the first bit must be a zero, then the first subexperiment of choosing the firstbit has only one outcome. For each of the next three bits, we have two choices. Inthis case, there are 1 2 2 2=8 ways of choosing a code word.(4) For the constant ratio code, we can specify a code word by choosingMof the bits tobe ones. The otherN Mbits will be zeroes. The number of ways of choosing sucha code word is NM .ForN=8 andM=3, there are 83 =56 code (1) In this problem,kbits received in error is the same askfailures in 100 trials. Thefailure Probability is =1 pand the success Probability is 1 =p. That is, theprobability ofkbits in error and 100 kcorrectly received bits isP Sk,100 k = 100k k(1 )100 k(1)6 For = ,P S0,100 =(1 )100=( )100= (2)P S1,99 =100( )( )99= (3)P S2,98 =4950( )2( )98= (4)P S3,97 =161,700( )3( )97= (5)(2) The Probability a packet is decoded correctly is justP[C]=P S0,100 +P S1,99 +P S2,98 +P S3,97 = (6)Quiz the chip works only if allntransistors work, the transistors in the chip are likedevices in series.

8 The Probability that a chip works isP[C]= module works if either 8 chips work or 9 chips work. LetCkdenote the event thatexactlykchips work. Since transistor failures are independent of each other, chip failuresare also independent. Thus eachP[Ck]has the binomial probabilityP[C8]= 98 (P[C])8(1 P[C])9 8=9p8n(1 pn),(1)P[C9]=(P[C])9=p9n.(2)The Probability a memory module works isP[M]=P[C8]+P[C9]=p8n(9 8pn)(3)Quiz (1,100);X=(R<= ) ..+ (2*(R> ).*(R<= )) ..+ (3*(R> ));Y=hist(X,1:3)For a MATLAB simulation, we first gen-erate a vectorRof 100 random , we generate vectorXas a func-tion ofRto represent the 3 possible out-comes of a flip. That is,X(i)=1if flipiwas heads,X(i)=2if flipiwas tails, andX(i)=3)is flipilanded on the see how this works, we note there are three cases: IfR(i) <= , thenX(i)=1. < R(i)andR(i)<= , thenX(i)=2.

9 < R(i), thenX(i)= three cases will have probabilities , and Lastly, we use thehistfunctionto count how many occurences of each possible value ofX(i).7 Quiz Solutions Chapter 2 Quiz sample space, probabilities and corresponding grades for the experiment areOutcomeP[ ] (1) To findc, we recall that the PMF must sum to 1. That is,3 n=1PN(n)=c 1+12+13 =1(1)This impliesc=6/11. Now that we have foundc, the remaining parts are straight-forward.(2)P[N=1]=PN(1)=c=6/11(3)P[N 2]=PN(2)+PN(3)=c/2+c/3=5/11(4)P[N>3]= n=4PN(n)=0 Quiz each transmitted bit is an independent trial where we call a bit error a suc-cess. Each bit is in error, that is, the trial is a success, with probabilityp. Now we caninterpret each experiment in the generic context of independent trials.(1) The random variableXis the number of trials up to and including the first to Example ,Xhas the geometric PMFPX(x)= p(1 p)x 1x=1,2.

10 0otherwise(1)(2) Ifp= , then the Probability exactly 10 bits are sent isP[X=10]=PX(10)=( )( )9= (2)8 The Probability that at least 10 bits are sent isP[X 10]= x=10PX(x). Thissum is not too hard to calculate. However, its even easier to observe thatX 10 ifthe first 10 bits are transmitted correctly. That is,P[X 10]=P[first 10 bits are correct]=(1 p)10(3)Forp= ,P[X 10]= (3) The random variableYis the number of successes in 100 independent trials. Just asin Example ,Yhas the binomial PMFPY(y)= 100y py(1 p)100 y(4)Ifp= , the Probability of exactly 2 errors isP[Y=2]=PY(2)= 1002 ( )2( )98= (5)(4) The Probability of no more than 2 errors isP[Y 2]=PY(0)+PY(1)+PY(2)(6)=( )100+100( )( )99+ 1002 ( )2( )98(7)= (8)(5) Random variableZis the number of trials up to and including the third success. ThusZhas the Pascal PMF (see Example )PZ(z)= z 12 p3(1 p)z 3(9)Note thatPZ(z)>0 forz=3,4,5.