Probability And Stochastic Processes : A Friendly Introduction For Electrical And Computer Engineers, 3rd Edition Solution Manual

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Probability and Stochastic ProcessesA Friendly Introduction for Electrical and Computer EngineersThird Edition’S SOLUTION MANUALRoy D. Yates, David J. Goodman, David FamolariSeptember 8, 2014Comments on this Solutions Manual•This solution manual is mostly complete. Please send error reports, suggestions, andcomments toryates@winlab.rutgers.edu.•To make solution sets for your class, use theSolution Set Constructorat the instruc-tors sitewww.winlab.rutgers.edu/probsolns.•Send email toryates@winlab.rutgers.edufor access to the instructors site.•Matlabfunctions written as solutions to homework problems can be found in thearchivematsoln3e.zip(available to instructors). OtherMatlabfunctions used inthe text or in these homework solutions can be found in the archivematcode3e.zip.The .m files inmatcode3eare available for download from the Wiley website. Twoother documents of interest are also available for download:–A manualprobmatlab3e.pdfdescribing thematcode3e .mfunctions–The quiz solutions manualquizsol.pdf.•This manual uses a page size matched to the screen of an iPad tablet.If you doprint on paper and you have good eyesight, you may wish to print two pages persheet in landscape mode. On the other hand, a “Fit to Paper” printing option willcreate “Large Print” output.1

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Problem Solutions – Chapter 1Problem 1.1.1 SolutionBased on the Venn diagram on the right, the complete Gerlandaspizza menu is•Regular without toppings•Regular with mushrooms•Regular with onions•Regular with mushrooms and onions•Tuscan without toppings•Tuscan with mushroomsMOTProblem 1.1.2 SolutionBased on the Venn diagram on the right, the answers are mostlyfairly straightforward. The only trickiness is that a pizza is eitherTuscan (T) or Neapolitan (N) so{N, T}is a partition but theyare not depicted as a partition.Specifically, the eventNis theregion of the Venn diagram outside of the “square block” of eventT. If this is clear, the questions are easy.MOT(a) SinceN=Tc,N∩M6=φ. ThusNandMare not mutually exclusive.(b) Every pizza is either Neapolitan (N), or Tuscan (T). HenceN∪T=SsothatNandTare collectively exhaustive. Thus its also (trivially) true thatN∪T∪M=S. That is,R,TandMare also collectively exhaustive.(c) From the Venn diagram,TandOare mutually exclusive.In words, thismeans that Tuscan pizzas never have onions or pizzas with onions are neverTuscan.As an aside, “Tuscan” is a fake pizza designation; one shouldn’tconclude that people from Tuscany actually dislike onions.(d) From the Venn diagram,M∩TandOare mutually exclusive.Thus Ger-landa’s doesn’t make Tuscan pizza with mushrooms and onions.(e) Yes. In terms of the Venn diagram, these pizzas are in the set (T∪M∪O)c.2

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Problem 1.1.3 SolutionAt Ricardo’s, the pizza crust is either Roman (R) or Neapolitan(N).To draw the Venn diagram on the right, we make the fol-lowing observations:RNMOW•The set{R, N}is a partition so we can draw the Venn diagram with thispartition.•Only Roman pizzas can be white. HenceW⊂R.•Only a Neapolitan pizza can have onions. HenceO⊂N.•Both Neapolitan and Roman pizzas can have mushrooms so that eventMstraddles the{R, N}partition.•The Neapolitan pizza can have both mushrooms and onions soM∩Ocannotbe empty.•The problem statement does not preclude putting mushrooms on a whiteRoman pizza. Hence the intersectionW∩Mshould not be empty.Problem 1.2.1 Solution(a) An outcome specifies whether the connection speed is high (h), medium (m),or low (l) speed, and whether the signal is a mouse click (c) or a tweet (t).The sample space isS={ht, hc, mt, mc, lt, lc}.(1)(b) The event that the wi-fi connection is medium speed isA1={mt, mc}.(c) The event that a signal is a mouse click isA2={hc, mc, lc}.(d) The event that a connection is either high speed or low speed isA3={ht, hc, lt, lc}.3

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(e) SinceA1∩A2={mc}and is not empty,A1,A2, andA3are not mutuallyexclusive.(f) SinceA1∪A2∪A3={ht, hc, mt, mc, lt, lc}=S,(2)the collectionA1,A2,A3is collectively exhaustive.Problem 1.2.2 Solution(a) The sample space of the experiment isS={aaa, aaf, af a, f aa, f f a, f af, af f, f f f}.(1)(b) The event that the circuit fromZfails isZF={aaf, af f, f af, f f f}.(2)The event that the circuit fromXis acceptable isXA={aaa, aaf, af a, af f}.(3)(c) SinceZF∩XA={aaf, af f} 6=φ,ZFandXAare not mutually exclusive.(d) SinceZF∪XA={aaa, aaf, af a, af f, f af, f f f} 6=S,ZFandXAare notcollectively exhaustive.(e) The event that more than one circuit is acceptable isC={aaa, aaf, af a, f aa}.(4)The event that at least two circuits fail isD={f f a, f af, af f, f f f}.(5)(f) Inspection shows thatC∩D=φsoCandDare mutually exclusive.(g) SinceC∪D=S,CandDare collectively exhaustive.4

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Problem 1.2.3 SolutionThe sample space isS={A♣, . . . , K♣, A♦, . . . , K♦, A♥, . . . , K♥, A♠, . . . , K♠}.(1)The eventHthat the first card is a heart is the setH={A♥, . . . , K♥}.(2)The eventHhas 13 outcomes, corresponding to the 13 hearts in a deck.Problem 1.2.4 SolutionThe sample space isS=1/1. . .1/31,2/1. . .2/29,3/1. . .3/31,4/1. . .4/30,5/1. . .5/31,6/1. . .6/30,7/1. . .7/31,8/1. . .8/31,9/1. . .9/31,10/1. . .10/31,11/1. . .11/30,12/1. . .12/31.(1)The eventHdefined by the event of a July birthday is given by the following etwith 31 sample outcomes:H={7/1,7/2, . . . ,7/31}.(2)Problem 1.2.5 SolutionOf course, there are many answers to this problem. Here are four partitions.1. We can divide students into engineers or non-engineers.LetA1equal theset of engineering students andA2the non-engineers. The pair{A1, A2}is apartition.2. To separate students by GPA, letBidenote the subset of students with GPAsGsatisfyingi−1≤G < i. At Rutgers,{B1, B2, . . . , B5}is a partition. NotethatB5is the set of all students with perfect 4.0 GPAs.Of course, otherschools use different scales for GPA.3. We can also divide the students by age. LetCidenote the subset of studentsof ageiin years. At most universities,{C10, C11, . . . , C100}would be an eventspace. Since a university may have prodigies either under 10 or over 100, wenote that{C0, C1, . . .}is always a partition.5

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4. Lastly, we can categorize students by attendance. LetD0denote the numberof students who have missed zero lectures and letD1denote all other students.Although it is likely thatD0is an empty set,{D0, D1}is a well definedpartition.Problem 1.2.6 SolutionLetR1andR2denote the measured resistances. The pair (R1, R2) is an outcomeof the experiment. Some partitions include1. If we need to check that neither resistance is too high, a partition isA1={R1<100, R2<100},A2={R1≥100} ∪ {R2≥100}.(1)2. If we need to check whether the first resistance exceeds the second resistance,a partition isB1={R1> R2}B2={R1≤R2}.(2)3. If we need to check whether each resistance doesn’t fall below a minimumvalue (in this case 50 ohms forR1and 100 ohms forR2), a partition isC1, C2, C3, C4whereC1={R1<50, R2<100},C2={R1<50, R2≥100},(3)C3={R1≥50, R2<100},C4={R1≥50, R2≥100}.(4)4. If we want to check whether the resistors in parallel are within an acceptablerange of 90 to 110 ohms, a partition isD1={(1/R1+ 1/R2)−1<90},(5)D2={90≤(1/R1+ 1/R2)−1≤110},(6)D2={110<(1/R1+ 1/R2)−1}.(7)6

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Problem 1.3.1 Solution(a)AandBmutually exclusive and collectively exhaustive imply P[A]+P[B] = 1.Since P[A] = 3 P[B], we have P[B] = 1/4.(b) Since P[A∪B] = P[A], we see thatB⊆A. This implies P[A∩B] = P[B].Since P[A∩B] = 0, then P[B] = 0.(c) Since it’s always true that P[A∪B] = P[A] + P[B]−P[AB], we have thatP[A] + P[B]−P[AB] = P[A]−P[B].(1)This implies 2 P[B] = P[AB]. However, sinceAB⊂B, we can conclude that2 P[B] = P[AB]≤P[B]. This implies P[B] = 0.Problem 1.3.2 SolutionThe roll of the red and white dice can be assumed to be independent.For eachdie, all rolls in{1,2, . . . ,6}have probability 1/6.(a) ThusP[R3W2] = P[R3] P[W2] =136.(b) In fact, each pair of possible rollsRiWjhas probability 1/36. To find P[S5],we add up the probabilities of all pairs that sum to 5:P[S5] = P[R1W4] + P[R2W3] + P[R3W2] + P[R4W1] = 4/36 = 1/9.Problem 1.3.3 SolutionAn outcome is a pair (i, j) whereiis the value of the first die andjis the value ofthe second die. The sample space is the setS={(1,1),(1,2), . . . ,(6,5),(6,6)}.(1)with 36 outcomes, each with probability 1/36 Note that the event that the absolutevalue of the difference of the two rolls equals 3 isD3={(1,4),(2,5),(3,6),(4,1),(5,2),(6,3)}.(2)Since there are 6 outcomes inD3, P[D3] = 6/36 = 1/6.7

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Problem 1.3.4 Solution(a) FALSE. Since P[A] = 1−P[Ac] = 2 P[Ac] implies P[Ac] = 1/3.(b) FALSE. SupposeA=Band P[A] = 1/2. In that case,P [AB] = P [A] = 1/2>1/4 = P [A] P [B].(1)(c) TRUE. SinceAB⊆A, P[AB]≤P[A], This impliesP [AB]≤P [A]<P [B].(2)(d) FALSE: For a counterexample, letA=φand P[B]>0 so thatA=A∩B=φand P[A] = P[A∩B] = 0 but 0 = P[A]<P[B].Problem 1.3.5 SolutionThe sample space of the experiment isS={LF, BF, LW, BW}.(1)From the problem statement, we know that P[LF]=0.5, P[BF]=0.2 andP[BW] = 0.2.This implies P[LW] = 1−0.5−0.2−0.2 = 0.1.The questionscan be answered using Theorem 1.5.(a) The probability that a program is slow isP [W] = P [LW] + P [BW] = 0.1 + 0.2 = 0.3.(2)(b) The probability that a program is big isP [B] = P [BF] + P [BW] = 0.2 + 0.2 = 0.4.(3)(c) The probability that a program is slow or big isP [W∪B] = P [W] + P [B]−P [BW] = 0.3 + 0.4−0.2 = 0.5.(4)8

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Problem 1.3.6 SolutionA sample outcome indicates whether the cell phone is handheld (H) or mobile (M)and whether the speed is fast (F) or slow (W). The sample space isS={HF, HW, M F, M W}.(1)The problem statement tells us that P[HF] = 0.2, P[M W] = 0.1 and P[F] = 0.5.We can use these facts to find the probabilities of the other outcomes. In particular,P [F] = P [HF] + P [M F].(2)This impliesP [M F] = P [F]−P [HF] = 0.5−0.2 = 0.3.(3)Also, since the probabilities must sum to 1,P [HW] = 1−P [HF]−P [M F]−P [M W]= 1−0.2−0.3−0.1 = 0.4.(4)Now that we have found the probabilities of the outcomes, finding any other prob-ability is easy.(a) The probability a cell phone is slow isP [W] = P [HW] + P [M W] = 0.4 + 0.1 = 0.5.(5)(b) The probability that a cell hpone is mobile and fast is P[M F] = 0.3.(c) The probability that a cell phone is handheld isP [H] = P [HF] + P [HW] = 0.2 + 0.4 = 0.6.(6)Problem 1.3.7 SolutionA reasonable probability model that is consistent with the notion of a shuffled deckis that each card in the deck is equally likely to be the first card. LetHidenotethe event that the first card drawn is theith heart where the first heart is the ace,9

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the second heart is the deuce and so on. In that case, P[Hi] = 1/52 for 1≤i≤13.The eventHthat the first card is a heart can be written as the mutually exclusiveunionH=H1∪H2∪ · · · ∪H13.(1)Using Theorem 1.1, we haveP [H] =13∑i=1P [Hi] = 13/52.(2)This is the answer you would expect since 13 out of 52 cards are hearts. The pointto keep in mind is that this is not just the common sense answer but is the resultof a probability model for a shuffled deck and the axioms of probability.Problem 1.3.8 SolutionLetsidenote the outcome that the down face hasidots.The sample space isS={s1, . . . , s6}.The probability of each sample outcome is P[si] = 1/6.FromTheorem 1.1, the probability of the eventEthat the roll is even isP [E] = P [s2] + P [s4] + P [s6] = 3/6.(1)Problem 1.3.9 SolutionLetsiequal the outcome of the student’s quiz. The sample space is then composedof all the possible grades that she can receive.S={0,1,2,3,4,5,6,7,8,9,10}.(1)Since each of the 11 possible outcomes is equally likely, the probability of receivinga grade ofi, for eachi= 0,1, . . . ,10 is P[si] = 1/11.The probability that thestudent gets an A is the probability that she gets a score of 9 or higher. That isP [Grade of A] = P [9] + P [10] = 1/11 + 1/11 = 2/11.(2)The probability of failing requires the student to get a grade less than 4.P [Failing] = P [3] + P [2] + P [1] + P [0]= 1/11 + 1/11 + 1/11 + 1/11 = 4/11.(3)10

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Problem 1.3.10 SolutionEach statement is a consequence of part 4 of Theorem 1.4.(a) SinceA⊂A∪B, P[A]≤P[A∪B].(b) SinceB⊂A∪B, P[B]≤P[A∪B].(c) SinceA∩B⊂A, P[A∩B]≤P[A].(d) SinceA∩B⊂B, P[A∩B]≤P[B].Problem 1.3.11 SolutionSpecifically, we will use Theorem 1.4(c) which states that for any eventsAandB,P [A∪B] = P [A] + P [B]−P [A∩B].(1)To prove the union bound by induction, we first prove the theorem for the case ofn= 2 events. In this case, by Theorem 1.4(c),P [A1∪A2] = P [A1] + P [A2]−P [A1∩A2].(2)By the first axiom of probability, P[A1∩A2]≥0. Thus,P [A1∪A2]≤P [A1] + P [A2].(3)which proves the union bound for the casen= 2.Now we make our inductionhypothesis that the union-bound holds for any collection ofn−1 subsets. In thiscase, given subsetsA1, . . . , An, we defineA=A1∪A2∪ · · · ∪An−1,B=An.(4)By our induction hypothesis,P [A] = P [A1∪A2∪ · · · ∪An−1]≤P [A1] +· · ·+ P [An−1].(5)This permits us to writeP [A1∪ · · · ∪An] = P [A∪B]≤P [A] + P [B](by the union bound forn= 2)= P [A1∪ · · · ∪An−1] + P [An]≤P [A1] +· · ·P [An−1] + P [An](6)which completes the inductive proof.11

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Problem 1.3.12 SolutionIt is tempting to use the following proof:SinceSandφare mutually exclusive, and sinceS=S∪φ,1 = P [S∪φ] = P [S] + P [φ].Since P[S] = 1, we must have P[φ] = 0.The above “proof” used the property that for mutually exclusive setsA1andA2,P [A1∪A2] = P [A1] + P [A2].(1)The problem is that this property is a consequence of the three axioms, and thusmust be proven. For a proof that uses just the three axioms, letA1be an arbitraryset and forn= 2,3, . . ., letAn=φ. SinceA1=∪∞i=1Ai, we can use Axiom 3 towriteP [A1] = P [∪∞i=1Ai] = P [A1] + P [A2] +∞∑i=3P [Ai].(2)By subtracting P[A1] from both sides, the fact thatA2=φpermits us to writeP [φ] +∞∑n=3P [Ai] = 0.(3)By Axiom 1, P[Ai]≥0 for alli.Thus,∑∞n=3P[Ai]≥0.This implies P[φ]≤0.Since Axiom 1 requires P[φ]≥0, we must have P[φ] = 0.Problem 1.3.13 SolutionFollowing the hint, we define the set of events{Ai|i= 1,2, . . .}such thati=1, . . . , m,Ai=Biand fori > m,Ai=φ.By construction,∪mi=1Bi=∪∞i=1Ai.Axiom 3 then impliesP [∪mi=1Bi] = P [∪∞i=1Ai] =∞∑i=1P [Ai].(1)12

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Fori > m, P[Ai] = P[φ] = 0, yielding the claim P[∪mi=1Bi] =∑mi=1P[Ai] =∑mi=1P[Bi].Note that the fact that P[φ] = 0 follows from Axioms 1 and 2. This problem ismore challenging if you just use Axiom 3. We start by observingP [∪mi=1Bi] =m−1∑i=1P [Bi] +∞∑i=mP [Ai].(2)Now, we use Axiom 3 again on the countably infinite sequenceAm, Am+1, . . .towrite∞∑i=mP [Ai] = P [Am∪Am+1∪ · · ·] = P [Bm].(3)Thus, we have used just Axiom 3 to prove Theorem 1.3:P [∪mi=1Bi] =m∑i=1P [Bi].(4)Problem 1.3.14 SolutionEach claim in Theorem 1.4 requires a proof from which we can check which axiomsare used.However, the problem is somewhat hard because there may still bea simpler proof that uses fewer axioms.Still, the proof of each part will needTheorem 1.3 which we now prove.For the mutually exclusive eventsB1, . . . , Bm, letAi=Bifori= 1, . . . , mandletAi=φfori > m. In that case, by Axiom 3,P [B1∪B2∪ · · · ∪Bm] = P [A1∪A2∪ · · ·]=m−1∑i=1P [Ai] +∞∑i=mP [Ai]=m−1∑i=1P [Bi] +∞∑i=mP [Ai].(1)13

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Now, we use Axiom 3 again onAm, Am+1, . . .to write∞∑i=mP [Ai] = P [Am∪Am+1∪ · · ·] = P [Bm].(2)Thus, we have used just Axiom 3 to prove Theorem 1.3:P [B1∪B2∪ · · · ∪Bm] =m∑i=1P [Bi].(3)(a) To show P[φ] = 0, letB1=Sand letB2=φ. Thus by Theorem 1.3,P [S] = P [B1∪B2] = P [B1] + P [B2] = P [S] + P [φ].(4)Thus, P[φ] = 0. Note that this proof uses only Theorem 1.3 which uses onlyAxiom 3.(b) Using Theorem 1.3 withB1=AandB2=Ac, we haveP [S] = P [A∪Ac] = P [A] + P [Ac].(5)Since, Axiom 2 says P[S] = 1, P[Ac] = 1−P[A]. This proof uses Axioms 2and 3.(c) By Theorem 1.8, we can write bothAandBas unions of mutually exclusiveevents:A= (AB)∪(ABc),B= (AB)∪(AcB).(6)Now we apply Theorem 1.3 to writeP [A] = P [AB] + P [ABc],P [B] = P [AB] + P [AcB].(7)We can rewrite these facts asP[ABc] = P[A]−P[AB],P[AcB] = P[B]−P[AB].(8)Note that so far we have used only Axiom 3. Finally, we observe thatA∪Bcan be written as the union of mutually exclusive eventsA∪B= (AB)∪(ABc)∪(AcB).(9)14

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Once again, using Theorem 1.3, we haveP[A∪B] = P[AB] + P[ABc] + P[AcB](10)Substituting the results of Equation (8) into Equation (10) yieldsP [A∪B] = P [AB] + P [A]−P [AB] + P [B]−P [AB],(11)which completes the proof. Note that this claim required only Axiom 3.(d) Observe that sinceA⊂B, we can writeBas the mutually exclusive unionB=A∪(AcB). By Theorem 1.3 (which uses Axiom 3),P [B] = P [A] + P [AcB].(12)By Axiom 1, P[AcB]≥0, hich implies P[A]≤P[B]. This proof uses Axioms 1and3.Problem 1.4.1 SolutionEach question requests a conditional probability.(a) Note that the probability a call is brief isP [B] = P [H0B] + P [H1B] + P [H2B] = 0.6.(1)The probability a brief call will have no handoffs isP [H0|B] = P [H0B]P [B]= 0.40.6 = 23.(2)(b) The probability of one handoff is P[H1] = P[H1B] + P[H1L] = 0.2.Theprobability that a call with one handoff will be long isP [L|H1] = P [H1L]P [H1]= 0.10.2 = 12.(3)(c) The probability a call is long is P[L] = 1−P[B] = 0.4. The probability thata long call will have one or more handoffs isP [H1∪H2|L] = P [H1L∪H2L]P [L]= P [H1L] + P [H2L]P [L]= 0.1 + 0.20.4= 34.(4)15

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