Probability and Stochastic ProcessesA Friendly Introduction for Electrical and Computer EngineersThird EditionSTUDENT’S SOLUTION MANUAL(Solutions to the odd-numbered problems)Roy D. Yates, David J. Goodman, David FamolariAugust 27, 20141

Comments on this Student Solutions Manual Matlab functions written as solutions to homework problems in this Student’s Solution Manual (SSM) can be found in the archive matsoln3student.zipOther Matlab functions used in the text or in these homework solutionscan be found in the archive The archives matcode3e.zipand are available for download from the John Wileycompanion website. Two other documents of interest are also available fordownload:– A manual probmatlab.pdf describing the .m functions in– The quiz solutions manual quizsol.pdf. This manual uses a page size matched to the screen of an iPad tablet. If youdo print on paper and you have good eyesight, you may wish to print twopages per sheet in landscape mode. On the other hand, a “Fit to Paper”printing option will create “Large Print” output. Send error reports, suggestions, or comments [email protected]

Problem Solutions – Chapter 1Problem 1.1.1 SolutionBased on the Venn diagram on the right, the complete Gerlandas pizza menu is Regular without toppings Regular with mushrooms Regular with onions Regular with mushrooms and onions Tuscan without toppings Tuscan with mushroomsOMTProblem 1.1.3 SolutionRAt Ricardo’s, the pizza crust is either Roman (R) or Neapolitan (N ). To draw the Venn diagram on the right, we makethe following observations:WNMO The set {R, N } is a partition so we can draw the Venn diagram withthis partition. Only Roman pizzas can be white. Hence W R. Only a Neapolitan pizza can have onions. Hence O N . Both Neapolitan and Roman pizzas can have mushrooms so that eventM straddles the {R, N } partition. The Neapolitan pizza can have both mushrooms and onions so M Ocannot be empty. The problem statement does not preclude putting mushrooms on awhite Roman pizza. Hence the intersection W M should not be empty.3

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 atweet (t). The sample space isS {ht, hc, mt, mc, lt, lc} .(1)(b) The event that the wi-fi connection is medium speed is A1 {mt, mc}.(c) The event that a signal is a mouse click is A2 {hc, mc, lc}.(d) The event that a connection is either high speed or low speed is A3 {ht, hc, lt, lc}.(e) Since A1 A2 {mc} and is not empty, A1 , A2 , and A3 are not mutuallyexclusive.(f) SinceA1 A2 A3 {ht, hc, mt, mc, lt, lc} S,(2)the collection A1 , A2 , A3 is collectively exhaustive.Problem 1.2.3 SolutionThe sample space isS {A , . . . , K , A , . . . , K , A , . . . , K , A , . . . , K } .(1)The event H that the first card is a heart is the setH {A , . . . , K } .The event H has 13 outcomes, corresponding to the 13 hearts in a deck.4(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. Let A1 equalthe set of engineering students and A2 the non-engineers. The pair{A1 , A2 } is a partition.2. We can also separate students by GPA. Let Bi denote the subset of students with GPAs G satisfying i 1 G i. At Rutgers, {B1 , B2 , . . . , B5 }is a partition. Note that B5 is the set of all students with perfect 4.0GPAs. Of course, other schools use different scales for GPA.3. We can also divide the students by age. Let Ci denote the subset ofstudents of age i in years. At most universities, {C10 , C11 , . . . , C100 }would be an event space. Since a university may have prodigies eitherunder 10 or over 100, we note that {C0 , C1 , . . .} is always a partition.4. Lastly, we can categorize students by attendance. Let D0 denote thenumber of students who have missed zero lectures and let D1 denote allother students. Although it is likely that D0 is an empty set, {D0 , D1 }is a well defined partition.Problem 1.3.1 Solution(a) A and B mutually 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 that B 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 havethatP[A] P[B] P[AB] P[A] P[B].(1)This implies 2 P[B] P[AB]. However, since AB B, we can concludethat 2 P[B] P[AB] P[B]. This implies P[B] 0.5

Problem 1.3.3 SolutionAn outcome is a pair (i, j) where i is the value of the first die and j is thevalue of the 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 theabsolute value 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 in D3 , P[D3 ] 6/36 1/6.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 0.5.6(4)

Problem 1.3.7 SolutionA reasonable probability model that is consistent with the notion of a shuffleddeck is that each card in the deck is equally likely to be the first card. Let Hidenote the event that the first card drawn is the ith heart where the first heartis the ace, the second heart is the deuce and so on. In that case, P[Hi ] 1/52for 1 i 13. The event H that the first card is a heart can be written asthe mutually exclusive unionH H1 H2 · · · H13 .(1)Using Theorem 1.1, we haveP [H] 13XP [Hi ] 13/52.(2)i 1This is the answer you would expect since 13 out of 52 cards are hearts. Thepoint to keep in mind is that this is not just the common sense answer butis the result of a probability model for a shuffled deck and the axioms ofprobability.Problem 1.3.9 SolutionLet si equal the outcome of the student’s quiz. The sample space is thencomposed of 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 ofreceiving a grade of i, for each i 0, 1, . . . , 10 is P[si ] 1/11. The probabilitythat the student gets an A is the probability that she gets a score of 9 orhigher. 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 4/11.7(3)

Problem 1.3.11 SolutionSpecifically, we will use Theorem 1.4(c) which states that for any events Aand B,P [A B] P [A] P [B] P [A B] .(1)To prove the union bound by induction, we first prove the theorem for thecase of n 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 case n 2.Now we make our inductionhypothesis that the union-bound holds for any collection of n 1 subsets. Inthis case, given subsets A1 , . . . , 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 for n 2) P [A1 · · · An 1 ] P [An ] P [A1 ] · · · P [An 1 ] P [An ](6)which completes the inductive proof.8

Problem 1.3.13 SolutionFollowing the hint, we define the set of events {Ai i 1, 2, . . .} such that i 1, . . . , m, Ai Bi and for i m, Ai φ. By construction, mi 1 Bi i 1 Ai .Axiom 3 then implies Xm P [ i 1 Bi ] P [ i 1 Ai ] P [Ai ] .(1)i 1PmmB] Fori m,P[A] P[φ] 0,yieldingtheclaimP[ iii 1i 1 P[Ai ] PmP[B].ii 1Note that the fact that P[φ] 0 follows from Axioms 1 and 2. This problemis more challenging if you just use Axiom 3. We start by observingP [ mi 1 Bi ] m 1XP [Bi ] i 1 XP [Ai ] .(2)i mNow, we use Axiom 3 again on the countably infinite sequence Am , Am 1 , . . .to write XP [Ai ] P [Am Am 1 · · ·] P [Bm ] .(3)i mThus, we have used just Axiom 3 to prove Theorem 1.3:mXmP [ i 1 Bi ] P [Bi ] .(4)i 1Problem 1.4.1 SolutionEach question requests a conditional probability.(a) Note that the probability a call is brief isP [B] P [H0 B] P [H1 B] P [H2 B] 0.6.(1)The probability a brief call will have no handoffs isP [H0 B] P [H0 B]0.42 .P [B]0.639(2)

(b) The probability of one handoff is P[H1 ] P[H1 B] P[H1 L] 0.2. Theprobability that a call with one handoff will be long isP [L H1 ] P [H1 L]0.11 .P [H1 ]0.22(3)(c) The probability a call is long is P[L] 1 P[B] 0.4. The probabilitythat a long call will have one or more handoffs isP [H1 L H2 L]P [L]0.1 0.23P [H1 L] P [H2 L] . P [L]0.44P [H1 H2 L] (4)Problem 1.4.3 SolutionSince the 2 of clubs is an even numbered card, C2 E so that P[C2 E] P[C2 ] 1/3. Since P[E] 2/3,P [C2 E] P [C2 E]1/3 1/2.P [E]2/3(1)The probability that an even numbered card is picked given that the 2 ispicked isP [E C2 ] P [C2 E]1/3 1.P [C2 ]1/3(2)Problem 1.4.5 SolutionThe first generation consists of two plants each with genotype yg or gy.They are crossed to produce the following second generation genotypes, S {yy, yg, gy, gg}. Each genotype is just as likely as any other so the probability of each genotype is consequently 1/4. A pea plant has yellow seeds if it10

possesses at least one dominant y gene. The set of pea plants with yellowseeds isY {yy, yg, gy} .(1)So the probability of a pea plant with yellow seeds isP [Y ] P [yy] P [yg] P [gy] 3/4.(2)Problem 1.4.7 SolutionThe sample outcomes can be written ijk where the first card drawn is i, thesecond is j and the third is k. The sample space isS {234, 243, 324, 342, 423, 432} .(1)and each of the six outcomes has probability 1/6. The events E1 , E2 , E3 , O1 ,O2 , O3 areE1 {234, 243, 423, 432} ,E2 {243, 324, 342, 423} ,E3 {234, 324, 342, 432} ,O1 {324, 342} ,O2 {234, 432} ,O3 {243, 423} .(2)(3)(4)(a) The conditional probability the second card is even given that the firstcard is even isP [E2 E1 ] P [243, 423]2/6P [E2 E1 ] 1/2.P [E1 ]P [234, 243, 423, 432]4/6(5)(b) The conditional probability the first card is even given that the secondcard is even isP [E1 E2 ] P [E1 E2 ]P [243, 423]2/6 1/2.P [E2 ]P [243, 324, 342, 423]4/611(6)

(c) The probability the first two cards are even given the third card is evenisP [E1 E2 E3 ] P [E1 E2 E3 ] 0.P [E3 ](7)(d) The conditional probabilities the second card is even given that the firstcard is odd isP [E2 O1 ] P [O1 E2 ]P [O1 ] 1.P [O1 ]P [O1 ](8)(e) The conditional probability the second card is odd given that the firstcard is odd isP [O2 O1 ] P [O1 O2 ] 0.P [O1 ](9)Problem 1.5.1 SolutionFrom the table we look to add all the mutually exclusive events to find eachprobability.(a) The probability that a caller makes no hand-offs isP [H0 ] P [LH0 ] P [BH0 ] 0.1 0.4 0.5.(1)(b) The probability that a call is brief isP [B] P [BH0 ] P [BH1 ] P [BH2 ] 0.4 0.1 0.1 0.6.(2)(c) The probability that a call is long or makes at least two hand-offs isP [L H2 ] P [LH0 ] P [LH1 ] P [LH2 ] P [BH2 ] 0.1 0.1 0.2 0.1 0.5.12(3)

Problem 1.5.3 Solution(a) For convenience, let pi P[F Hi ] and qi P[V Hi ]. Using this shorthand, the six unknowns p0 , p1 , p2 , q0 , q1 , q2 fill the table asH0 H1 H2F p0 p1 p2 .V q0 q1 q2(1)However, we are given a number of facts:p0 q0 1/3,p2 q2 1/3,p1 q1 1/3,p0 p1 p2 5/12.(2)(3)Other facts, such as q0 q1 q2 7/12, can be derived from these facts.Thus, we have four equations and six unknowns, choosing p0 and p1 willspecify the other unknowns. Unfortunately, arbitrary choices for eitherp0 or p1 will lead to negative values for the other probabilities. In termsof p0 and p1 , the other unknowns areq0 1/3 p0 ,q1 1/3 p1 ,p2 5/12 (p0 p1 ),q2 p0 p1 1/12.(4)(5)Because the probabilities must be nonnegative, we see that0 p0 1/3,0 p1 1/3,1/12 p0 p1 5/12.(6)(7)(8)Although there are an infinite number of solutions, three possible solutions are:p0 1/3,q0 0,p1 1/12,q1 1/4,13p2 0,q2 1/3.(9)(10)

andp0 1/4,q0 1/12,p1 1/12,q1 3/12,p2 1/12,q2 3/12.(11)(12)p0 0,q0 1/3,p1 1/12,q1 3/12,p2 1/3,q2 0.(13)(14)and(b) In terms of the pi , qi notation, the new facts are p0 1/4 and q1 1/6.These extra facts uniquely specify the probabilities. In this case,p0 1/4,q0 1/12,p1 1/6,q1 1/6,p2 0,q2 1/3.(15)(16)Problem 1.6.1 SolutionThis problems asks whether A and B can be independent events yet satisfyA B? By definition, events A and B are independent if and only if P[AB] P[A] P[B]. We can see that if A B, that is they are the same set, thenP [AB] P [AA] P [A] P [B] .(1)Thus, for A and B to be the same set and also independent,P [A] P [AB] P [A] P [B] (P [A])2 .There are two ways that this requirement can be satisfied: P[A] 1 implying A B S. P[A] 0 implying A B φ.14(2)

Problem 1.6.3 SolutionLet Ai and Bi denote the events that the ith phone sold is an Apricot or aBanana respectively. The works “each phone sold is twice as likely to be anApricot than a Banana” tells us thatP [Ai ] 2 P [Bi ] .(1)However, since each phone sold is