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Probability and Stochastic ProcessesA Friendly Introduction for Electrical and Computer EngineersSECOND EDITIONProblem SolutionsJuly 26, 2004 DraftRoy D. Yates and David J. GoodmanJuly 26, 2004 This solution manual remains under construction. The current count is that 575 out of 695problems in the text are solved here, including all problems through Chapter 5. At the moment, we have not confirmed the correctness of every single solution. If you finderrors or have suggestions or comments, please send email to [email protected] M ATLAB functions written as solutions to homework probalems can be found in the archivematsoln.zip (available to instructors) or in the directory matsoln. Other M ATLAB functions used in the text or in these hoemwork solutions can be found in the archive matcode.zipor directory matcode. The .m files in matcode are available for download from the Wileywebsite. Two oter documents of interest are also available for download:– A manual probmatlab.pdf describing the matcode .m functions is also available.– The quiz solutions manual quizsol.pdf. A web-based solution set constructor for the second edition is also under construction. A major update of this solution manual will occur prior to September, 2004.1

Problem Solutions – Chapter 1Problem 1.1.1 SolutionBased on the Venn diagramOMTthe answers are fairly straightforward:(a) Since T M φ, T and M are not mutually exclusive.(b) Every pizza is either Regular (R), or Tuscan (T ). Hence R T S so that R and T arecollectively exhaustive. Thus its also (trivially) true that R T M S. That is, R, T andM are also collectively exhaustive.(c) From the Venn diagram, T and O are mutually exclusive. In words, this means that Tuscanpizzas never have onions or pizzas with onions are never Tuscan. As an aside, “Tuscan” isa fake pizza designation; one shouldn’t conclude that people from Tuscany actually dislikeonions.(d) From the Venn diagram, M T and O are mutually exclusive. Thus Gerlanda’s doesn’t makeTuscan pizza with mushrooms and onions.(e) Yes. In terms of the Venn diagram, these pizzas are in the set (T M O)c .Problem 1.1.2 SolutionBased on the Venn diagram,OMTthe complete Gerlandas pizza menu is Regular without toppings Regular with mushrooms Regular with onions Regular with mushrooms and onions Tuscan without toppings Tuscan with mushrooms2

Problem 1.2.1 Solution(a) An outcome specifies whether the fax is high (h), medium (m), or low (l) speed, and whetherthe fax has two (t) pages or four ( f ) pages. The sample space isS {ht, h f, mt, m f, lt, l f } .(1)(b) The event that the fax is medium speed is A1 {mt, m f }.(c) The event that a fax has two pages is A2 {ht, mt, lt}.(d) The event that a fax is either high speed or low speed is A3 {ht, h f, lt, l f }.(e) Since A1 A2 {mt} and is not empty, A1 , A2 , and A3 are not mutually exclusive.(f) SinceA1 A2 A3 {ht, h f, mt, m f, lt, l f } S,(2)the collection A1 , A2 , A3 is collectively exhaustive.Problem 1.2.2 Solution(a) The sample space of the experiment isS {aaa, aa f, a f a, f aa, f f a, f a f, a f f, f f f } .(1)(b) The event that the circuit from Z fails isZ F {aa f, a f f, f a f, f f f } .(2)The event that the circuit from X is acceptable isX A {aaa, aa f, a f a, a f f } .(3)(c) Since Z F X A {aa f, a f f } φ, Z F and X A are not mutually exclusive.(d) Since Z F X A {aaa, aa f, a f a, a f f, f a f, f f f } S, Z F and X A are not collectivelyexhaustive.(e) The event that more than one circuit is acceptable isC {aaa, aa f, a f a, f aa} .(4)The event that at least two circuits fail isD { f f a, f a f, a f f, f f f } .(f) Inspection shows that C D φ so C and D are mutually exclusive.(g) Since C D S, C and D are collectively exhaustive.3(5)

Problem 1.2.3 SolutionThe sample space isS {A , . . . , K , A , . . . , K , A , . . . , K , A , . . . , K } .(1)The event H is the setH {A , . . . , K } .(2)Problem 1.2.4 SolutionThe sample space is 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,S 9/1 . . . 9/31, 10/1 . . . 10/31, 11/1 . . . 11/30, 12/1 . . . 12/31 .(1)The event H defined by the event of a July birthday is described by following 31 sample points.H {7/1, 7/2, . . . , 7/31} .(2)Problem 1.2.5 SolutionOf course, there are many answers to this problem. Here are four event spaces.1. We can divide students into engineers or non-engineers. Let A1 equal the set of engineeringstudents and A2 the non-engineers. The pair {A1 , A2 } is an event space.2. We can also separate students by GPA. Let Bi denote the subset of students with GPAs Gsatisfying i 1 G i. At Rutgers, {B1 , B2 , . . . , B5 } is an event space. Note that B5 isthe set of all students with perfect 4.0 GPAs. Of course, other schools use different scales forGPA.3. We can also divide the students by age. Let Ci denote the subset of students of age i in years.At most universities, {C10 , C11 , . . . , C100 } would be an event space. Since a university mayhave prodigies either under 10 or over 100, we note that {C0 , C1 , . . .} is always an event space4. Lastly, we can categorize students by attendance. Let D0 denote the number of students whohave missed zero lectures and let D1 denote all other students. Although it is likely that D0 isan empty set, {D0 , D1 } is a well defined event space.Problem 1.2.6 SolutionLet R1 and R2 denote the measured resistances. The pair (R1 , R2 ) is an outcome of the experiment.Some event spaces include1. If we need to check that neither resistance is too high, an event space isA1 {R1 100, R2 100} ,A2 {either R1 100 or R2 100} .4(1)

2. If we need to check whether the first resistance exceeds the second resistance, an event spaceisB2 {R1 R2 } .(2)B1 {R1 R2 }3. If we need to check whether each resistance doesn’t fall below a minimum value (in this case50 ohms for R1 and 100 ohms for R2 ), an event space isC1 {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 acceptable range of 90 to110 ohms, an event space is D1 (1/R1 1/R2 ) 1 90 ,(5) D2 90 (1/R1 1/R2 ) 1 110 ,(6) 1D2 110 (1/R1 1/R2 ).(7)Problem 1.3.1 SolutionThe sample space of the experiment isS {L F, B F, L W, BW } .(1)From the problem statement, we know that P[L F] 0.5, P[B F] 0.2 and P[BW ] 0.2. Thisimplies P[L W ] 1 0.5 0.2 0.2 0.1. The questions can be answered using Theorem 1.5.(a) The probability that a program is slow isP [W ] P [L W ] P [BW ] 0.1 0.2 0.3.(2)(b) The probability that a program is big isP [B] P [B F] 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)Problem 1.3.2 SolutionA sample outcome indicates whether the cell phone is handheld (H ) or mobile (M) and whetherthe speed is fast (F) or slow (W ). The sample space isS {H F, H W, M F, M W } .(1)The problem statement tells us that P[H F] 0.2, P[M W ] 0.1 and P[F] 0.5. We can usethese facts to find the probabilities of the other outcomes. In particular,P [F] P [H F] P [M F] .5(2)

This impliesP [M F] P [F] P [H F] 0.5 0.2 0.3.(3)Also, since the probabilities must sum to 1,P [H W ] 1 P [H F] 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 probability is easy.(a) The probability a cell phone is slow isP [W ] P [H W ] 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 [H F] P [H W ] 0.2 0.4 0.6.(6)Problem 1.3.3 SolutionA reasonable probability model that is consistent with the notion of a shuffled deck is that each cardin the deck is equally likely to be the first card. Let Hi denote the event that the first card drawn isthe ith heart where the first heart is the ace, the second heart is the deuce and so on. In that case,P[Hi ] 1/52 for 1 i 13. The event H that the first card is a heart can be written as the disjointunion(1)H H1 H2 · · · H13 .Using Theorem 1.1, we have13P [H ] P [Hi ] 13/52.(2)i 1This is the answer you would expect since 13 out of 52 cards are hearts. The point to keep in mindis that this is not just the common sense answer but is the result of a probability model for a shuffleddeck and the axioms of probability.Problem 1.3.4 SolutionLet si denote the outcome that the down face has i dots. The sample space is S {s1 , . . . , s6 }. Theprobability of each sample outcome is P[si ] 1/6. From Theorem 1.1, the probability of the eventE that the roll is even is(1)P [E] P [s2 ] P [s4 ] P [s6 ] 3/6.Problem 1.3.5 SolutionLet si equal the outcome of the student’s quiz. The sample space is then composed of all the possiblegrades that she can receive.S {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} .6(1)

Since each of the 11 possible outcomes is equally likely, the probability of receiving a grade of i, foreach i 0, 1, . . . , 10 is P[si ] 1/11. The probability that the student gets an A is the probabilitythat she gets a score of 9 or higher. That isP [Grade of A] P  P  1/11 1/11 2/11.(2)The probability of failing requires the student to get a grade less than 4.P Failing P  P  P  P  1/11 1/11 1/11 1/11 4/11.(3)Problem 1.4.1 SolutionFrom the table we look to add all the disjoint events that contain H0 to express the probability thata caller makes no hand-offs asP [H0 ] P [L H0 ] P [B H0 ] 0.1 0.4 0.5.(1)In a similar fashion we can express the probability that a call is brief byP [B] P [B H0 ] P [B H1 ] P [B H2 ] 0.4 0.1 0.1 0.6.(2)The probability that a call is long or makes at least two hand-offs isP [L H2 ] P [L H0 ] P [L H1 ] P [L H2 ] P [B H2 ] 0.1 0.1 0.2 0.1 0.5.(3)(4)Problem 1.4.2 Solution(a) From the given probability distribution of billed minutes, M, the probability that a call isbilled for more than 3 minutes isP [L] 1 P [3 or fewer billed minutes](1) 1 P [B1 ] P [B2 ] P [B3 ](2) 1 α α(1 α) α(1 α)2(3) (1 α) 0.57.(4)3(b) The probability that a call will billed for 9 minutes or less is9P [9 minutes or less] α(1 α)i 1 1 (0.57)3 .i 17(5)

Problem 1.4.3 SolutionThe first generation consists of two plants each with genotype yg or gy. They are crossed to producethe following second generation genotypes, S {yy, yg, gy, gg}. Each genotype is just as likelyas any other so the probability of each genotype is consequently 1/4. A pea plant has yellow seedsif it possesses at least one dominant y gene. The set of pea plants with yellow seeds 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 1.4.4 SolutionEach statement is a consequence of part 4 of Theorem 1.4.(a) Since A A B, P[A] P[A B].(b) Since B A B, P[B] P[A B].(c) Since A B A, P[A B] P[A].(d) Since A B B, P[A B] P[B].Problem 1.4.5 SolutionSpecifically, we will use Theorem 1.7(c) which states that for any events A and 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 the case of n 2 events. Inthis case, by Theorem 1.7(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 induction hypothesis that theunion-bound holds for any collection of n 1 subsets. In this case, given subsets A1 , . . . , An , wedefineB An .(4)A A1 A2 · · · An 1 ,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](6) P [A] P [B](by the union bound for n 2)(7) P [A1 · · · An 1 ] P [An ](8) P [A1 ] · · · P [An 1 ] P [An ](9)which completes the inductive proof.8

Problem 1.4.6 Solution(a) For convenience, let pi P[F Hi ] and qi P[V Hi ]. Using this shorthand, the six unknownsp0 , p1 , p2 , q0 , q1 , q2 fill the table asFVH0 H1 H2p0 p1 p2 .q0 q1 q2(1)However, we are given a number of facts:p0 q0 1/3,p1 q1 1/3,p2 q2 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 havefour equations and six unknowns, choosing p0 and p1 will specify the other unknowns. Unfortunately, arbitrary choices for either p0 or p1 will lead to negative values for the otherprobabilities. In terms of p0 and p1 , the other unknowns areq0 1/3 p0 ,p2 5/12 ( p0 p1 ),(4)q1 1/3 p1 ,q2 p0 p1 1/12.(5)Because the probabilities must be nonnegative, we see that0 p0 1/3,(6)0 p1 1/3,(7)1/12 p0 p1 5/12.(8)Although there are an infinite number of solutions, three possible solutions are:p0 1/3,p1 1/12,p2 0,q0 0,q1 1/4,q2 1/3.(10)p0 1/4,p1 1/12,p2 1/12,(11)q0 1/12,q1 3/12,q2 3/12.(12)p0 0,p1 1/12,p2 1/3,(13)q0 1/3,q1 3/12,q2 0.(14)(9)andand(b) In terms of the pi , qi notation, the new facts are p0 1/4 and q1 1/6. These extra factsuniquely specify the probabilities. In this case,p0 1/4,p1 1/6,p2 0,(15)q0 1/12,q1 1/6,q2 1/3.(16)9

Problem 1.4.7 SolutionIt is tempting to use the following proof:Since S and φ are mutually exclusive, and since S S φ,1 P [S φ] P [S] P [φ] .(1)Since P[S] 1, we must have P[φ] 0.The above “proof” used the property that for mutually exclusive sets A1 and A2 ,P [A1 A2 ] P [A1 ] P [A2 ] .(2)The problem is that this property is a consequence of the three axioms, and thus must be proven. Fo