CHAPTER ONE Instructor’s Solutions Manual Anna Amirdjanova University of Michigan, Ann Arbor Neil A Weiss Arizona State University FOR A Course In Probability Neil A Weiss Arizona State University Boston London Mexico City Toronto Munich San Francisco Sydney New York Tokyo Singapore Madrid Paris Cape Town Hong Kong Montreal Publisher: Greg Tobin Editor-in-Chief: Deirdre Lynch Associate Editor: Sara Oliver Gordus Editorial Assistant: Christina Lepre Production Coordinator: Kayla Smith-Tarbox Senior Author Support/Technology Specialist: Joe Vetere Compositor: Anna Amirdjanova and Neil A Weiss Accuracy Checker: Delray Schultz Proofreader: Carol A Weiss Copyright © 2006 Pearson Education, Inc All rights reserved No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher Chapter Probability Basics 1.1 From Percentages to Probabilities Basic Exercises 1.1 a) The population under consideration consists of the five senators specified in the problem statement, namely, Graham, Baucus, Conrad, Murkowski, and Kyl b) Because three of the five senators are Democrats and each senator is equally likely to be the one selected, the probability is 3/5, or 0.6, that the chosen senator is a Democrat 1.2 We represent each possible subcommittee as a subset of two of the the five senators Also, for convenience, we use the first letter of each senator’s last name to represent the senator a) The 10 possible subcommittees are as follows: {G, B}, {G, C}, {G, M}, {G, K}, {B, C}, {B, M}, {B, K}, {C, M}, {C, K}, {M, K} b) Of the 10 possible subcommittees, exactly three consist of two Democrats, namely, {G, B}, {G, C}, and {B, C} Hence, the probability that the committee will consist of two Democrats is 3/10, or 0.3 c) Of the 10 possible subcommittees, exactly one consists of two Republicans, namely, {M, K} Hence, the probability that the committee will consist of two Republicans is 1/10, or 0.1 d) Of the 10 possible subcommittees, exactly six consist of one Democrat and one Republican, namely, {G, M}, {G, K}, {B, M}, {B, K}, {C, M}, and {C, K} Hence, the probability that the committee will consist of one Democrat and one Republican is 6/10, or 0.6 1.3 We represent each possible chair/vice-chair possibility as an ordered pair of two of the five senators, the first entry being the chair and the second the vice-chair Also, for convenience, we use the first letter of each senator’s last name to represent the senator a) The 20 possibilities are (G, B), (B, G), (G, C), (C, G), (G, M), (M, G), (G, K), (K, G), (B, C), (C, B), (B, M), (M, B), (B, K), (K, B), (C, M), (M, C), (C, K), (K, C), (M, K), (K, M) b) Equally probable Indeed, in the scenario described in Exercise 1.1, the probability that Graham will be chosen as chair is 1/5, or 0.2 In the current scenario, of the 20 possibilities, exactly four have Graham as chair, namely, (G, B), (G, C), (G, M), and (G, K); thus, the probability that Graham will be chosen as chair is 4/20, or 0.2 c) Of the 20 possibilities, exactly 12 have a Democrat as chair, namely, (G, B), (B, G), (G, C), (C, G), (G, M), (G, K), (B, C), (C, B), (B, M), (B, K), (C, M), and (C, K) Hence, the probability that the senator chosen to be chair is a Democrat is 12/20, or 0.6 1.4 We work in number of thousands of housing units From the table, we see that the total number of housing units is 471 + 1,470 + · · · + 15,647 = 112,356 1-1 1-2 CHAPTER Probability Basics a) The number of housing units with (exactly) four rooms is 23,468 Hence, the probability that the housing unit selected has four rooms is 23,468/112,356 ≈ 0.209 b) The number of housing units with more than four rooms is 24,476 + 21,327 + 13,782 + 15,647 = 75,232 Thus, the probability that the housing unit selected has more than four rooms is 75,232/112,356 ≈ 0.670 c) The number of housing units with either one or two rooms is 471 + 1,470 = 1,941 Hence, the probability that the housing unit selected has either one or two rooms is 1,941/112,356 ≈ 0.0173 d) None of the housing units have fewer than one room Hence, the probability that the housing unit selected has fewer than one room is 0/112,356 = e) All of the housing units have one or more rooms Hence, the probability that the housing unit selected has one or more rooms is 112,356/112,356 = f) The population under consideration consists of all U.S housing units 1.5 From the table, we see that the total number of murder cases during the year in question in which the person murdered was between 20 and 59 years old, inclusive, is 2,916 + 2,175 + · · · + 372 = 11,527 a) The number of these murder cases in which the person murdered was between 40 and 44 years old, inclusive, is 1213 Hence, the probability that the murder victim of the case selected was between 40 and 44 years old, inclusive, is 1,213/11,527 ≈ 0.105 b) We see that the number of these murder cases in which the person murdered was 25 years old or older is 11,527 − 2,916 = 8,611 Consequently, the probability that the murder victim of the case selected was 25 years old or older is 8,611/11,527 ≈ 0.747 c) The number of these murder cases in which the person murdered was between 45 and 59 years old, inclusive, is 888 + 540 + 372 = 1800 Hence, the probability that the murder victim of the case selected was between 45 and 59 years old, inclusive, is 1,800/11,527 ≈ 0.156 d) The number of these murder cases in which the person murdered was either under 30 or over 54 is 2916 + 2175 + 372 = 5463 Consequently, the probability that the murder victim of the case selected was either under 30 or over 54 is 5,463/11,527 ≈ 0.474 e) The population under consideration consists of all murder cases during the year in question in which the person murdered was between 20 and 59 years old, inclusive 1.6 Answers will vary, depending on the students in your class Let us, however, denote by ℓ1 and ℓ2 the number of left-handed females and left-handed males in your class, respectively, and by n1 and n2 the number of females and males in your class, respectively Note that the total number of students in your class is represented by n1 + n2 a) The number of females in your class is n1 Hence, the probability that a randomly selected student in your class is female is n1 /(n1 + n2 ) b) The number of left-handed students in your class is ℓ1 + ℓ2 Hence, the probability that a randomly selected student in your class is left-handed is (ℓ1 + ℓ2 )/(n1 + n2 ) c) The number of left-handed females in your class is ℓ1 Hence, the probability that a randomly selected student in your class is a left-handed female is ℓ1 /(n1 + n2 ) 1.1 From Percentages to Probabilities 1-3 d) A student is neither female nor left-handed if and only if the student is a right-handed male The number of right-handed males in your class is the number of males who are not left-handed, which is n2 − ℓ2 Hence, the probability that a randomly selected student in your class is neither female nor left-handed is (n2 − ℓ2 )/(n1 + n2 ) 1.7 Because we are selecting at random from a finite population (the students at the university in question), probabilities are the same as percentages a) We know that 62% of the students are bilingual and that 80% of those students speak Spanish Now, 80% of 62% is 49.6% Hence, 49.6% of the students speak Spanish, so that 50.4% don’t Thus, the probability that a randomly selected student at this university doesn’t speak Spanish is 0.504 b) From part (a), 49.6% of the students speak Spanish and we know that 10% of the Spanish-speaking students also speak French Now, 10% of 49.6% is 4.96% Hence, 4.96% of the students speak both Spanish and French Consequently, the probability that a randomly selected student at this university speaks both Spanish and French is 0.0496 1.8 Answers will vary However, to illustrate, we tossed a balanced coin 20 times and obtained the following table Here E denotes the event that, on any particular toss, the coin comes up a head Toss n Outcome (H or T) Number of heads n(E ) Proportion of heads n(E )/n 10 11 12 13 14 15 16 17 18 19 20 T H H T H T T T H H T T T H H H H T T T 2 3 3 5 5 9 9 0.000 0.500 0.667 0.500 0.600 0.500 0.429 0.375 0.444 0.500 0.455 0.417 0.385 0.429 0.467 0.500 0.529 0.500 0.474 0.450 a) See the first five rows of the preceding table b) Referring to the frequentist interpretation of probability on page and the fifth row of the preceding table, we estimate, based on the first five tosses of the coin, that the probability of a head is 3/5 = 0.6 c) Referring to the frequentist interpretation of probability and the 10th row of the preceding table, we estimate, based on the first 10 tosses of the coin, that the probability of a head is 5/10 = 0.5 d) Referring to the frequentist interpretation of probability and the 20th row of the preceding table, we estimate, based on the first 20 tosses of the coin, that the probability of a head is 9/20 = 0.45 e) We see from parts (b)–(d) that, based on the frequentist interpretation of probability, our estimate of the probability of a head will fluctuate This phenomenon will be the case regardless of how many times we toss the coin CHAPTER Probability Basics 1-4 1.9 Note that 2004 was a leap year, so it had 366 days Now, consider the random experiment of selecting one 2004 U.S state governor at random, and let E denote the event that the governor chosen is a Republican We know that P (E) = 28/50 = 0.56 Thus, from the frequentist interpretation of probability, if we independently repeat the random experiment n times, then the proportion of times that event E occurs will approximately equal 0.56 Consequently, in 366 repetitions of the random experiment, we would expect event E to occur roughly 0.56 · 366 = 204.96 times In other words, if on each of the 366 days of 2004, one U.S state governor was randomly selected to read the invocation on a popular radio program, then we would expect a Republican to be chosen on approximately 205 of those days 1.10 In each case, we refer to the frequentist interpretation of probability on page a) Approximately 31.4% of human gestation periods exceed months b) The favorite finishes in the money in roughly two-thirds of horse races c) About 40% of traffic fatalities involve an intoxicated or alcohol-impaired driver or nonoccupant 1.11 Let E be an event and let P (E) denote its probability For n independent repetitions of the random experiment, let n(E) denote the number of times that event E occurs Then, it follows from Relation (1.1) on page that n(E) ≈ nP (E), for large n (∗) a) The random experiment consists of observing the duration of a human gestation period and event E is that the gestation period exceeds months We know that P (E) = 0.314 Here n = 4000 so that from Relation (∗), n(E) ≈ 4000 · 0.314 = 1256 Hence, of 4000 human gestation periods, roughly 1256 will exceed months b) The random experiment consists of observing a horse race and event E is that the favorite finishes in the money We know that P (E) = 2/3 Here n = 500 so that from Relation (∗), 1000 = ≈ 333.3 3 Therefore, in 500 horse races, the favorite will finish in the money about 333 times c) The random experiment consists of observing a traffic fatality (or reading a traffic-fatality report) and event E is that the traffic fatality involves an intoxicated or alcohol-impaired driver or nonoccupant We know that P (E) = 0.40 Here n = 389 so that from Relation (∗), n(E) ≈ 500 · n(E) ≈ 389 · 0.40 = 155.6 Therefore, in 389 traffic fatalities, approximately 156 will involve an intoxicated or alcohol-impaired driver or nonoccupant Advanced Exercises 1.12 The probability that the ball lands on a red number is 18/38, so that the odds against that event are − 18/38 to 18/38 or 20/38 to 18/38 or 20 to 18 or 10 to Hence, to make the bet fair, the house should pay 10 to for a bet on red; that is, for a winning bet on red, the house should pay the gambler $10 for each $9 bet 1.13 a) As the odds against Fusaichi Pegasus were to 5, the probability that Fusaichi Pegasus would win the race is 5/(5 + 3) = 0.625 b) Because the odds against Red Bullet were to 2, the probability that Red Bullet would win the race is 2/(2 + 9) ≈ 0.182 1.2 Set Theory 1-5 1.2 Set Theory Note: For convenience, we use “iff” to represent the phrase “if and only if.” Basic Exercises 1.14 To avoid trivialities, we assume that there are at least two sets in the collection Hence, let A1 , A2 , be pairwise disjoint sets Then, in particular, we have A1 ∩ A2 = ∅ Consequently, An ⊂ A1 ∩ A2 = ∅ n As the empty set is a subset of every set, we conclude that n An = ∅ Therefore, if the sets in a collection are pairwise disjoint, then the intersection of all the sets in the collection must be empty 1.15 Answers will vary, but here is one possibility: 1.16 Answers will vary, but here is one possibility: 1.17 Answers will vary, but the subsets of R2 portrayed in the solution to Exercise 1.16 provide one possibility For another possibility, let U = {a, b, c, d, e, f } and consider the following four subsets of U : {a, b, e}, {b, c, f }, {c, d, e}, {a, d, f } We note that each pairwise intersection is nonempty and, hence, in particular, the four sets are not pairwise disjoint; however, every three sets have an empty intersection 1.18 a) We note that A1 ⊃ A2 ⊃ · · · Therefore, 4 An = A4 = [0, 1/4] n=1 An = A1 = [0, 1] and n=1 CHAPTER Probability Basics 1-6 b) We have x ∈ ∞ n=1 An iff x ∈ An for all n ∈ N iff x ∈ [0, 1/n] for all n ∈ N iff ≤ x ≤ 1/n for all n ∈ N iff x = Hence, ∞ An = {0} n=1 As A1 ⊃ A2 ⊃ · · · , we have ∞ An = A1 = [0, 1] n=1 1.19 In each part, let An denote the set in the union or intersection It will be useful to note the following facts, which hold for all real numbers c ● x < c iff x ≤ c − 1/n for some n ∈ N ● x > c iff x ≥ c + 1/n for some n ∈ N ● x ≤ c iff x < c + 1/n for all n ∈ N ● x ≥ c iff x > c − 1/n for all n ∈ N a) We have x ∈ ∞ n=1 An iff x ∈ An for some n ∈ N iff + 1/n ≤ x ≤ − 1/n for some n ∈ N iff < x < Hence, ∞ [1 + 1/n, − 1/n] = (1, 2) n=1 b) We have x ∈ Hence, ∞ n=1 An iff x ∈ An for some n ∈ N iff ≤ x ≤ − 1/n for some n ∈ N iff ≤ x < ∞ [1, − 1/n] = [1, 2) n=1 c) We note that x ∈ iff ≤ x ≤ Hence, ∞ n=1 An iff x ∈ An for all n ∈ N iff − 1/n < x < + 1/n for all n ∈ N ∞ (1 − 1/n, + 1/n) = [1, 2] n=1 d) We note that x ∈ iff ≤ x ≤ Hence, ∞ n=1 An iff x ∈ An for all n ∈ N iff − 1/n < x < + 1/n for all n ∈ N ∞ (3 − 1/n, + 1/n) = {3} n=1 e) We have x ∈ ∞ n=1 An iff x ∈ An for all n ∈ N iff x > n for all n ∈ N However, there are no real numbers x that satisfy x > n for all n ∈ N Hence, ∞ (n, ∞) = ∅ n=1 f) We have x ∈ ∞ n=1 An iff x ∈ An for all n ∈ N iff − 1/n < x < for all n ∈ N iff ≤ x < 5, which is impossible Hence, ∞ (5 − 1/n, 5) = ∅ n=1 1.2 g) We have x ∈ Hence, ∞ n=1 An Set Theory 1-7 iff x ∈ An for all n ∈ N iff − 1/n < x < for all n ∈ N iff ≤ x < ∞ (5 − 1/n, 6) = [5, 6) n=1 1.20 a) From Definition 1.5 on page 16, we see that {1, 2, 3} × {3, 4, 5} = {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5)} Thus, the members of {1, 2, 3} × {3, 4, 5} are the nine ordered pairs inside the curly braces on the right of the preceding display b) We determined {1, 2, 3} × {3, 4, 5} in part (a) Proceeding similarly, we find that {3, 4, 5} × {1, 2, 3} = {(3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)} Noting that the only element common to {1, 2, 3} × {3, 4, 5} and {3, 4, 5} × {1, 2, 3} is (3, 3), we see that the members of the union of those two sets are (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), and (5, 3) c) We note that U has 25 members, namely, all ordered pairs of the form (x, y) where x, y ∈ {1, 2, 3, 4, 5} Hence, A consists of the eight elements in U not listed in the solution to part (b), namely, (1, 1), (1, 2), (2, 1), (2, 2), (4, 4), (4, 5), (5, 4), and (5, 5) d) We have {(1, 1), (1, 2), (2, 1), (2, 2)} = {1, 2} × {1, 2} and {(4, 4), (4, 5), (5, 4), (5, 5)} = {4, 5} × {4, 5} Therefore, from part (c), A = {1, 2} × {1, 2} ∪ {4, 5} × {4, 5} 1.21 Refer to Definition 1.5 on page 16 a) We have {0, 1}3 = {0, 1} × {0, 1} × {0, 1} = {(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)} Hence, the members of {0, 1}3 are the eight ordered triplets inside the curly braces on the second line of the preceding display b) We have {0, 1} × {0, 1} × {1, 2} = {(0, 0, 1), (0, 0, 2), (0, 1, 1), (0, 1, 2), (1, 0, 1), (1, 0, 2), (1, 1, 1), (1, 1, 2)} Thus, the members of {0, 1} × {0, 1} × {1, 2} are the eight ordered triplets inside the curly braces on the right of the preceding display c) We have {a, b} ∪ {c, d, e} × {f, g, h} = {a, b, c, d, e} × {f, g, h} = {(a, f ), (a, g), (a, h), (b, f ), (b, g), (b, h), (c, f ), (c, g), (c, h), (d, f ), (d, g), (d, h), (e, f ), (e, g), (e, h)} Hence, the members of {a, b} ∪ {c, d, e} × {f, g, h} are the 15 ordered pairs shown in the preceding display CHAPTER Probability Basics 1-8 d) We have {a, b} × {f, g, h} = {(a, f ), (a, g), (a, h), (b, f ), (b, g), (b, h)} and {c, d, e} × {f, g, h} = {(c, f ), (c, g), (c, h), (d, f ), (d, g), (d, h), (e, f ), (e, g), (e, h)} Therefore, {a, b} × {f, g, h} ∪ {c, d, e} × {f, g, h} = {(a, f ), (a, g), (a, h), (b, f ), (b, g), (b, h), (c, f ), (c, g), (c, h), (d, f ), (d, g), (d, h), (e, f ), (e, g), (e, h)} Hence, the members of {a, b} × {f, g, h} ∪ {c, d, e} × {f, g, h} are the 15 ordered pairs shown in the preceding display Note that the answer here is identical to that in part (c) This fact is due to the identity (A ∪ B) × C = (A × C) ∪ (B × C) 1.22 We first establish the identity (A × B) ∩ (C × D) = (A ∩ C) × (B ∩ D) Indeed, (x, y) ∈ (A × B) ∩ (C × D) iff (x, y) ∈ A × B and (x, y) ∈ C × D iff x ∈ A and y ∈ B and x ∈ C and y ∈ D iff x ∈ A ∩ C and y ∈ B ∩ D iff (x, y) ∈ (A ∩ C) × (B ∩ D) Applying the preceding identity, we get [0, 2] × [0, 2] ∩ [1, 3] × [1, 3] = [0, 2] ∩ ([1, 3] × [0, 2] ∩ [1, 3] = [1, 2] × [1, 2] 1.23 Answers will vary One possibility is to take In = (0, 1/n) for n ∈ N Clearly, we have In ⊂ [0, 1] for all n ∈ N Also, for j, k ∈ N , Ij ∩ Ik = (0, 1/j ) ∩ (0, 1/k) = 0, 1/ max{j, k} = ∅ Moreover, x ∈ ∞ n=1 In iff x ∈ In for all n ∈ N iff < x < 1/n for all n ∈ N iff < x ≤ 0, which is impossible Hence, ∞ n=1 In = ∅ 1.24 Answers will vary One possibility is to take In = [0, 1/n) for n ∈ N Indeed, x ∈ ∞ n=1 In iff x ∈ In for all n ∈ N iff ≤ x < 1/n for all n ∈ N iff ≤ x ≤ iff x = Hence, ∞ I = {0} n=1 n 1.25 Answers will vary One possibility is to take In = (n − 1, n] for n ∈ Z Clearly, these intervals constitute a countably infinite collection of intervals of R, each of length Also, suppose that j, k ∈ Z with j = k, say, j < k If x ∈ Ij , then x ≤ j ≤ k − and, hence, x ∈ Ik Consequently, we see that Ij ∩ Ik = ∅ Now, let x ∈ R and set n = min{j ∈ Z : x ≤ j } Then n − < x ≤ n so that x ∈ In Therefore, we have shown that ∞ R= In = n∈Z (n − 1, n] n=−∞ Theory Exercises 1.26 a) By observing that the last Venn diagrams in the following two rows are identical, we conclude that (A ∩ B)c = Ac ∪ B c 1.2 Set Theory 1-9 / A ∩ B so that x ∈ / A or x ∈ / B But then x ∈ Ac or x ∈ B c , b) Suppose that x ∈ (A ∩ B)c Then x ∈ c c c c c which means x ∈ A ∪ B Consequently, (A ∩ B) ⊂ A ∪ B Conversely, suppose that x ∈ Ac ∪ B c Then x ∈ Ac or x ∈ B c so that x ∈ / A or x ∈ / B But then x ∈ / A ∩ B, which means x ∈ (A ∩ B)c Conc c c sequently, A ∪ B ⊂ (A ∩ B) We have shown that (A ∩ B)c ⊂ Ac ∪ B c and Ac ∪ B c ⊂ (A ∩ B)c Therefore, (A ∩ B)c = Ac ∪ B c c c) We first note that E c = E for any set E Now we apply part (a) of De Morgan’s laws to the sets Ac and B c to obtain (A ∩ B)c = Ac c ∩ Bc c c = Ac ∪ B c c c = Ac ∪ B c Hence, we have verified part (b) of De Morgan’s laws 1.27 a) We have the following Venn diagrams: As the last Venn diagrams in the two preceding rows are identical, we conclude that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) CHAPTER Probability Basics 1-10 We have the following Venn diagrams: As the last Venn diagrams in the two preceding rows are identical, we conclude that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) b) For part (a) of the distributive laws, we have x ∈ A ∩ (B ∪ C) iff x ∈ A and either x ∈ B or x ∈ C iff either x ∈ A and x ∈ B or x ∈ A and x ∈ C iff either x ∈ A ∩ B or x ∈ A ∩ C iff x ∈ (A ∩ B) ∪ (A ∩ C) Hence, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) We can use a similar argument to establish part (b) of the distributive laws Alternatively, we can apply De Morgan’s laws and part (a) of the distributive laws (which we just verified), as follows: A ∪ (B ∩ C) = Ac ∩ (B ∩ C)c = Ac ∩ B c c ∩ c = Ac ∩ B c ∪ C c Ac ∩ C c c = Ac c = c c Ac ∩ B c ∪ Ac ∩ C c ∪ Bc c ∩ Ac c ∪ Cc c = (A ∪ B) ∩ (A ∪ C) 1.28 a) We observe that the two commutative laws [i.e., parts (a) and (b) of Proposition 1.3] are clear from the following two Venn diagrams: 1.2 Set Theory 1-11 For the first associative law [i.e., part (c) of Proposition 1.3], we have the following Venn diagrams: As the last Venn diagrams in the two preceding rows are identical, we conclude that A ∩ (B ∩ C) = (A ∩ B) ∩ C For the second associative law [i.e., part (d) of Proposition 1.3], we have the following Venn diagrams: As the last Venn diagrams in the two preceding rows are identical, we conclude that A ∪ (B ∪ C) = (A ∪ B) ∪ C CHAPTER Probability Basics 1-12 b) First we prove part (a) of Proposition 1.3 We have x ∈ A ∩ B iff x ∈ A and x ∈ B iff x ∈ B and x ∈ A iff x ∈ B ∩ A Hence, A ∩ B = B ∩ A We can use a similar argument to establish part (b) of Proposition 1.3 Alternatively, we can apply part (a) of that proposition (which we just verified) and De Morgan’s laws, as follows: A ∪ B = Ac c ∪ Bc c = Ac ∩ B c c = B c ∩ Ac c = Bc c ∪ Ac c = B ∪ A Next we prove part (c) of Proposition 1.3 We have x ∈ A ∩ (B ∩ C) iff x ∈ A, and x ∈ B and x ∈ C iff x ∈ A and x ∈ B, and x ∈ C iff x ∈ A ∩ B and x ∈ C iff x ∈ (A ∩ B) ∩ C Hence, A ∩ (B ∩ C) = (A ∩ B) ∩ C We can use a similar argument to establish part (d) of Proposition 1.3 Alternatively, we can apply part (c) of that proposition (which we just verified) and De Morgan’s laws, as follows: A ∪ (B ∪ C) = Ac = c ∪ Bc c = Ac c ∪ Bc ∩ Cc = Ac ∩ B c c ∪ Cc ∪ Cc Ac ∩ B c ∩ C c c c c = c = Ac ∩ B c ∩ C c Ac c ∪ Bc c ∪ Cc c c = (A ∪ B) ∪ C 1.29 a) By definition, A ∪ ∅ consists of all elements that are either in A or ∅ But, ∅ contains no elements Hence, we must have A ∪ ∅ = A b) If x ∈ A, then x ∈ A or x ∈ B, which means that x ∈ A ∪ B Hence, A ⊂ A ∪ B c) Suppose that A = A ∪ B Then, from part (b), B ⊂ B ∪ A = A ∪ B = A Hence, B ⊂ A Conversely, suppose that B ⊂ A Let x ∈ A ∪ B Then either x ∈ A or x ∈ B However, as B ⊂ A, if x ∈ B, then x ∈ A Hence, in either case, we have x ∈ A Thus, we have shown that A ∪ B ⊂ A From part (b), A ⊂ A ∪ B Consequently, we have A = A ∪ B 1.30 a) By definition, A ∩ ∅ consists of all elements that are both in A and ∅ But, ∅ contains no elements Hence, we must have A ∩ ∅ = ∅ b) If x ∈ A ∩ B, then x ∈ A and x ∈ B, so that, in particular, x ∈ A Consequently, A ∩ B ⊂ A or, equivalently, A ⊃ A ∩ B c) Suppose that A = A ∩ B Then, from part (b), B ⊃ B ∩ A = A ∩ B = A Hence, B ⊃ A Conversely, suppose that B ⊃ A Let x ∈ A Then, as B ⊃ A, we must have x ∈ B Therefore, x ∈ A ∩ B Thus, we have shown that A ⊂ A ∩ B From part (b), A ⊃ A ∩ B Consequently, A = A ∩ B 1.31 a) Let x ∈ A Now, either x ∈ B or x ∈ B c In the former case, we then have x ∈ A ∩ B, whereas, in the latter case, x ∈ A ∩ B c Thus, x ∈ (A ∩ B) ∪ (A ∩ B c ) Hence, A ⊂ (A ∩ B) ∪ (A ∩ B c ) From Exercise 1.30(b), A⊃A∩B and A ⊃ A ∩ Bc Hence, A ⊃ (A ∩ B) ∪ (A ∩ B c ) Therefore, we have now shown that A = (A ∩ B) ∪ (A ∩ B c ) 1.2 Set Theory 1-13 b) Suppose that A ∩ B = ∅ Let x ∈ A Then x ∈ / B so that x ∈ B c Hence, A ⊂ B c c) Suppose that A ⊂ B Let x ∈ B c Then x ∈ / B so that x ∈ / A (because if x ∈ A, then x ∈ B) Hence, x ∈ Ac Consequently, we have shown that B c ⊂ Ac c 1.32 We have x ∈ / n An iff x ∈ / An for some n iff x ∈ Acn for some n iff x ∈ n An iff x ∈ Hence, Proposition 1.4(a) holds Applying that result to the sets Ac1 , Ac2 , , we get c An c Acn = n c n An c c c Acn = n Acn = n n Hence, Proposition 1.4(b) holds 1.33 We have x ∈ B ∩ n An iff x ∈ B and x ∈ n An iff x ∈ B and x ∈ An for some n iff x ∈ B ∩ An for some n iff x ∈ n (B ∩ An ) Hence, Proposition 1.5(a) holds Applying that result to the sets B c and Ac1 , Ac2 , and using De Morgan’s laws, we get B∪ An = B c c c c ∪ n c c c = B ∩ An n c B c ∩ Acn = n = B c ∩ Acn = c n Bc = n Acn = B ∩ An n c c c ∪ Acn c n (B ∪ An ) n Hence, Proposition 1.5(b) holds 1.34 a) Referring first to Exercise 1.30(c) and then to Proposition 1.5(a) on page 14, we get B=B∩ An = n (B ∩ An ) = n (An ∩ B) n b) Let E be a subset of U Then E ⊂ U = n An and, hence, from part (a), we have E = c) Referring to Exercise 1.30(a), we get, for m = n, n (An ∩ E) (Am ∩ E) ∩ (An ∩ E) = Am ∩ E ∩ An ∩ E = (E ∩ E) ∩ (Am ∩ An ) = E ∩ (Am ∩ An ) = E ∩ ∅ = ∅ Hence, A1 ∩ E, A2 ∩ E, are pairwise disjoint d) Let E be a subset of U Because A1 , A2 , form a partition of U , we have n An = U Hence, from part (b), we have E = n (An ∩ E) And, because A1 , A2 , form a partition of U , the sets A1 , A2 , are pairwise disjoint, which, by part (c), implies that A1 ∩ E, A2 ∩ E, are pairwise disjoint Therefore, we see that E can be expressed as a disjoint union of the sets A1 ∩ E, A2 ∩ E, Advanced Exercises 1.35 ∞ ∞ a) Let x ∈ ∞ k=n0 Ak , which means that x ∈ Ak k=n Ak Then there is an n0 ∈ N such that x ∈ n=1 for all k ≥ n0 Now let n ∈ N and set m = max{n, n0 } Then x ∈ Am ⊂ ∞ k=n Ak Thus, we have shown ∞ ∞ A A for all n ∈ N , which means that x ∈ that x ∈ ∞ k=n k Consequently, k=n k n=1 ∞ ∞ ∞ Ak n=1 k=n ∞ ⊂ Ak n=1 k=n CHAPTER Probability Basics 1-14 b) We see that x ∈ lim inf n→∞ An iff there is an n ∈ N such that x ∈ ∞ k=n Ak , which is the case iff x ∈ Ak for all k ≥ n Thus, lim inf n→∞ An consists of all elements of U that belong to all but a finite number of the An s Also, we see that x ∈ lim supn→∞ An iff for each n ∈ N , we have x ∈ ∞ k=n Ak , which is the case iff for each n ∈ N , there is a k ≥ n such that x ∈ Ak Thus, lim supn→∞ An consists of all elements of U that belong to an infinite number of the An s c) We first determine lim inf n→∞ An To that end, we begin by noting that An ∩ An+1 = {0} for all n ∈ N Hence, as ∞ k=n Ak ⊂ An ∩ An+1 for all n ∈ N , we deduce that ∞ n→∞ ∞ ∞ lim inf An = Ak {0} = {0} ⊂ n=1 k=n n=1 However, from the definition of the An s, we see that ∈ An for all n ∈ N and, consequently, it is clear that {0} ⊂ lim inf n→∞ An We have thus shown that lim inf n→∞ An = {0} Next we find lim supn→∞ An We first note that A1 ⊃ A3 ⊃ A5 · · · and A2 ⊃ A4 ⊃ A6 · · · Therefore, ∞ k=n Ak = An ∪ An+1 for all n ∈ N and, hence, ∞ lim sup An = n→∞ ∞ ∞ Ak n=1 k=n = (An ∪ An+1 ) n=1 Now, from the definition of the An s, we see that An ⊃ [−1, 0] for all odd n ∈ N and An ⊃ [0, 1] for all even n ∈ N Therefore, An ∪ An+1 ⊃ [−1, 1] for all n ∈ N and, hence, lim supn→∞ An ⊃ [−1, 1] Conversely, suppose that x ∈ / [−1, 1] If x < −1, then there is an odd n ∈ N such that x < −1 − 1/n Thus, x ∈ / An ∪ An+1 , which implies that x ∈ / lim supn→∞ An If x > 1, then there is an even n ∈ N such that x > + 1/n Thus, x ∈ / An ∪ An+1 , which implies that x ∈ / lim supn→∞ An Consequently, no number outside of [−1, 1] is a member of lim supn→∞ An or, equivalently, lim supn→∞ An ⊂ [−1, 1] We have thus shown that lim supn→∞ An = [−1, 1] 1.36 Define f : N → Z by f (n) = n/2, −(n − 1)/2, if n is even; if n is odd, We claim that f is one-to-one and onto Z, which will show that Z is countably infinite Let z ∈ Z If z is a positive integer, then 2z is a even positive integer and, hence, f (2z) = 2z/2 = z If z is a nonpositive integer, then − 2z is an odd positive integer and, hence, f (1 − 2z) = − (1 − 2z) − /2 = z Thus, we have shown that f is onto Z Next, we verify that f is one-to-one Suppose then that f (m) = f (n) Then either both m and n are even or both are odd If they are both even, then we have m/2 = n/2 and, hence, m = n; if they are both odd, then we have −(m − 1)/2 = −(n − 1)/2 and, hence, m = n 1.37 Define f : N → N by f (m, n) = 2m−1 (2n − 1) We claim that f is one-to-one and onto N , which will show that N is countably infinite Let k ∈ N If k is even, let j denote the largest positive integer such that 2j divides k Then k = 2j (2n − 1) for some n ∈ N , and we have f (j + 1, n) = 2(j +1)−1 (2n − 1) = 2j (2n − 1) = k If k is odd, then (k + 1)/2 ∈ N , and we have f 1, (k + 1)/2 = 21−1 · k+1 − = · (k + 1) − = k Thus, we have shown that f is onto N Next, we establish that f is one-to-one Suppose then that f (m, n) = f (j, k), where, say, m ≥ j Then, we have 2m−1 (2n − 1) = 2j −1 (2k − 1) or, equivalently, 2m−j (2n − 1) = 2k − As 2k − is odd, we must have m − j = 0, or m = j , in which case, we have 2n − = 2k − and, hence, n = k Therefore, (m, n) = (j, k) 1.2 Set Theory 1-15 1.38 For convenience, let us use the terminology 1-1 correspondence as an abbreviation for a function that is one-to-one and onto Suppose A is countable Then, by definition, it is either finite or countably infinite If it is countably infinite, then, by definition, it is equivalent to N , which means there is a one-to-one and onto function, f , from N to A Letting sn = f (n), n ∈ N , we have that A is the range of the infinite sequence {sn }∞ n=1 If A is finite (and nonempty), then, by definition, there is an N ∈ N such that A is equivalent to the first N positive integers Let g be a one-to-one and onto function from {1, 2, , N} to A Select x ∈ A and define sn = g(n) if n = 1, 2, , N, and sn = x if n > N Then A is the range of the infinite sequence {sn }∞ n=1 Conversely, suppose A is the range of an infinite sequence, {sn }∞ n=1 We claim that A is countable If A is finite, there is nothing to prove So, assume that A is infinite We will construct a 1-1 correspondence from N to A, thereby proving that A is countably infinite and, hence, countable Let n1 = As A is infinite, A \ {sn1 } = ∅ Therefore, because the range of {sn }∞ n=1 is A, the set { n ∈ N : sn = sn1 } is not empty Denote by n2 the smallest integer in that set Note that n1 < n2 Proceeding inductively, note again that, because A is infinite, we have that A \ {sn1 , sn2 , , snk } = ∅ Therefore, as the range of {sn }∞ n=1 is A, the set { n ∈ N : sn = snj , ≤ j ≤ k } is not empty Denote by nk+1 the smallest integer in that set and note that nk < nk+1 We claim that the function f : N → A defined by f (k) = snk is a 1-1 correspondence By construction, f is one-to-one So it remains to show that f is onto Let x ∈ A Because the range of {sn }∞ n=1 is A, the set { n ∈ N : sn = x } is not empty Let m be the smallest integer in that set If m = 1, then x = s1 = sn1 = f (1) Otherwise, let k be the smallest integer such that m ≤ nk As sn = x = sm for n < m, we have that sm = snj for ≤ j ≤ k − 1, which implies that m ≥ nk Therefore, m = nk and, consequently, x = snk = f (k) 1.39 Let A be a countable set and f a function defined on A If A = ∅, then f (A) = ∅ and, hence, is countable So, assume that A = ∅ We know from Exercise 1.38 that A is the range of an infinite sequence, {sn }∞ n=1 For each n ∈ N , define tn = f (sn ) Now, let y ∈ f (A) Then there is an x ∈ A such that f (x) = y Because A is the range of the infinite sequence {sn }∞ n=1 , there is an n ∈ N such that sn = x Therefore, y = f (x) = f (sn ) = tn This result shows that f (A) is the range of the infinite sequence {tn }∞ n=1 Hence, by Exercise 1.38, f (A) is countable 1.40 Let A be a countable set and let B ⊂ A We claim that B is countable If B = ∅, there is nothing to prove; so, assume that B = ∅ This assumption implies that A is nonempty and, hence, by Exercise 1.38, A is the range of an infinite sequence, {sn }∞ / B n=1 Choose x ∈ B Let tn = sn if sn ∈ B, and tn = x if sn ∈ Then B is the range of the infinite sequence {tn }∞ Applying Exercise 1.38 again, we conclude that B n=1 is countable 1.41 Let A and B be two countable sets If either A or B is empty, then A × B = ∅ and, hence, is countable Therefore, let us assume that both A and B are nonempty By Exercise 1.38, each of A and B ∞ is the range of an infinite sequence, say, {an }∞ n=1 and {bn }n=1 , respectively Define f : N → A × B by f (m, n) = (am , bn ) From Exercise 1.37, we know that N is countable and, clearly, f is onto A × B Thus, A × B is the image of N under f Applying Exercise 1.39, we conclude that A × B is countable We next use mathematical induction to prove that the Cartesian product of a finite number of countable sets is countable We have just verified that result for n = Assuming its truth for n − 1, we now prove it for n Thus, let A1 , , An be countable sets For convenience, set Bk = ×jk=1 Aj for each k ∈ N From the induction assumption, we know that Bn−1 is countable and, hence, so is Bn−1 × An Let us define f : Bn−1 × An → Bn by f (a1 , , an−1 ), an = (a1 , , an−1 , an ) Clearly, f is one-to-one and onto and, hence, as Bn−1 × An is countable, so is Bn = ×jn=1 Aj 1.42 We know from Exercise 1.36 that Z is countable Hence, by Exercise 1.41, so is Z × N Define f : Z × N → Q by f (z, n) = z/n Clearly, f is onto Q, and, hence, Q is the image of Z × N under f Therefore, we conclude from Exercise 1.39 that Q is countable 1-16 CHAPTER Probability Basics 1.43 a) Suppose to the contrary that [0, 1) is countable Let {xn }∞ n=1 be an enumeration of its elements, that is, the function f : N → [0, 1) defined by f (n) = xn is one-to-one and onto For each n ∈ N , let 0.dn1 dn2 denote the unique decimal expansion of xn not containing only finitely many digits −n = 0.y y , where y = if d differing from Now consider the number y = ∞ n nn = 0, n=1 yn 10 and yn = otherwise Note that y ∈ [0, 1) and, hence, as {xn }∞ is an enumeration of [0, 1), we must n=1 have y = xn for some n ∈ N However, for each n ∈ N , we have y = xn because the decimal expansions of y and xn differ at the nth decimal digit and neither decimal expansion contains only finitely many digits differing from Thus, we have a contradiction Therefore, [0, 1) is uncountable b) Suppose to the contrary that (0, 1) is countable Let {xn }∞ n=1 be an enumeration of its elements Define f : N → [0, 1) by f (1) = and f (n) = xn−1 if n ≥ Then f is a one-to-one function from N onto [0, 1), which implies that [0, 1) is countable, a contradiction to part (a) c) Define f : (0, 1) → (a, b) by f (x) = a + (b − a)x Now, let y ∈ (a, b) Upon solving the equation y = a + (b − a)x for x, we get the unique number x = (y − a)/(b − a), which, clearly, is in (0, 1) Consequently, f is a one-to-one function from (0, 1) onto (a, b), which, by part (b), implies that (a, b) is uncountable d) From Exercise 1.40, we know that any subset of a countable set is countable or, equivalently, if a set contains an uncountable subset, then it must be uncountable Because any nondegenerate interval contains a bounded interval of the form (a, b), it follows from part (c) that any nondegenerate interval is uncountable 1.44 Denote the countable collection of sets by C If C is empty, then its union is empty and, hence, countable So, we can assume that C is nonempty Without loss of generality, we can also assume that each member of C is nonempty; otherwise, we simply discard all empty members of C, which does not affect the union Because C is nonempty and countable, Exercise 1.38 shows that it is the range of an ∞ infinite sequence, say, {An }∞ n=1 An is countable Because each member n=1 We need to prove that A = (n) ∞ of C is countable, Exercise 1.38 shows that each An is the range of an infinite sequence, say, xm m=1 (n) Now, define f : N → A by f (m, n) = xm We note that f is onto Furthermore, by Exercise 1.37, we know that N is countable Therefore, in view of Exercise 1.39, we conclude that A is countable, being the image of N under f 1.45 a) Suppose that I is finite, say, I = {1, 2, , N} On the one hand, by Definition 1.5, ×N n=1 An is the set of all ordered N-tuples (a1 , a2 , , aN ), where an ∈ An for ≤ n ≤ N On the other hand, by the definition of the Cartesian product of an indexed collection, ×N n=1 An is the set of all functions x on {1, 2, , N} such that x(n) ∈ An for ≤ n ≤ N Identifying each such function x with the ordered N -tuple x(1), x(2), , x(N ) , we obtain a one-to-one correspondence between the members of the Cartesian product ×N n=1 AN as defined by Definition 1.5 and the members of the Cartesian N product ×n=1 An of the indexed collection {Aι }ι∈I b) False, it is not necessarily the case that the Cartesian product of a countable number of countable sets is countable For instance, let D denote the set of decimal digits and let E = D ∞ = D × D × D × · · ·, the Cartesian product of a countable number of countable (actually, finite) sets Define f : E → [0, 1] −n by f (d1 , d2 , ) = ∞ n=1 dn 10 We have f (E) = [0, 1]; that is, [0, 1] is the image of f under E If E were countable, then, by Exercise 1.39, so would be [0, 1] However, by Exercise 1.43(d), we know that [0, 1] is uncountable Hence, E must be uncountable Review Exercises for Chapter 1-17 Review Exercises for Chapter Basic Exercises 1.46 a) Referring to Table 1.1 on page 4, we see that 16 of the states are in the South and, of those, 11 seceded from the Union Therefore, the probability that a randomly selected state in the South seceded from the union is 11/16, or 0.6875 b) The population under consideration consists of the 16 states in the South 1.47 From the table, the total number of winners for the years 1901–1997 is 190 + 71 + · · · + 87 = 448 a) The number of winners from Japan is Hence, the probability that the recipient selected is from Japan is 4/448 ≈ 0.00893 b) The number of winners from either France or Germany is 25 + 61 = 86 Hence, the probability that the recipient selected is from either France or Germany is 86/448 ≈ 0.192 c) The number of winners from any country other than the United States is 448 − 190 = 258 Hence, the probability that the recipient selected is from any country other than the United States is 258/448 ≈ 0.576 1.48 a) Referring to the frequentist interpretation of probability on page 5, we see that, in a large number of tosses of the die, the result will be about 1/6 of the time b) Again referring to the frequentist interpretation of probability, we see that, in a large number of tosses of the die, the result will be or more about 2/3 of the time c) From part (a), we see that, in 10,000 tosses, the die will come up about 10,000 · (1/6) times, or roughly 1667 times d) From part (b), we see that, in 10,000 tosses, the die will come up or more about 10,000 · (2/3) times, or roughly 6667 times 1.49 Consider the random experiment of selecting one voter (at random) from the population and let E denote the event that the voter chosen will vote yes on the proposition Because 60% of the voters will vote yes, we know that P (E) = 0.6 Now, choosing n voters at random with replacement is equivalent to independently repeating the random experiment n times Hence, from the frequentist interpretation of probability, specifically Relation (1.1) on page 5, we have, for large n, that n(Y ) = n(E) ≈ P (E) · n = 0.6n 1.50 a) Squaring each element of the set S = {−2, −1, 0, 1, 2}, we find that the elements of { x : x ∈ S } are 0, 1, and b) If −2 < x < 2, then ≤ x < and, vice-versa Hence, { x : −2 < x < } = [0, 4) 1.51 The eight members of {0, 1}3 are (0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), and (1, 1, 1) By associating with a tail and with a head, we can regard the eight members of {0, 1}3 as the eight possible outcomes of the experiment of tossing a coin three times We now consider the eight members of {0, 1}3 a finite population If we select a member at random from that population, then each member is equally likely to be the one obtained Hence, that random experiment can be regarded as tossing a balanced coin three times Note that there are three members of the population corresponding to getting two heads and one tail, namely, (0, 1, 1), (1, 0, 1), and (1, 1, 0) Therefore, the probability that you get two heads and one tail when you toss a balanced coin three times equals the probability that a member selected at random from {0, 1}3 consists of exactly two 1s and one 0, which is 3/8, or 0.375 CHAPTER Probability Basics 1-18 1.52 We can proceed in several ways to obtain the simplification One way is to apply the distributive law of Proposition 1.2(a) on page 12, as follows: (A ∩ B) ∪ (A ∩ B c ) = A ∩ (B ∪ B c ) = A ∩ U = A 1.53 For each n ∈ N , let An = 1/(n + 1), 1/n and Bn = 1/(n + 1), 1/n a) Set A = ∞ n=1 An If x ∈ A, then there is an n ∈ N such that x ∈ An and, so, 1/(n + 1) < x ≤ 1/n] But then we have 1 < x ≤ ≤ 0< n+1 n Thus, A ⊂ (0, 1] Conversely, suppose that x ∈ (0, 1] Then x > Let n be the smallest positive integer such that 1/(n + 1) < x Then 1/(n + 1) < x ≤ 1/n so that x ∈ An , which, in turn, implies that x ∈ A Thus, (0, 1] ⊂ A We have shown that A ⊂ (0, 1] and (0, 1] ⊂ A Consequently, A = (0, 1] b) Set B = ∞ n=1 Bn If x ∈ B, then there is an n ∈ N such that x ∈ Bn and, so, 1/(n + 1) ≤ x ≤ 1/n] But then we have 1 ≤ x ≤ ≤ 0< n+1 n Thus, B ⊂ (0, 1] Conversely, suppose that x ∈ (0, 1] Then x > Let n be the smallest positive integer such that 1/(n + 1) < x Then 1/(n + 1) < x ≤ 1/n so that x ∈ Bn , which, in turn, implies that x ∈ B Thus, (0, 1] ⊂ B We have shown that B ⊂ (0, 1] and (0, 1] ⊂ B Consequently, B = (0, 1] c) The sets in the union in part (a) are pairwise disjoint Indeed, let m = n, say, m < n Then, if x ∈ An , we have x ≤ 1/n ≤ 1/(m + 1) and, hence, x ∈ / Am Consequently, An ∩ Am = ∅ The sets in the union in part (b) are not pairwise disjoint For instance, we have B1 ∩ B2 = [1/2, 1] ∩ [1/3, 1/2] = {1/2} = ∅ However, ∞ Bn ⊂ B1 ∩ B3 = [1/2, 1] ∩ [1/4, 1/3] = ∅ n=1 Consequently, ∞ n=1 Bn = ∅ 1.54 You may find it helpful to refer to the discussion of finite and infinite sets on pages 19 and 20 Also, note that other methods can be used to obtain the following results a) We have {1, 2, 3} × {2, 3, 4} = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4)} Thus, {1, 2, 3} × {2, 3, 4} consists of nine elements and, hence, is finite b) Let A = {1, 2, 3} × [2, 4] We have A = (1, y) : y ∈ [2, 4] ∪ (2, y) : y ∈ [2, 4] ∪ (3, y) : y ∈ [2, 4] Denote by B the first set in the union of the preceding display Define f : [2, 4] → B by f (y) = (1, y) Clearly, f is one-to-one and onto Hence, from Exercise 1.43, we deduce that B is uncountable As B ⊂ A, it follows from Exercise 1.40 that A is uncountable c) Let A be as in part (b) and set C = [2, 4] × {1, 2, 3} Define f : A → C by f (x, y) = (y, x) Clearly, f is one-to-one and onto From part (b), we know that A is uncountable; hence, so is C d) Let D = {1, 2, 3} × {1, 2, 3, } From Exercise 1.41, we see that D is countable, being the Cartesian product of two countable sets (one finite and the other countably infinite) Obviously, D is infinite and, therefore, we conclude that it is countably infinite Review Exercises for Chapter 1-19 e) Let E = [1, 3] × [2, 4] and let B be as in part (b) As we have seen, B is uncountable Therefore, because E ⊃ B, we deduce from Exercise 1.40 that E is uncountable f) We note that ∞ n=1 {n, n + 1, n + 2} = N , which is countably infinite g) Let An = {n, n + 1, n + 2} and F = ∞ n=1 An We have F ⊂ A1 ∩ A4 = {1, 2, 3} ∩ {4, 5, 6} = ∅ Therefore, F = ∅ and, hence, is finite 1.55 a) From De Morgan’s laws, {3, 4}c ∩ {4, 5}c c = {3, 4}c c ∪ {4, 5}c c = {3, 4} ∪ {4, 5} = {3, 4, 5} b) We have {3, 4}c ∩ {4, 5}c c = { , −2, −1, 0, 1, 2, 5, 6, } ∩ { , −2, −1, 0, 1, 2, 3, 6, 7, } = { , −2, −1, 0, 1, 2, 6, 7, } c c = {3, 4, 5} Theory Exercises 1.56 Let B1 = A1 and, for k ≥ 2, set k−1 Bk = Ak ∩ c Aj = Ac1 ∩ · · · ∩ Ack−1 ∩ Ak j =1 Observe that, for all k ∈ N , we have Bk ⊂ Ak and Bk ⊂ Ajc for ≤ j ≤ k − Let m = n, say, m < n Then Bm ⊂ Am and Bn ⊂ Acm , so that Bm ∩ Bn ⊂ Am ∩ Acm = ∅ Consequently, Bm ∩ Bn = ∅, and we see that B1 , B2 , are pairwise disjoint sets We use these sets in parts (a) and (b) a) For convenience, set A(n) = jn=1 Aj and B (n) = jn=1 Bj As Bj ⊂ Aj for all j ∈ {1, 2, , n}, we have B (n) ⊂ A(n) Conversely, suppose that x ∈ A(n) Then x ∈ Aj for some j ∈ {1, 2, , n} Let k be the smallest such j Then x ∈ Ak and x ∈ / Aj for ≤ j ≤ k − 1, which means that x ∈ Bk and, consequently, that x ∈ B (n) Therefore, A(n) ⊂ B (n) We have now shown that B (n) ⊂ A(n) and A(n) ⊂ B (n) Hence, A(n) = B (n) b) We use the notation of part (a) and recall that A(n) = B (n) for all n ∈ N Let A = ∞ n=1 An ∞ ∞ ∞ (n) (n) and B = n=1 Bn We observe that A = n=1 A and B = n=1 B Hence, ∞ ∞ B (n) = B A(n) = A= n=1 n=1 1.57 a) From the distributive law of Proposition 1.2(a) on page 12 and the fact that A ∩ C ⊂ A, we get (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C) ⊂ A ∪ (B ∩ C) b) Answers will vary Note, however, from the solution to part (a) that any choice of A, B, and C in which A ∩ C = A will the trick For instance, take A = {1} and B = C = ∅ Then (A ∪ B) ∩ C = ({1} ∪ ∅) ∩ ∅ = ∅ = {1} = {1} ∪ ∅ = {1} ∪ (∅ ∩ ∅) = A ∪ (B ∩ C) 1-20 CHAPTER Probability Basics c) If A ⊂ C, then A ∩ C = A Hence, from the distributive law of Proposition 1.2(a), (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C) = A ∪ (B ∩ C) d) If (A ∪ B) ∩ C = A ∪ (B ∩ C), then A ⊂ A ∪ (B ∩ C) = (A ∪ B) ∩ C ⊂ C 1.58 a) Let C = { C ⊂ U : A ⊂ C and B ⊂ C } We want to prove that A ∪ B = C∈C C To begin, we note that, because A ⊂ A ∪ B and B ⊂ A ∪ B, we have A ∪ B ∈ C Thus, A ∪ B ⊃ C∈C C Conversely, let C ∈ C Then A ⊂ C and B ⊂ C, which implies that A ∪ B ⊂ C Therefore, A ∪ B ⊂ C∈C C b) If A ⊂ D and B ⊂ D, then A ∪ B ⊂ D Hence, we see that A ∪ B is the smallest set that contains both A and B as subsets c) Let D = { C ⊂ U : A ⊃ C and B ⊃ C } We want to prove that A ∩ B = C∈D C To begin, we note that, because A ⊃ A ∩ B and B ⊃ A ∩ B, we have A ∩ B ∈ D Thus, A ∩ B ⊂ C∈D C Conversely, let C ∈ D Then A ⊃ C and B ⊃ C, which implies that A ∩ B ⊃ C Therefore, A ∩ B ⊃ C∈D C d) If A ⊃ D and B ⊃ D, then A ∩ B ⊃ D Hence, we see that A ∩ B is the largest set that is contained as a subset by both A and B e) Let A1 , A2 , be a countable sequence of subsets of U Generalization of part (a): Let E = { C ⊂ U : An ⊂ C for all n } We want to prove that An = n C C∈E For convenience, set A = n An Because An ⊂ A for all n, we have A ∈ E Thus, A ⊃ C∈E C Conversely, let C ∈ E Then An ⊂ C for all n, which implies that A ⊂ C Therefore, A ⊂ C∈E C Generalization of part (b): If An ⊂ D for all n, then smallest set that contains all An s as subsets n An ⊂ D Hence, we see that n An is the Generalization of part (c): Let F = { C ⊂ U : An ⊃ C for all n } We want to prove that An = n C C∈F For convenience, set A = n An Because An ⊃ A for all n, we have A ∈ F Thus, A ⊂ C∈F C Conversely, let C ∈ F Then An ⊃ C for all n, which implies that A ⊃ C Therefore, A ⊃ C∈F C Generalization of part (d): If An ⊃ D for all n, then largest set that is contained as a subset of all An s n An ⊃ D Hence, we see that n An is the 1.59 a) This result is true Indeed, we have (x, y) ∈ A × (B ∪ C) iff x ∈ A and y ∈ B ∪ C iff x ∈ A and either y ∈ B or y ∈ C iff either x ∈ A and y ∈ B or x ∈ A and y ∈ C iff either (x, y) ∈ A × B or (x, y) ∈ A × C iff (x, y) ∈ (A × B) ∪ (A × C) Hence, A × (B ∪ C) = (A × B) ∪ (A × C) b) This result is true Indeed, we have (x, y) ∈ A × (B ∩ C) iff x ∈ A and y ∈ B ∩ C iff x ∈ A and y ∈ B and y ∈ C iff x ∈ A and y ∈ B and x ∈ A and y ∈ C iff (x, y) ∈ A × B and (x, y) ∈ A × C iff (x, y) ∈ (A × B) ∩ (A × C) Hence, A × (B ∩ C) = (A × B) ∩ (A × C) c) This result is not always true For instance, let A = {0} and B = {1} Then A × B = {(0, 1)} = {(1, 0)} = B × A Review Exercises for Chapter 1-21 Advanced Exercises 1.60 Refer to the table in the problem statement a) The number of left-handed individuals is 71 + 92 = 163 Hence, the probability that the person obtained is left-handed is 163/525 ≈ 0.310 b) The number of colorblind individuals who are not ambidextrous is 71 + 61 = 132 Hence, the probability that the person obtained is colorblind but not ambidextrous is 132/525 ≈ 0.251 c) The number of left-handed individuals is 71 + 92 = 163 of which 71 are colorblind Hence, if the person selected is left-handed, the probability that he or she is colorblind is 71/163 ≈ 0.436 d) The number of colorblind individuals is 71 + 61 + 37 = 169 of which 71 are left-handed Hence, if the person selected is colorblind, the probability that he or she is left-handed is 71/169 ≈ 0.420 e) The number of left-handed, colorblind women is 24 Hence, the probability that the person obtained is a left-handed, colorblind woman is 24/525 ≈ 0.0457 1.61 The probability is 0.75 that a randomly selected adult female believes that having a "cyber affair" is cheating Therefore, the odds against an adult female believing that having a "cyber affair" is cheating are − 0.75 to 0.75 or 0.25 to 0.75 or to 1.62 a) We have / B } = { x : x ∈ U and x ∈ / B } = U \ B Bc = { x : x ∈ b) We have x ∈ A \ B iff x ∈ A and x ∈ / B iff x ∈ A and x ∈ B c iff x ∈ A ∩ B c Consequently, we have c shown that A \ B = A ∩ B c) Applying part (b) and De Morgan’s law of Proposition 1.1(b) on page 11, we get (A \ B)c = A ∩ B c c = Ac ∪ B c c = Ac ∪ B 1.63 We note that, by definition, x ∈ A △ B iff either x ∈ A or x ∈ B, and x ∈ / A ∩ B, which is the c c case iff x ∈ A ∪ B and x ∈ (A ∩ B) iff x ∈ (A ∪ B) ∩ (A ∩ B) Thus, A △ B = (A ∪ B) ∩ (A ∩ B)c Applying De Morgan’s laws, the distributive laws, and Exercise 1.62(b), we get A △ B = (A ∪ B) ∩ (A ∩ B)c = (A ∪ B) ∩ Ac ∪ B c = (A ∪ B) ∩ Ac ∪ (A ∪ B) ∩ B c = (A ∩ Ac ) ∪ (B ∩ Ac ) ∪ (A ∩ B c ) ∪ (B ∩ B c ) = ∅ ∪ (B \ A) ∪ (A \ B) ∪ ∅ = (A \ B) ∪ (B \ A) 1.64 a) We note that x ∈ B △ C iff x is a member of exactly one of B and C Hence, x ∈ / B △ C iff either x is in both B and C or x is in neither B nor C In other words, (B △ C)c = (B ∩ C) ∪ (B c ∩ C c ), a result that we could also obtain by applying properties of set operations and results from Exercises 1.62 and 1.63 Referring now to those two exercises, we get A △ (B △ C) = A ∩ (B △ C)c ∪ (B △ C) ∩ Ac = A ∩ (B ∩ C) ∪ (B c ∩ C c ) ∪ (B ∩ C c ) ∪ (C ∩ B c ) ∩ Ac = A ∩ B ∩ C ∪ A ∩ B c ∩ C c ∪ B ∩ C c ∩ Ac ∪ C ∩ B c ∩ Ac Replacing A by C, B by A, and C by B in the previous display, we find that C △ (A △ B) = A △ (B △ C) The required result now follows from the easily established fact that E △ F = F △ E CHAPTER Probability Basics 1-22 b) We have A △ U = A ∩ U c ∪ U ∩ Ac = (A ∩ ∅) ∪ Ac = ∅ ∪ Ac = Ac c) We have A △ ∅ = A ∩ ∅c ∪ ∅ ∩ Ac = (A ∩ U ) ∪ ∅ = A ∪ ∅ = A d) We have A △ A = A ∩ Ac ∪ A ∩ Ac = ∅ ∪ ∅ = ∅ 1.65 a) Applying properties of set operations, we get A ∩ (B △ C) = A ∩ (B ∩ C c ) ∪ (C ∩ B c ) = A ∩ (B ∩ C c ) ∪ A ∩ (C ∩ B c ) = A ∩ B ∩ C c ∪ A ∩ C ∩ B c = (A ∩ B) ∩ (Ac ∪ C c ) ∪ (A ∩ C) ∩ (Ac ∪ B c ) = (A ∩ B) ∩ (A ∩ C)c ∪ (A ∩ C) ∩ (A ∩ B)c = (A ∩ B) △ (A ∩ C) b) Again applying properties of set operations, we get (A ∪ B) △ (A ∪ C) = (A ∪ B) ∩ (A ∪ C)c ∪ (A ∪ C) ∩ (A ∪ B)c = (A ∪ B) ∩ (Ac ∩ C c ) ∪ (A ∪ C) ∩ (Ac ∩ B c ) = B ∩ Ac ∩ C c ∪ C ∩ Ac ∩ B c = Ac ∩ (B ∩ C c ) ∪ (C ∩ B c ) = Ac ∩ (B △ C) Therefore, (A ∪ B) △ (A ∪ C) = Ac ∩ (B △ C) ⊂ B △ C ⊂ A ∪ (B △ C) Thus, we have shown that (A ∪ B) △ (A ∪ C) ⊂ A ∪ (B △ C) c) Suppose that A = ∅ Then A ∪ (B △ C) = ∅ ∪ (B △ C) = B △ C = (∅ ∪ B) △ (∅ ∪ C) = (A ∪ B) △ (A ∪ C) Conversely, suppose that A ∪ (B △ C) = (A ∪ B) △ (A ∪ C) Then, referring to the solution to part (b), we find that A ⊂ A ∪ (B △ C) = (A ∪ B) △ (A ∪ C) = Ac ∩ (B △ C) ⊂ Ac From this result, we conclude that A = A ∩ A ⊂ A ∩ Ac = ∅; that is, A = ∅ We have therefore shown that A ∪ (B △ C) = (A ∪ B) △ (A ∪ C) precisely when A = ∅ 1.66 Suppose that B = ∅ Then, from Exercise 1.64(c), we have A △ B = A △ ∅ = A Conversely, suppose that A = A △ B Then, applying in turn parts (c), (d), (a), and (d) of Exercise 1.64, we find that B = B △ ∅ = ∅ △ B = (A △ A) △ B = A △ (A △ B) = A △ A = ∅