Solution manual for a course in probability by weiss

Hence, the probability that a randomly selected student inyour class is female is n1/n1+ n2.. Hence, the probability that a randomlyselected student in your class is left-handed is ℓ1+ ℓ

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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 thesenator 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 ofhousing units is

471 + 1,470 + · · · + 15,647 = 112,356

1-1

a) The number of housing units with (exactly) four rooms is 23,468 Hence, the probability that thehousing 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, theprobability 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 unitselected has fewer than one room is 0/112,356 = 0

e) All of the housing units have one or more rooms Hence, the probability that the housing unit selectedhas one or more rooms is 112,356/112,356 = 1

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 whichthe 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 selectedwas 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

a) The number of females in your class is n1 Hence, the probability that a randomly selected student inyour 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 randomlyselected 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 selectedstudent in your class is a left-handed female is ℓ1/(n1+ n2)

d) A student is neither female nor left-handed if and only if the student is a right-handed male Thenumber 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 norleft-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-speakingstudents also speak French Now, 10% of 49.6% is 4.96% Hence, 4.96% of the students speak bothSpanish and French Consequently, the probability that a randomly selected student at this universityspeaks both Spanish and French is 0.0496

1.8 Answers will vary However, to illustrate, we tossed a balanced coin 20 times and obtained thefollowing table Here E denotes the event that, on any particular toss, the coin comes up a head

Toss

n

Outcome (H or T)

a) See the first five rows of the preceding table

b) Referring to the frequentist interpretation of probability on page 5 and the fifth row of the precedingtable, 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, weestimate, 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, weestimate, 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 ofthe probability of a head will fluctuate This phenomenon will be the case regardless of how many times

we toss the coin

1.9 Note that 2004 was a leap year, so it had 366 days Now, consider the random experiment ofselecting 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 ofprobability, if we independently repeat the random experiment n times, then the proportion of times thatevent 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 popularradio 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 5

a) Approximately 31.4% of human gestation periods exceed 9 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 randomexperiment, let n(E) denote the number of times that event E occurs Then, it follows from Relation (1.1)

on page 5 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 9 months We know that P (E) = 0.314 Here n = 4000 so thatfrom Relation (∗),

n(E) ≈4000 · 0.314 = 1256

Hence, of 4000 human gestation periods, roughly 1256 will exceed 9 months

b) The random experiment consists of observing a horse race and event E is that the favorite finishes inthe money We know that P (E) = 2/3 Here n = 500 so that from Relation (∗),

n(E) ≈500 ·2

3 =

1000

3 ≈333.3.

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) andevent E is that the traffic fatality involves an intoxicated or alcohol-impaired driver or nonoccupant Weknow that P (E) = 0.40 Here n = 389 so that from Relation (∗),

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,

n

An⊂ A1∩ A2 = ∅

As the empty set is a subset of every set, we conclude that

nAn= ∅ Therefore, if the sets in a collectionare 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 onepossibility 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 emptyintersection

● x < c iff x ≤ c − 1/n for some n ∈ N

● x > c iff x ≥ c + 1/n for some n ∈ N

● x ≤ ciff x < c + 1/n for all n ∈ N

● x ≥ ciff x > c − 1/n for all n ∈ N

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 thatthe 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),

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 precedingdisplay

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 ∈ Cand y ∈ D iff x ∈ A ∩ C and y ∈ B ∩ D iff (x, y) ∈ (A ∩ C) × (B ∩ D) Applying the precedingidentity, we get

b) Suppose that x ∈ (A ∩ B)c Then x /∈ A ∩ B so that x /∈ Aor x /∈ B But then x ∈ Acor x ∈ Bc,which means x ∈ Ac∪ Bc Consequently, (A ∩ B)c⊂ Ac∪ Bc Conversely, suppose that x ∈ Ac∪ Bc.Then x ∈ Acor x ∈ Bcso that x /∈ Aor x /∈ B But then x /∈ A ∩ B, which means x ∈ (A ∩ B)c Con-sequently, Ac∪ Bc ⊂ (A ∩ B)c We have shown that (A ∩ B)c⊂ Ac∪ Bc and Ac∪ Bc⊂ (A ∩ B)c.Therefore, (A ∩ B)c= Ac∪ Bc.

c) We first note thatEcc

= Efor any set E Now we apply part (a) of De Morgan’s laws to the sets Ac

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)

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 iffeither 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:

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

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 ∈ Aiff 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:

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:

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 ever, as B ⊂ A, if x ∈ B, then x ∈ A Hence, in either case, we have x ∈ A Thus, we have shownthat A ∪ B ⊂ A From part (b), A ⊂ A ∪ B Consequently, we have A = A ∪ B

1.31

a) Let x ∈ A Now, either x ∈ B or x ∈ Bc In the former case, we then have x ∈ A ∩ B, whereas, inthe latter case, x ∈ A ∩ Bc Thus, x ∈ (A ∩ B) ∪ (A ∩ Bc) Hence, A ⊂ (A ∩ B) ∪ (A ∩ Bc) FromExercise 1.30(b),

A ⊃ A ∩ B and A ⊃ A ∩ Bc.Hence, A ⊃ (A ∩ B) ∪ (A ∩ Bc) Therefore, we have now shown that A = (A ∩ B) ∪ (A ∩ Bc)

b) Suppose that A ∩ B = ∅ Let x ∈ A Then x /∈ B so that x ∈ Bc Hence, A ⊂ Bc.

c) Suppose that A ⊂ B Let x ∈ Bc Then x /∈ B so that x /∈ A (because if x ∈ A, then x ∈ B).Hence, x ∈ Ac Consequently, we have shown that Bc⊂ Ac

1.32 We have x ∈

nAn

c

iff x /∈

nAniff x /∈ Anfor some n iff x ∈ Ac

nfor some n iff x ∈

nAcn.Hence, Proposition 1.4(a) holds Applying that result to the sets Ac1, Ac2, , we get

c c

=

1.33 We have x ∈ B ∩

nAn iff x ∈ B and x ∈ nAniff x ∈ B and x ∈ Anfor 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 Bc

and Ac1, Ac2, and using De Morgan’s laws, we get

b) Let E be a subset of U Then E ⊂ U =

nAnand, hence, from part (a), we have E =

n(An∩ E)

c) Referring to Exercise 1.30(a), we get, for m = n,

(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

nAn= U Hence, frompart (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,

k=n0Ak, which means that x ∈ Ak

for all k ≥ n0 Now let n ∈ N and set m = max{n, n0} Then x ∈ Am ⊂∞

k=nAk Thus, we have shownthat x ∈∞

k=nAkfor all n ∈ N , which means that x ∈∞

n=1

∞ k=nAk Consequently,

(A< small>m∩ E) ∩ (A< small>n∩ E) = A< small>m∩ E ∩ A< small>n∩ E = (E ∩ E) ∩ (A< small>m∩ A< small>n)

= E ∩ (A< small>m∩ A< small>n)...

nA< small>n= U Hence, frompart (b), we have E =

n (A< small>n∩ E) And, because A< small>1, A< small>2, form a partition...

Hence, A< small>1∩ E, A< small>2∩ E, are pairwise disjoint

d) Let E be a subset of U Because A< small>1, A< small>2, form a partition of U , we have