Chapter 4: Preparation of Graphene suspensions in volatile solvents
III. Preparation of THF-water inks
m−1)
n=1
P (An) + P (Am) = )m n=1
P (An),
where we used the fact that if A1, A2,· · · , Am are mutually exclusive, then the events (∪m−1i=1 Ai) and Am are mutually exclusive, and the induction assumption.
b) For mutually exclusive A1, A2,· · · ,
P ( ,∞ n=1
An) = P 9 ∞
,
N =1
( ,N n=1
An) :
= lim
N →∞P ( ,N n=1
An)
= lim
N →∞
)N n=1
P (An) = )∞ n=1
P (An),
where the 2nd equality holds by Proposition 2.11(a), while the 3rd equality holds by the addi-tivity axiom for a finite number of events.
c) Clearly, axiom of countable additivity implies the axiom of finite additivity, since A1∪ A2= A1∪ A2∪ ∅ ∪ ∅ ∪ · · · , thus,
P (A1∪ A2) = P (A1∪ A2∪ ∅ ∪ ∅ ∪···) = P (A1) + P (A2) + 0 + 0 +···= P (A1) + P (A2) for arbitrary mutually exclusive events A1 and A2. Taking into account that the axiom of count-able additivity implies the continuity property of probability measures (by Proposition 2.11, which was proved in Exercise 2.73), then the axiom of countable additivity implies both the axiom of finite additivity and the continuity. Conversely, the axioms of finite additivity and con-tinuity together imply the countable additivity of probability measures by part (b). Therefore, the axioms of finite additivity and continuity together are equivalent to the axiom of countable additivity.
2.5 Review Exercises
2.79 {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)}.
2.80 a)
Ω ={{a, b}, {a, c}, {a, d}, {a, e}, {a, f}, {b, c}, {b, d}, {b, e}, {b, f}, {c, d}, {c, e}, {c, f}, {d, e}, {d, f}, {e, f}}.
b){{a, b}, {a, c}, {a, d}, {a, e}, {a, f}, {b, f}, {c, f}, {d, f}, {e, f}}.
c) Ω ={{a, d}, {a, e}, {a, f}, {b, d}, {b, e}, {b.f}, {c, d}, {c, e}, {c, f}}.
The event of interest is {{a, d}, {a, e}, {a, f}, {b, f}, {c, f}}.
d) The sample space is given by:
Ω ={(a, b), (a, c), (a, d), (a, e), (a, f), (b, a), (b, c), (b, d), (b, e), (b, f ), (c, a), (c, b), (c, d), (c, e), (c, f ), (d, a), (d, b), (d, c), (d, e), (d, f ), (e, a), (e, b), (e, c), (e, d), (e, f )}.
The event of interest is equal to:
{(a, d), (d, a), (a, e), (e, a), (a, f), (f, a), (b, f), (f, b), (c, f), (f, c)}.
2.81 The cars are classified into 6 categories. Let xi represent the number of cars (in a given set of 100 cars) that are in category i, where i = 1,· · · , 6.
Ω ={(x1, x2, x3, x4, x5, x6) : xi∈ N ∪ {0} for i ∈ {1,···, 6} and )6 i=1
xi= 100}.
2.82 a) Not mutually exclusive, since the intersection of the two given events is not empty (in fact, the intersection contains the following outcomes: 6 men and 6 women; 5 men and 7 women; 7 men and 5 women).
b) Yes, the events are mutually exclusive, since a jury of 12 people cannot have at least 5 men and at least 8 women as members, thus, the intersection of the two events is empty.
c) The three events are mutually exclusive, since the number of men (and women) selected in each case is different, thus, every pairwise intersection is empty.
d) The events are not mutually exclusive, since the intersection of the three events is not empty.
2.83 a) Ω ={(A, B, C), (A, C, B), (B, A, C), (B, C, A), (C, A, B), (C, B, A)}.
b){(A, B, C), (A, C, B)}.
c) Ac∩ Bc∩ Cc=“A is not the first alternate, B is not the second and C is not the third”
={(C, A, B), (B, C, A)}.
d) A∩ B ∩ C =“A is the first alternate, B is the second and C is the third”=(A, B, C)}.
e) A∩ C=“A is the first alternate and C is the third”= {(A, B, C)}.
f ) The equal-likelihood model is appropriate, since who is the first alternate, who is the second and who is the third is determined by chance (and the sample space is finite). Since the sample space contains six outcomes, each outcome is assigned probability 1/6.
g) P (Ac ∩ Bc∩ Cc) = 2/6 = 1/3 and P (A∩ B ∩ C) = P (A ∩ C) = 1/6.
2.84 Answers will vary.
2.85 A∪ Bc=“A occurs or B does not occur”;
Ac∩ Bc =“Neither A nor B occurs”;
(A∩ Bc)∪ (Ac∩ B) =“Exactly one of the events A,B occurs”;
A∩ B ∩ C =“Events A,B, and C occur”;
A∩ (B ∪ C) =“A occurs and either B or C occurs”;
A∪ (B ∩ C) =“Either A occurs or both B and C occur”;
A∪ B ∪ C =“At least one of the events A, B, C occurs”;
Ac∪ Bc∪ Cc =“At least one of the events A, B and C does not occur”;
∩∞n=1An=“All events A1, A2,· · · occur”;
∪∞n=1An=“At least one of A1, A2,· · · occurs.”
2.86 (A∩Bc)∪(Ac∩B). (Other equivalent descriptions are also possible, say, (A∪B)∩(A∩B)c).
2.87 a) True, by definition.
b) Not true. The following provides a simple counter-example: Take Ω ={1, 2, 3}, A = {1, 3}, B ={2} and C = {2, 3}. Clearly, A and B are mutually exclusive, but events A, B, C are not mutually exclusive since A∩ C ̸= ∅.
2.88 a) The assignment determines a unique probability measure on Ω by Proposition 2.3, since each outcome is assigned a non-negative probability and the sum of probabilities of all individual outcomes in Ω sum to one.
b) The probabilities of the 16 events of the random experiment are give by:
Event Probability Event Probability
∅ 0 {HT, TH} 2p(1− p)
{HH} p2 {HT, TT} p(1− p) + (1 − p)2 = 1− p
{HT} p(1− p) {TH, TT} p(1− p) + (1 − p)2 = 1− p
{TH} p(1− p) {HH, HT, TH} p + p(1− p) = 2p − p2
{TT} (1− p)2 {HH, HT, TT} p + (1− p)2= p2− p + 1
{HH, HT} p2+ p(1− p) = p {HH, TH, TT} p + (1− p)2= p2− p + 1 {HH, TH} p2+ p(1− p) = p {HT, TH, TT} 2p(1 − p) + (1 − p)2= 1− p2
{HH, TT} p2+ (1− p)2 = 2p2− 2p + 1 Ω 1
2.89 a) The sample space is given by:
Ω ={(w, w, w), (w, w, b), (w, b, w), (b, w, w), (w, b, b), (b, w, b), (b, b, w), (b, b, b)}, and each outcome in Ω is assigned probability 1/8.
b) P ({(b, b, b)}) = 1/8;
c) P ({(w, w, w), (w, w, b), (w, b, w), (b, w, w)}) = 4/8 = 0.5;
2.90 a) For Ω ={(x, y) : x2+ y2 < 1} (with |Ω| = π, where | · | denotes the two-dimensional volume)
P ({(x, y) ∈ Ω :6
x2+ y2> 1
2}) = 1 − P 'G
(x, y) : x2+ y2 ≤ 1 22
I(
= 1−π(1/2)2
π = 3
4 = 0.75;
b) Let C be the cube with vertices at (0, 1/2), (1/2, 0), (0,−1/2), (−1/2, 0). Then P ({(x, y) ∈ Ω :|x| + |y| > 1
2}) = 1 −|C|
π = 1− 1
2π ≈ 0.8408 c) P%
{(x, y) ∈ Ω : max{|x|, |y|} > 12&
= 1− P (max{|x|, |y|} ≤ 12)
= 1− P ({(x, y) : |x| ≤ 12,|y| ≤ 12}) = 1 − 1/π ≈ 0.6817.
2.91 a) Each pn= P ({n}) = 1/2n is non-negative and )∞
n=1
pn= )∞ n=1
1
2n = 1
1− (1/2) − 1 = 2 − 1 = 1, thus the probability assignment is legitimate.
b) )∞ k=0
p2k+1= )∞ k=0
1 22k+1 = 1
2 )∞ k=0
1 4k = 1
2 × 1
1− (1/4) = 2 3.
2.92 a) Note that Bn+1 ⊂ Bn for all n = 3, 4,· · · , implying that P (Bn+1) ≤ P (Bn) by the domination principle. In particular, it follows that P (B6)≤ P (B5).
b) P (+∞
n=3Bn) = limn→∞P (Bn), by Proposition 2.11(b) (in the textbook).
c) Let B denote the event that you will never reach your goal. The probability of reaching the goal after just three tosses is equal to (1/2)3 = 1/8; the probability of first reaching the goal after exactly six tosses is equal to P (THHHHH)+P (HTHHHH)+ P (HHTHHH)=3(1/2)6 > 0.
Clearly these two events are contained in Bc, thus P (Bc) ≥ (1/8) + 3(1/2)6 > 1/8, implying that P (B) = 1− P (Bc) < 1− (1/8) = 7/8.
2.93 Note that Ω ={(x, y) : 0 ≤ x ≤ 120, 0 ≤ y ≤ 120}, where x and y denote the arrival times of woman #1 and woman #2 respectively, measured in minutes after 3 P.M.. Then
P ({(x, y) ∈ [0, 120]2 :|x − y| ≤ 40}) = 1 − P ({(x, y) ∈ [0, 120]2 :|x − y| > 40})
= 1−(120− 40)2
1202 = 1− 22 32 = 5
9,
where we used the fact that {(x, y) ∈ [0, 120]2 : |x − y| > 40} is the union of two (mutually exclusive) equal right triangles (with legs equal to 120− 40 = 80 each).
2.94 Subjective. South Carolina is a large state, with different types of microclimate in different parts of the state, thus, no single mathematical probability model can be realistically constructed for the entire state. The empirical method does not apply here because a particular day is chosen.
2.95 a) Empirical method is most likely here. Manufacturing processes typically involve large numbers of almost identical experiments, and a relative frequency of defective bolts can be used to assign probability of occurrence of a defective bolt.
b) Subjective. Since the event involves a particular race (and a particular horse), it cannot be placed in the context of a repeatable experiment, thus the empirical method cannot be used.
Clearly, mathematical probability modeling is also not possible here.
c) Empirical. One can collect information on a large number of races, look at the proportion of races won by the favorite, and use it to assign the probability.
d) The answers will vary.
2.96 a) P (A∩ Hc) + P (H ∩ Ac) = (0.8− 0.35) + (0.45 − 0.35) = 0.55 b) P (Ac∩ Hc) = 1− P (A ∪ H) = 1 − (0.8 + 0.45 − 0.35) = 0.1 2.97 Total frequencies (in thousands) are given by:
V1: 151, 473; V2: 4, 461; V3: 87, 160; C1: 209, 539; C2: 20, 391; C3: 13, 164; Total: 243, 094.
a) 243, 094− 151, 473 = 91, 621;
b) 20, 391;
c) 4, 461;
d) 320;
e) 4, 461 + 20, 391− 320 = 24, 532;
f ) C1=“A randomly selected vehicle is from the U.S.”; V3=“A randomly selected vehicle is a truck”; C1∩ V3=“A randomly selected vehicle is a truck from the U.S.”;
g) P (C1) = 209, 539/243, 094 ≈ 0.862; P (V3) = 87, 160/243, 094 ≈ 0.359;
P (C1∩ V3) = 75, 940/243, 094≈ 0.312;
h) P (C1∪ V3) = (129, 728 + 3, 871 + 75, 940 + 6, 933 + 4, 287)/243, 094≈ 0.908;
i) P (C1∪ V3)≈ 0.862 + 0.359 − 0.312 = 0.909;
j) The joint probability distribution table is given by:
C1 C2 C3 P (Vi) V1 0.534 0.054 0.035 0.623 V2 0.016 0.001 0.001 0.018 V3 0.312 0.029 0.018 0.359 P (Cj) 0.862 0.084 0.054 1
2.98 a) Note that Ω ={(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)}, where, say, outcome (2, 3, 1) denotes that ball #2 was drawn first, ball #3 was drawn second and ball #1 was drawn third. Then
P (A1) = P ({(1, 2, 3), (1, 3, 2)}) = 2/6 = 1/3, P (A2) = P ({(1, 2, 3), (3, 2, 1)}) = 2/6 = 1/3, P (A3) = P ({(1, 2, 3), (2, 1, 3)}) = 2/6 = 1/3.
b) A1, A2 and A3 are not mutually exclusive, since (1, 2, 3) ∈ A1∩ A2∩ A3, for example.
2.99 a) P ({THHH, THHT, HTHH, HHHH}) = 4/(24) = 4/16 = 0.25, where the event in question should be understood as: “The first tail, if there is one at all in four tosses, is followed by (at least) two consecutive heads.”
b) P ({HHHH,HHHT,THHH}) = 3/(24) = 3/16 = 0.1875 2.100 The sample space can be taken to be
Ω ={(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)},
where, for example, (2, 3, 1) denotes the event that husband #2 (i.e. from the 2nd couple) dances with wife #1 and husband #3 dances with wife #2 and husband #1 dances with wife #3. Then a) P ({(1, 2, 3)}) = 1/6;
b) P ({(2, 3, 1), (3, 1, 2)}) = 2/6 = 1/3;
c) 1− (1/3) = 2/3.
2.101 a) P (Wc∩Dc∩Ic) = 1−P (W ∪D∪I) = 1−[0.4+0.3+0.23+(−0.15−0.13−0.09)+0.05] = 0.39, where we applied the complementation rule, de Morgan’s law and the inclusion-exclusion principle.
b) By the inclusion-exclusion principle,
P ((W ∩ D) ∪ (W ∩ I) ∪ (D ∩ I)) = P (W ∩ D) + P (W ∩ I) + P (D ∩ I)
−3P (W ∩ D ∩ I) + P (W ∩ D ∩ I)
= P (W ∩ D) + P (W ∩ I) + P (D ∩ I) − 2P (W ∩ D ∩ I)
= 0.15 + 0.13 + 0.09− 2(0.05) = 0.27 c) 1− 0.39 − 0.27 = 0.34, where we used the answers to parts (a),(b).
2.102 Let N be the event that some non-legitimate email arrives during a given hour and let L be the event that some legitimate email arrives during the hour. Then
P (Nc∩ Lc) = 1− P (N ∪ L) = 1 − (P (N) + P (L) − P (N ∩ L))
= 1− (0.5 + 0.7 − 0.4) = 0.2,
where we applied the complementation rule, de Morgan’s law and the inclusion-exclusion prin-ciple.
Theory Exercises
2.103 Finite additivity follows from equation (2.4) (in the textbook) by induction, since for mutually exclusive events A1,· · · , An, if finite additivity holds for unions of (n− 1) (or less) events, then
P ( ,n i=1
Ai) = P ((
n−1,
i=1
Ai)∪ An) = P (
n−1,
i=1
Ai) + P (An) =
n−1)
i=1
P (Ai) + P (An).
On the other hand, the countable additivity property cannot be obtained from finite additivity.
2.104 a) If P (A) = P (B) = 0, then P (A∪ B) = 0 by Boole’s inequality, since 0≤ P (A ∪ B) ≤ P (A) + P (B) = 0 + 0 = 0.
b) If P (A) = P (B) = 1, then P (Ac) = P (Bc) = 0, which, by part (a), implies that P (A∩ B) = 1 − P ((A ∩ B)c) = 1− P (Ac∪ Bc) = 1− 0 = 1.
c) Yes, parts (a) and (b) remain valid for countably many events. To see that (a) holds for countably many events, apply Exercise 2.75(c) (i.e. Boole’s inequality for a countably infinite number of events) and proceed as in part (a). To prove part (b) for countably many events, use part (a) for countably many events and the de Morgan’s law (Proposition 1.4).
d) For uncountably many events, the results are not valid. Indeed, take Ω = (0, 1), where for each event E ⊂ Ω, P (E) = |E| (where | · | denotes the one-dimensional volume). Then, if (a) were true for uncountably many events, then one would have that
P (Ω) = P
⎛
⎝ ,
x∈(0,1)
{x}
⎞
⎠= 0,
since|{x}| = 0 for each point x ∈ (0, 1). But this contradicts P (Ω) = 1. On the other hand, if (b) were true for an uncountable number of events, one would have that
P (∅) = P
⎛
⎝ H
x∈(0,1)
((0, 1)\ {x})
⎞
⎠ = 1,
since |(0, 1) \ {x}| = |(0, x)| + |(x, 1)| = x + 1 − x = 1. But then there is a contradiction with P (∅) = 1 − P (Ω) = 0.
Advanced Exercises
2.105 a) 4/52. Note that Ω = {((x, y) : x ∈ {1, 2, · · · , 52}, y ∈ {1, 2, · · · , 52}}, where we enumerated all the cards in the deck starting from aces. Then
P ({(x, y) : x ∈ {1, 2, · · · , 52}, y ∈ {1, 2, 3, 4}) = 52× 4 52× 52 = 4
52.
b) 4/52. Note that Ω ={(x, y) : x ∈ {1, 2, · · · , 52}, y ∈ {1, 2, · · · , 52}, x ̸= y}, where the cards are enumerated starting from aces. Then
P ({(x, y) : x ∈ {1, 2, · · · , 52}, y ∈ {1, 2, 3, 4}, x ̸= y}) = 52× 4 − 4
52× 52 − 52 = 4× 51 52× 51 = 4
52. 2.106 a) A\ B=“A occurs but B does not.”
b) P (A\ B) = P (A ∩ Bc) = P (A)− P (A ∩ B), by Exercise 1.62(b) and the law of partitions.
c) By Exercise 1.30(c), if B⊂ A, then A ∩ B = B, which by (b) implies that P (A\ B) = P (A) − P (A ∩ B) = P (A) − P (B).
2.107 a) A△ B=“Exactly one of the events A and B occurs.”
b) Using the answer to Exercise 1.63 and the law of partitions, one obtains that
P (A△ B) = P ((A \ B) ∪ (B \ A)) = P (A \ B) + P (B \ A) = P (A ∩ Bc) + P (B∩ Ac)
= (P (A)− P (A ∩ B)) + (P (B) − P (B ∩ A)) = P (A) + P (B) − 2P (A ∩ B).
2.108 a) “All but finitely many of A1, A2,· · · occur.”
b) “Infinitely many of A1, A2,· · · occur.”
c) The event in part (a) is contained in the event given in (b) by Exercise 1.35(a).
2.109 Let Ai be the event: “5” on ith throw, where i = 1, 2,· · · . Since P (Ai∩Aj)̸= 0 for i ̸= j, then P (∪8i=1Ai)̸=/8
i=1P (Ai). (The result is impossible since probability cannot exceed 1).
2.110 In the notation of Exercise 2.109, P (Ai) = 1/6, P (Ai ∩ Aj) = 1/(62) for i ̸= j, and P (Ai∩ Aj∩ Ak) = 1/(63) for distinct i, j, k. Then
P (A1∪ A2∪ A3) = 1 6 +1
6 +1 6− 1
62 − 1 62 − 1
62 + 1 63 = 91
216 ≈ 0.421
2.111 a) Ω ={(x1,· · · , xn) : n ∈ N , xi ∈ {1, 2, 3, 4, 5} for all i ̸= n, xn = 6}*{(x1, x2,· · · ) : xi ∈ {1, 2, 3, 4, 5} for all i ∈ N }.
b) For each n∈ N , assign
P ({(x1,· · · , xn)}) = 5n−1 6n . Note that/∞
n=1P ({(x1,· · · , xn)}) =/∞
n=15n−1
6n = 16×1−(5/6)1 = 1; thus, for all xj ∈ {1, 2, 3, 4, 5}, j∈ N , assign P ({(x1, x2,· · · )}) = 0.
c) 1, by solution to (b).
d) P (Tom) = )∞ k=0
5(3k+1)−1 63k+1 = 1
6 × 1
1− (5/6)3 = 62
63− 53 = 36
91 ≈ 0.3956;
P (Dick) = )∞ k=0
5(3k+2)−1 63k+2 = 5
62 × 1
1− (5/6)3 = 30
91 ≈ 0.3297;
P (Harry) = 1− 36 91−30
91 = 25
91 ≈ 0.2747.