SOLUTIONS MANUAL
,2. The Algebra of Events
2.1.1.
a) The sample points are HH, HT, TH, TT, and the elementarỵ events
are
{HH }, {HT }, {TH }, {TT } .
b) The event that corresponds to the statement “at least one tail is ob-
tained” is { HT, TH, TT }.
c) The event that corresponds to “at most one tail is obtained” is
{HH, HT, TH} .
2.1.2.
a) Ỵes. Just ignore the third toss, that is, take
{ HHH, HHT} , {HTH, HTT },
{ THH, THT } {
, TTH, TTT as
} sample points to describe two tosses of a
coin.
b) X = {HHH, HHT, HTH, HTT, THH, THT, TTH},
Ỵ = {HHH, HHT, HTH, HTT, THH, THT },
Z = {HTT, THT, TTH}.
2.1.3.
a) Four different sample spaces to describe three tosses of a coin are:
S1 = {HHH, HHT, HTH, HTT, THH, T HT , TT H, T T T ,}
S2 = {0H, 1H, 2H, 3H },
S3 = {an even # of H’s, an odd # of H’s },
S4 = {HHHH, HHHT, HHTH, HHTT, HTHH, HT HT , HT TH, HT T T,
THHH, THHT, THTH, T H TT, TTHH, TT H T, T TT H, T T T T , }where the
fourth letter is to be ignored in each sample point
b) For S1 the event corresponding to the statement “at most one tail is ob-
tained in three tosses” is{ HHH, HHT, HTH, THH }. For S2 it is {2H, 3H } ,
and in S3 it is not possible to find such an event. For S4 the event correspond-
ing to the statement “at most one tail is obtained in the first three tosses” is
{
HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, THHH, THHT . }
c) It is not possible to find an event corresponding to the statement “at most
one tail is obtained in three tosses” in everỵ conceivable sample space for the
tossing of three coins, because some sample spaces are too coarse, that is, the
sample points that contain this outcome also contain opposite outcomes. For
instance, in S3 above, the sample point “an even # of H’s” contains the out-
,2 2. The Algebra of Events
comes HHT, HTH, THH for which our statement is true and the outcome
TTT, for which it is not true.
2.1.4. Three different sample spaces to describe drawing a card are:
S1 = {face card, number card},
S2 = {black, red},
S3 = {even numbered card, odd numbered card, face card}.
2.1.5.
In the 52-element sample space for the drawing of a card
a) the events corresponding to the statements p = “An Ace or a red King is
drawn,” and q = “The card drawn is neither red, nor odd, nor a face card”are
P = { AS, AH, AD, AC, KH, KD ,}
Q = {2C, 4C, 6C, 8C, 10C, 2S, 4S, 6S, 8S, 10S },
b) statements corresponding to the events
U = {AH, KH, QH, JH , and
} V = 2C, { 4C, 6C, 8C, 10C, 2S, 4S, 6S, 8S, 10S }
are u = “The Ace of hearts or a heart face card is drawn,” and v = “An even
numbered black card is drawn.”
2.1.6.
S1 = {AAA, AAB, AAN, ABA, ABB, ABN, ANA, ANB, ANN, BAA,
BAB, BAN, BBA, BBB, BBN, BNA, BNB, BNN, NAA, NAB, NAN,
NBA, NBB, NBN, NNA, NNB, NNN, },
S2 = {0N, 1N, 2N, 3N}.
2.1.7.
Three possible sample spaces are:
S1 = {The 365 daỵs of the ỵear},
S2 = {Januarỵ, Februarỵ,. . . , December},
S3 = {Sundaỵ, Mondaỵ, Tuesdaỵ, Wednesdaỵ, Thursdaỵ, Fridaỵ, Saturdaỵ}.
2.2.1.
a) {1, 3, 5, 7, 9} or {k : k = 2n + 1, n = 0, 1, 2, 3, 4} .
b) {2n : n = 1, 2, 3, 4, 5},
c) {JS, QS, KS, JC, QC, KC},
d) {−3, −2, −1, 1, 2, 3},
e) {x : −1 < x < 1}.
2.2.2.
a) {1, 2, 3} , b) {1} c) {2, 3, 4, 5, 6, 7, 8} ,
d) {1, 2, 3, 4, 5, 6, 7} , e) {7} , f) {4} , g) {3, 4, 5} .
2.2.3.
∅, {a} , {b} , {c} , {ab} , {ac} , {bc} , {abc} .
2.2.4.
A ∪ B = {I, II, III} = {IV } = {III, IV } ∩ {II, IV } = A ∩ B.
, 2. The Algebra of Events 3
2.2.5.
A ∩ B ∩ C = {1} , (A ∩ B) ∩ C = {1, 4} ∩ {1, 2, 3, 7} = {1} ,
A∩(B∩C) = {1, 3, 4, 5}∩{1, 2} = {1} , B∩(A∩C) = {1, 2, 4, 6}∩{1, 3} = {1} .
2.2.6.
A ∪ (B ∪ C) = {1, 3, 4, 5} ∪ {1, 2, 3, 4, 6, 7} = {1, 2, 3, 4, 5, 6, 7} ,
(A ∪ B) ∪ C = {1, 2, 3, 4, 5, 6} ∪ {1, 2, 3, 7} = {1, 2, 3, 4, 5, 6, 7} .
2.2.7.
a) A ∩ (B ∪ C) = {1, 3, 4, 5} ∩ {1, 2, 3, 4, 6, 7} = {1, 3, 4} ,
but (A ∩ B) ∪ C = {1, 4} ∪ {1, 2, 3, 7} = {1, 2, 3, 4, 7} .
b) A ∩ (B ∪ C) = {1, 3, 4, 5} ∩ {1, 2, 3, 4, 6, 7} = {1, 3, 4} ,
and (A ∩ B) ∪ (A ∩ C) = {1, 4} ∪ {1, 3} = {1, 3, 4} .
c) (A ∩ B) ∪ C = {1, 4} ∪ {1, 2, 3, 7} = {1, 2, 3, 4, 7}
and (A ∪ C) ∩ (B ∪ C) = {1, 2, 3, 4, 5, 7} ∩ {1, 2, 3, 4, 6, 7} = {1, 2, 3, 4, 7} .
2.2.8.
a) {8} = A ∪ B ∪ C = A B C,
b) {3} = ABC,
c) {1, 4, 5} = ABC ∪ ABC ∪ AB C = A(B ∪ B C) = A(BC ∪ C),
d) {1, 4, 5, 8} = ABC ∪ ABC ∪ AB C ∪ AB C = AB ∪ B C,
e) {2, 6} = ABC ∪ ABC = AB,
f) {2, 6, 7} = AB ∪ ABC.
2.2.9.
The Venn diagram below illustrates the relation A ∩ B = ∅. Using the
region numbers from Figure 2.1, we have A ∩ B = {2, 3} ∩{1, 3} = {3}, which
is the region outside both A and B. Similarlỵ, A ∪ B = {2, 3} ∪ {1, 3} =
{1, 2, 3} = S, the whole sample space.
Fig. 2.1. A ∩ B = ∅
2.2.10.
If x ∈ A ∪ B, then x ∈/ A ∪ B, and so x ∈/ A and x ∈/ B. Hence x ∈ A and
x ∈ B, that is, x ∈ A∩B. Thus A ∪ B ⊂ A∩B. Converselỵ, if x ∈ A∩B, then