Chapter 2: Problems Some notation: N denotes the set of natural numbers, i.e., the set of positive integers. Z denotes the set of all integers. Z+ denotes the set of all nonnegative integers.
1.
1◦ . (with replacement) Ω = {(r, r), (r, b), (r, g), (g, r), (g, b), (g, g), (b, r), (b, g), (b, b)} = {r, g, b}2 2◦ . (without replacement) Ω = {(r, b), (r, g), (g, r), (g, b), (b, r), (b, g)}
2. Ω = {(n, x1 , . . . , xn−1 )|n ∈ N and for k = 1, 2, . . . , n − 1, xk ∈ {1, 2, 3, 4, 5}} ∪{(x1 , x2 , . . .)|xk ∈ {1, 2, 3, 4, 5} for k = 1, 2, . . .} The interpretation of an element of the form (n, x1 , . . . , xn−1 ) is that the first 6 appeared on roll n and roll k had outcome xk for k = 1, . . . , n−1, and the interpretation of an element of the form (x1 , x2 , . . .) is that a 6 never appears. We are using a notational convention here: if n = 1 or n = 2, i.e., if n − 1 ≤ 1, (n, x1 , . . . , xn−1 ) cannot be taken literally. If n = 1, the only element is (1); if n = 2, the elements have the form (2, x1 ), where x1 ∈ {1, 2, 3, 4, 5}. The event that the experiment ends on roll n, where n ∈ N, is En = {(n, x1 , . . . , xn−1 )|(n, x1 , . . . , xn−1 ) ∈ Ω}. The event
!c
∞ [
En
n=1
1
is the event that 6 never appears. One might also describe the sample space in the following way. Ω = {(x1 , . . . , xn )|n ∈ N, for k = 1, 2, . . . , n − 1, xk ∈ {1, 2, 3, 4, 5}, and xn = 6} ∪{(x1 , x2 , . . .)|xk ∈ {1, 2, 3, 4, 5} for k = 1, 2, . . .}
3. EF is the event that one die shows 1 and the sum is odd. Equivalently, it is the event that one shows 1 and the other shows an even number. Explicitly, EF = {(1, 2), (1, 4), (1, 6), (2, 1), (4, 1), (6, 1)}. E ∪ F is the event that the sum is odd or one die shows 1. Explicitly, E ∪ F = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5), (1, 1), (1, 3), (1, 5), (3, 1), (5, 1)}. F G is the event that one die shows 1 and the sum is 5. Equivalently, it is the event that one die shows 1 and the other shows 4. Explicitly, F G = {(1, 4), (4, 1)}. EF c is the event that neither die shows 1, and the sum is odd. Explicitly, EF c = {(2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 3), (6, 5)} EF G = F G 4. a. There are two kinds of sample points: one point is an infinite string of 0s; the other points have a string of zero or more 0s followed by a 1. The infinite string of 0s represents the outcome that a head never appears. A string of n − 1 zeros, where n ∈ N means that the first n − 1 tosses resulted in tails, and the n-th toss yielded a head. b. A is the set of all sample points of the form 0| ·{z · · 0} 1. 3k
where k ∈ Z+ . That is, A = {1, 0001, 0000001, . . .}. 2
B is the set of all sample points of the form 0| ·{z · · 0} 1. 3k+1
where k ∈ Z+ . That is, B = {01, 00001, 00000001, . . .}. (A ∪ B)c is the event that neither A nor B wins. Equivalently, it is the set of all sample points of the form 0| ·{z · · 0} 1. 3k+2
together with the sample point 0000 · · · . That is, (A ∪ B)c = {001, 000001, 000000001, . . .} ∪ {0000 · · · }
6. a. Ω = {(0, g), (0, f ), (0, s), (1, g), (1, f ), (1, s)} = 0, 1 × {g, f, s} b. A = {(0, s), (1, s)} c. B = {(0, g), (0, f ), (0, s)} d. B c ∪ A = {(1, g), (1, f ), (1, s), (0, s)}
7. a. For each team member there are 2 possibilities for type of job and 3 possibilities for political affiliation, thus there are 6 possibilities in all. There are 15 team members, so the choice must be made 15 times. Hence there are 615 elements in the sample space.
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b. There are three possible classifications of a team member who is not a bluecollar worker: namely, that the member is a white-collar worker and has one of the three possibile poitical affiliations. There are thus 315 sample points in the event that no team member is a blue-collar worker. Thus there are 615 − 315 sample points in the event that at least one team member is a blue-collar worker. c. There are four possible classifications of a team member who is not an Independent. Thus there are 415 sample points in the event that no team member is an Independent.
8. a. Since A and B are mutually exclusive, P (A ∪ B) = P (A) + P (B)) = 0.5 + 0.3 = 0.8 b. In general, the event that A occurs but B does not occur is AB c . Since A and B are mutually exclusive, A ⊂ B c . Thus P (AB c ) = P (A) = 0.3 c. Since A and B are mutually exclusive, AB = ∅, so P (AB) = 0.
9. Let A denote the event that a customer carries an American Express card, and let V denote the event that a customer carries a Visa card. Thus, P (A ∪ V ) = P (A) + P (V ) − P (AV ) = 0.24 + 0.61 − 0.11 = 0.74 So 74% of the customers carry a card the establishment accepts.
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Chapter 2: Theoretical Exercises 6. There are different ways of expressing many of these. Sometimes I will give more than one. a. EF c Gc = E \ (F ∪ G) b. EGF c = EG \ F c. E ∪ F ∪ G d. F G ∪ EG ∪ EF e. EF G f. (E ∪ F ∪ G)c = E c F c Gc g. E c F c Gc ∪ EF c Gc ∪ E c F Gc ∪ E c F c G = (F G ∪ EG ∪ EF )c See parts (a) and (f) for the lhs and part (d) for the rhs. h. (EF G)c i. E c F G ∪ EF c G ∪ EF Gc j. S, where S is the universal set.
7. a.
b.
(E ∪ F )(E ∪ F c ) = = = = =
(E ∪ F )E ∪ (E ∪ F )F c EE ∪ F E ∪ EF c ∪ F F c E ∪ (EF ∪ EF c ) ∪ ∅ E∪E E
(E ∪ F )(E c ∪ F )(E ∪ F c ) = F (E ∪ F c ) = EF
where we have used the result in part (a).
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c.
(E ∪ F )(F ∪ G) = (E ∪ F )F ∪ (E ∪ F )G = EF ∪ F ∪ EG ∪ F G = F ∪ EG
because EF ∪ F G ⊂ F .
8. For each n in N, let Sn = {1, 2, . . . , n}. First we find all partitions of the set S3 , using what we know about partitions of S1 and S2 . Fix a subset A of S2 , and let B = S2 \ A. For each partition P of B, Q = {A ∪ {3}} ∪ P is a partition of S3 . As we shall see below, every partition of S3 is obtained this way. Here is what we get: A B P Q {1, 2} ∅ {{1, 2, 3}} {1} {2} {{2}} {{1, 3}, {2}} {2} {1} {{1}} {{2, 3}, {1}} ∅ {1, 2} {{1}, {2}} {{3}, {1}, {2}} ” ” {{1, 2}} {{3}, {1, 2}} Thus, T3 = 5.
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We find all the partitions of S4 by the same method: A B {1, 2, 3} ∅
P
Q {{1, 2, 3, 4}}
{1, 2} {1, 3} {2, 3}
{3} {2} {1}
{{3}} {{2}} {{1}}
{{1, 2, 4}, {3}} {{1, 3.4}, {2}} {{2, 3, 4}, {1}}
{1} ” {2} ” {3} ”
{2, 3} ” {1, 3} ” {1, 2} ”
{{2}, {3}} {{2, 3}}} {{1}, {3}} {{1, 3}}} {{1}, {2}} {{1, 2}}}
{{1, 4}, {2}, {3}} {{1, 4}, {2, 3}} {{2, 4}, {1}, {3}} {{2, 4}, {1, 3}} {{3, 4}, {1}, {2}} {{3, 4}, {1, 2}}
∅ ” ” ” ”
{1, 2, 3} ” ” ” ”
{{1}, {2}, {3}} {{1}, {2, 3}} {{2}, {1, 3}} {{3}, {1, 2}} {{1, 2, 3}}
{{4}, {1}, {2}, {3}} {{4}, {1}, {2, 3}} {{4}, {2}, {1, 3}} {{4}, {3}, {1, 2}} {{4}, {1, 2, 3}}
Thus T4 = 15. This method works in general. We could use it to find all partitions of S5 , S6 , etc. (It isn’t really clear why one would want to, but, in principle, it is possible.)
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b. Constructing the Partitions. For each integer n ≥ 1, we describe how to construct a partition Q of Sn+1 from a subset A of Sn and a partition P of the relative complement B = Sn \ A. If A = Sn , then B = ∅, and we have not defined partitions of the emptyset. By convention, we shall take P = ∅ in this case. We form Q by first adding the new element n + 1 to A and then adding the set A ∪ {n + 1} to P: Q = {A ∪ n + 1} ∪ P. We must show that this construction gives us each partition of Sn+1 once and only once. Let R be a partition of Sn+1 . Since the union of the sets in R is Sn+1 , the new element n + 1 belongs to exactly one of them. Since the sets in R are pairwise disjoint, n + 1 belongs to only one of them. The set containing n + 1 is of the form A ∪ {n + 1}, where A ⊂ Sn . Note that A is completely determined by R. If A = Sn , then R = {Sn+1 }. In this case, R is the partition constructed from (Sn , ∅). Otherwise, the other sets in R form a partition P of Sn \ A, and R is the partition constructed from (A, P). Observe that the partition P is also completely determined by R. Thus each partition R of Sn+1 arises from this construction, and since R uniquely determines A and P, we obtain R only once. Counting the Partitions. Now fix an integer k such that 1 ≤ k ≤ n. Let A be a subset of Sn containing n − k elements. Then B = Sn \ A contains k elements and so has Tk partitions. So there are Tk partitions arising from A by our construction. n n There are n−k = k subsets A of Sn that have n − k elements. By the BPC there are nk · Tk partitions arising from the subsets of Sn of size n − k. Thus letting A run over all proper subsets A of Sn , i.e., letting k take the values 1, . . . , n, we find that the proper subsets of Sn generate n X n Tk k k=1
partitions of Sn+1 . There is one more partition of Sn+1 , namely, the partition {Sn+1 }. Thus, n X n Tn+1 = 1 + Tk . k k=1
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Calculating T10 . Beginning with T1 = 1, we calculate T2 = 2 T3 = 5 T4 = 15 T5 = 52 T6 = 203 T7 = 877 T8 = 4140 T9 = 21147 T10 = 115975
9. Verification of Axiom 1. For any event E, 0 ≤ n(E) ≤ n, for E cannot occur more times than the experiment is performed. Dividing through by the positive number n, we find 0 ≤ f (E) ≤ 1.
Verification of Axiom 2. If S is the sample space, then S is an event and S occurs each time the experiment is performed. Thus n(S) = n, so n(S) n n = n = 1
f (S) =
Verification of Axiom 3. For each integer m = 0, . . . , n, and each event A, we let νm (A) be the number of times that the event A occurs in the first m repetitions of the experiment. Now, ν0 (A) = 0 and νn (A) = n(A) for all events A. ∞ Let hEk i∞ k=1 be a sequence of pairwise mutually exclusive events and let E = ∪k=1 Ek be their union. Now ! ∞ ∞ [ X ν0 Ek = ν0 (Ek ) k=1
k=1
since each side of the equation has the value zero. 9
We next show that ν1
∞ [
! Ek
=
k=1
∞ X
ν1 (Ek )
k=1
(Logically speaking, this is unnecessary, but it might help with understanding the inductive step below.) Run the experiment for the first time. Since the events Ek , k = 1, 2, . . ., are pairwise mutually exclusive, at most one of them occurs. If none of these events occurs, then E also does not occur and so ν1 (Ek ) = 0 for all k, k = 1, 2, . . ., and ν1 (E) = 0. Thus the equation holds in this case. If one of the events occurs, then exactly one term in the sum is equal to 1. All the other terms are equal to 0. Thus the value of the sum is 1. The union also occurs, and so the left hand side of the equation also has value 1. Thus, the equation holds in both cases. Now, we are ready for the induction. Suppose that for some integer m, with 0 ≤ m < n, we have ! ∞ ∞ [ X νm Ek = νm (Ek ) (1) k=1
We show that then νm+1
∞ [
k=1
! =
Ek
∞ X
νm+1 (Ek )
(2)
k=1
k=1
For when the experiment is run for the m + 1st time, either (i) none of the events Ek , k = 1, 2, . . . occurs or (ii) exactly one, say Ej , occurs. In case (i), we have νm+1 (Ek ) = νm (Ek ) for k = 1, 2, . . .. Also E does not occur, so νm+1 (E) = νm (E). Thus the two sides of equation 2 are equal, respectively, to the two sides of equation 1. Thus the last equation holds in case (i). In case (ii) we have νm+1 (Ek ) = νm (Ek ) for k = 1, 2, . . . , k 6= j, νm+1 (Ej ) = νm (Ej ) + 1, and νm+1 (E) = νm (E) + 1. Thus, we obtain each side of equation 2 by adding 1 to the corresponding side of equation 1. Thus the equation 2 holds in this case also. Therefore equation 1 implies equation 2. Thus we have, by induction, ! ∞ ∞ [ X νn Ek = νn (Ek ). k=1
k=1
That is, f
∞ [
! Ek
=
k=1
∞ X k=1
10
f (Ek ).
Chapter 1 Problems 1. a. There are 26 possibilities for each of the first two places, and there are 10 possibilities for each of the 5 following places. By the GBPC there are 262 · 105 possible license plates. b. There are 26 possible choices for the first letter, 25 for the second letter, 10 for the first digit, 9 for the second digit, and so on. Thus there are 26 · 25 · 10 · 9 · 8 cdot7 · 6 = 19, 656, 000. possible license plates.
4. If each can play any of the four instruments, then 4! = 24 arrangements are possible, by GBPC. For John can choose any of the four instruments, Jim can then choose any of the remaiuning three, Jay can choose either of the remaining two, and Jack will take the remaining one. If John and Jim can play all four instruments, but Jay and Jack can each play only piano and drums, then there are two ways that Jay and Jack can be given piano and drums. For each of these there are two ways to assign the other two instruments to John and Jim. Thus in this case there are 4 possibilities.
8. a. 5! = 120, for there are 5 distinct letters. b.
7! = 1260, for there are 7 letters, but o and p occur twice. 2!2!
11! c. = 34, 650 for there are 11 letters, but i and s each occur four times, and 4!2!4! p occurs twice. d.
7! = 1260, for there are 7 letters, but a and r occur twice. 2!2!
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11. a. 6! = 720 b. First arrange the books by kind. Since there are three kinds, this can be done in 3! = 6 ways. For each of these ways permute the books of each kind within their sections of the shelf. Thus by the GBPC, 3! · 3! · 2! · 1! = 72 arrangements are possible. c. First think of the novels as being one unit. There are now 4 units to arrange. This can be done in 4! ways. Then the unit corresponding to the three novels can be expanded to an ordering of the novels in 3! ways. Thus by the BPC, 3! · 4! = 144 arrangements are possible.
13. There will be one handshake for each 2-element subset of the set of 20 people. Thus there will be 20 = 190 2 handshakes.
52 14. because a poker hand is just a 5-element subset of the set of 52 cards in 5 the deck.
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10 15. The set of 5 women can be chosen from the 10 in the class in ways. 5 12 Similarly the 5 men can be chosen in 5 ways. The set of partners can be formed in 5! ways. For we can order the women arbitrarily, and then the first woman can choose any of the 5 men as her partner. The second woman can choose any of the remaining 4 men as her partner, etc. Thus, by the GBPC, 10 12 · · 5! 5 5 different results are possible.
18. By the GBPC, 5 6 4 · · = 600 2 2 3 distinct committees are possible.
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