SolutionsChapter3_CS40

SolutionsChapter3_CS40 - Chapter 3 Set Theory Section...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 3 Set Theory Section 3.1—p. 134 Section 3.2~p. N16 1. 3. :49- ll. 13. 17. 19. They are all the same set. Parts (b) and (d) are false: the remaining parts are true. a) {0, 2} b} {2,2535%15 e) {0,2,1236‘801 a} Vx{.rEA—->A'EBlAEleIEBAJ‘éA} b) 3x[xEAA.tEB]vi[xEBVJ.-EAJ 012,3;-[.tEAAxEB]VVx{xEB—>_rEAJ a) |A| = 6 b) |Bl = "l e) If B has 2” subsets ofodd cardinality then {Bl = i: + l. a) N b) so or 28 13. a} a“) b) (r) o (r) + (2:) + (r) + (in Let W : {l}, X = {{l}, 2}. and Y = {X. 3}. e) If): E A, then A g B => 1' E B. and B C C 2:» x E C. Hence A E C. Since B C C. there exists y E C with y E B. Also, A g B and y E B :5 li- E A. Consequently. A <_: C and _r E C" withygé A :>A c: C. d) Since A C B‘ it follows that A g B. The result then follows from part (e). a) For n. l' E Z"r with n 3 k + 1. consider the hexagon centered at Thin has the form {2:0 in») m (on where the two alternating triplesflaiatnely, fl , (k _" (H ') and ("t-— I}: l): (i.- i: !)—'Satisry : l}(l::l)(n J) : (H; b) For”, k E Z" with n 3: k + l, (til) (ll—l) n )(an+l)_[ (it—l)! ] n! ] (n+1)! lt—l (k+l k (k— l}!(ri—l(}! [fli-l—lflUt—k- I)! [l(l{H-l-l-~ it}! _ (it — I)! [ (:1 +1)! it! _‘ (n —l (H ~l-- I) n k!(n—l—-k)ll (k+l)!{n-—lr)l fk-—l)l(n—k+l)!l it ) k—i—l (li‘lj- 21. rt = 20 23. The fifth, sixth‘ and seventh entries in the row for n = M provide the unique solution. 25. As an ordered set, A = {x. u, w. z‘ y}. . a) If S E S, then sinee S = {Al/l E A} we have 3 gé 3. b} [l’ S E 3, then by the definition of 5' it follows that S E 3. a) {1. 2. 3,5} b) A e) and 6]) 6ll. — {2} e) {4‘ 8} f) {1, ,3,4‘5.8} g) it) h) {2,4,8} i) {1.14.5.8} a) A = {1,3, 4, 7.9, 11} B = {2,4, 6, 8. 9} b)C={l,2,4,5.9} D={5‘?.8,9} a) True b} True I?) True 11) False e) True f } True g) True h) False i) FalSe . a} Leigh: {l‘2.3},A ={l},B={2},andC= {3}.ThenAflC= BflC =lifithtA¢ B. b) FOI'OH. = {1. 2}. A = {l}, B = {2},andC = E"ll. we haVeA UC = B UC butA 79 8. [From parts (a) and (b) we see that we do not have cancellation laws for H or U. This differs from what we know about R, where fora, b. t' E R (Dab = rte and a # 02> b = c; and (ii) a + b = a + 6 => 33 = 6.] e} x E (12:: ,r E A U C 2> x E B U C. 50): E 8 orx E C. Ifx E B. then we are finished. If x E C, then .r E A n C = B m C andx E B. In eilhercase, .t- E B so A g B, Likewise, S-14 Solutions Section 3.3—p. 150 Section 3.4 — p. 156 Section 3.5—p. 164 13. 15. 17. 19. 9”!" 1-w—t 11. 13. .55 3. 29+28—25=736 . a) 24! + 24! — 22! . [Bl/(203] ! 3[12!/(2!)2‘l + 3(1 ll/2!) — 10! snows yeB:>y€BUC=AUC,soyeAoryeC.InyC,thenyeBflC=AflC.Ineither case, )1 e A and B Q A. Hence A = B. d) Letx e A. Considertwo cases: (1)x E C :3 x g? A A C :>.\‘ 6% B A C :>.x E B. (2) x 92 C :> E A A C :> .r E B A C :> x E B (because 11’ 6! C). In either case. .1“ E B, so A g B. In 21 similar way it follows that B Q A and A = B. .7;l .a)1x1:(AUBm(AUE)mAUBm(AUE) b) A=AU<AflB1 c) AmB=(AUBm(Au§m(AuB1 d) A :(Amnmmmnn a) Letoll = {1, 2, 3}, A = {1}. and B = {2}.Then {1. 2} ECWA U B) but {1. 2} §é @(A) UOJ’(B). b) Xe@(AflB)<:>X§AflB<=>XgAanngB<:¢Xe§P(A)and X E @(B) <:> X E @(A) HQNB), so @(A (1 B) = 9PM) 095(8). 3) 2!1 b) 2“ c) In the—membership table, A g B ifthe columns for A, B are such that whenever a 1 occurs in the co umn ’or A, there is a corresponding 1 in the column for B. d) A B C AUE (AnB)U(BnC) 0 0 0 l 1 O 0 l l 1 0 1 O 0 l O 1 1 O 0 l 0 0 l 1 1 0 1 l 1 l l 0 l l 1 1 l l l a)Afl(B~A)=Afl(BflA)=Bfl(AflA)=Bfl17l=l/l b) [(A m B) u (A n B n E n 0)] u (A m B) = (A n B) U (A m B) by the Absorption Law =(AUAmB=61mB:B d)AofiwAanE)=(AnB)u[(AmBmE]= [(AmB)0(AnB)]n[(AmB)UC]2[(AnB)uC]=AUBuC 601—61911 095 e) A7 g)R 5. 9! + 9! — 8! = 685,440 b) 26! — [24! + 24! — 23!] a) 3/8 b) 1/2 c) 1/4 (1) 5/8 e) 5/8 f) 7/8 g) l/8 6 5. a) (23)/(‘§) =5/22 1)) 7/22 7. 49/99 a) 1/64 1))3/32 c) 15/64 d) 1/2 0) 11/32 11. 3) 55/216 1)) 5/54 .a) = % b) 2/15 c) 3/35 PI‘(A)= 1/3, [313(13): 7/15, Pr-(A m B) = 2/15, Pr(A U B) = 2/3; P1'(A U B) = 2/3 : 1/3 17/15 2/15» P1‘(A) 1 Pm?) Pr(AflB) . P1~(”A) = 0.6; Pm?) 2 0.7; Pr(A o B) = 0.5; Pr(A u B) = 0.5;P1~(A mE):(1.2; PNAO B) = 0.1-, P1-(A U?) = 0.9; PHAU B) = 0.3 .a) S: {(.x,y)1x,ye11,2, lO].x 791} b) 1/2 c) 5/9 .0.4 7.11) 11/21 b) 12/21 c) 9/21 9. 3/16 a) (1) 27/38 (11) 27/38 b) (081/361 (11) 18/361 11/14 15. (9)4870) = 330/3,176,716.400 Section 3.6 — p. 173 Section 3.7—p. 185 Solutions 5-15 17. Since A U B Q if. it follows from the result ofthe preceding exercise that FHA U B) : Pr(51’)= 1.8013 P1‘(A U B) = FHA) + P7'(B) — P/‘(A fl B).11n(l Pr(A 0 B) 3 Pr(A) + Pr(B) — l = 0.7 —l— 0.5 *1: 0.2. 1. 1/4 3. (0.80)(0.75) : 0.60 5. In general, Pr(A U B) = Pr(A) + Pr(B) — Pl‘(A H B). Since A, B are independent, Pr(A D B) = Pr(A)Pr(B). So ll Pr(A u B) Pr(A) — PI“(B) — Pr(A)Pr(B) = FHA) +11 — P1‘(A,l.lPI'(B) Pr(A) » Pump/<13). l| The proof for P/‘(B) + Pr(B)Pr(A) is similar. 7. a) 52/85 1)) ll/26 9. 3/7 11. Pr(A F) B) = 1/4 = (1/2)('l/2) = P1‘(A)P1‘(B), so the events A, B are independent. 13. 1/5 15. (0.05)(0.02) =0.001 17. 5/2] 19. Any two of the events are independent. However, Pr(A F) B 1) C) : l/4 # l/8 = (l/2)(1/2)(1/2) : Pr(A)Pr(B)Pr(C), so the events A, B, C are not independent. 21. a) 5/16 b) ll/32 c) ll/32 23. 0.6 25. a) 25 — (3) — (i) = 26 b) 2'1—(;;)—(';) = 2” — (11+ 1) 27. 30/77 29. 0.15 1. a) 1/4 b) 1 c) 7/8 d) 3/4 e) 2/7 f) 1/2 (9)141“) 3. a) Pr(X =x) = :0,1,2,3,4,5. IN (10;(5112)> b) Pr(X = 4) = 4(120)‘ = 275/2,268,786 5 c) 139/1,134,393 d) 2675/8796 5. a) 2/3 b) 2/3 c) 1/4 (1) 7/2 e) 35/12 7. a) 2:1/15 b) 3/5 c) 7/3 d) 14/9 9. n =200, p=0.35 11. a) (0.75)?5 0.100113 b) (f3)(0.25)3(0.75)5 : 0.207642 0) [$316 §)(0.25)x(0.75)8-X 2 0.004227 d) 0.037139 (approximately) e) 2 f) 1.5 13. c =10 15. a) Pr(X = 1) = 1/5; Pr(X = 2) = 16/95; Pr(X = 3) = 12/19 b) 7/19 c) 19/35 d) 231/95 22431579 e) 5824/9025 20.645319 17. a) E(X(X —1))= Zxoc —1)Pr(X = x) = 2x01 —1)Pr(X = x) x:0 .\'=Z '1 n , " r1! . >pan—.\ :Z ‘ ‘x(x_])pan—1 X22 x FZ x.(n ~ x). ” 77! , _, 7 " (77 — 2)! ,_2 7 _ A ll A _ -1 V 1 .\ Il .\ 1120—2107—5011) ‘7 ‘0'!” (x—Z)!(n.—X)!p q 77—2! . . . wall/(n 1): ( ) ,Iflq’!“"'+2)1 s11bst1tut1ngx~2=>n (11 — 2)! . _ _. __ 2 y (11 2) V1 [217(77 1)))![01— 2) _ ypp q = 1221707 ~ l)(/) —l— q)”'2, by the Binomial Theorem = [9277.01 ~ l)(l)"’3 = 1121707 — 1) 2112122 w 11])2 b) Va1‘(X) = 15002 — [1300!2 = [E(X(X —1))+ 15(X)J—lE(X)lI : [(172];2 17122) l 171)] (up)2 “7721)1 77.122 l 77/) 173/23 — 71/) 1'1/2 “7712(1 p) 7 npq. 5-15 Solutions 19. a) me = 2) = 1/4; P1'(X = 3) = 1/8; Pr(X = 4) = 1/4; Pr(X = 5) = 1/4; Pr(X=6)=1/8 b) 31/8 c) 119/64 21. E(X) = 4; ox :1 Supplementary Exercises—p. 189 1. Suppose that (A — B) (_2 C and x E A ~ C. Then x E A but x ¢ C. If x ¢ 8, then [x E A /\ x gt B] :> x E (A — B) _C_ C. So now we have X g? C and x e C. This contradiction gives us x E B, so (A — C) 9 B. Conversely, if (A — C) Q B, let )1 E A — B. Then y E A but y gé B. If y ¢ C, then [)1 E A /\ y g2 C] :> y E (A — C) g B. This contradiction—that is, y ¢ B and y E B —yields ‘ yeC,so(A—B)§C. 3. a) The sets 611,: {1, 2, 3}, A = {1, 2}, B = {l}, and C = {2} provide a counterexample. b) A=Afl01L=Afl(CUC)=(AflC)U(AflC)=(AflC)U(A~C) =(BflC)U(B—C)=(BflC)U(BflC)= Bfl(CUC)= BnOn=B 5. a) 126 (if teams wear different uniforms); 63 (if teams are not distinguishable) 1 12 (if teams wear different uniforms); 56 (if teams are not distinguishable) b) 2” — 2; (l/2)(2" — 2). 2” — 2 * 2n; (1/2)(2” — 2 # 2n). 7. a) 128 b) {Al=8 9. Suppose that (A f) B) U C = A O (B U C) and that x e C. Then x€C:>xe(AflB)UC:>xEAfl(BUC)§A,sox€A,andC§A. Conversely, suppose that C C; A. (l) Ifye(AflB)UC,thenyEAflBoryEC. (i) yeAflB:>ye(AflB)U(AflC):>yeAfl(BUC). (ii) )1EC:y6A,becauseCgA.Also,yeC:>yEBUC,SO)'€AF)(BUC), In either case(i) or case (ii), we have ye Afl(BUC),so (AflB)UC E Afl(BUC). (2) NowletzeAfl(BUC).Thenz€Afl(BUC)=(AflB)U(AflC)§(AflB)UC, since A r) C Q C. From parts (1) and (2) it follows that (A F) B) U C = A r) (B U C). 11. a) ,0, 14/3] 1)) {O} U (6, 12] c) [0, +00) (1) V1 13- a) A B A {'1 B Since A g B, consider only rows 1, 2, and 4. For these rows, A r) B = A. 0 0 0 O 1 O l 0 0 l 1 l 0 A B C (A n?) u (B n?) A n? For C Q B g A, consider only 0 h 0 O 0 0 rows 1, 5, 7, and 8. Here 0 0 I 0 O (AflB)U(BflC) AflC. 0 l 0 l O 0 1 1 0 O 1 0 O l 1 l 0 1 l 0 1 1 O l 1 1 l 1 0 O 15. 17. 23. 25. 29. 31. 35. 37. 41. 43. S- 17 Solutions d) A B C A A B A A C B A C When A A B : C, we consider 0 0 0 rows 1, 4. 6, and 7. In these cases. 0 0 0 AAC=BundBAC=A. 0 0 1 0 1 1 O 1 O 1 O 1 0 1 1 1 1 0 1 O 0 1 1 0 1 O 1 1 O 1 1 1 0 0 1 1 1 1 1 0 0 O a) (if) (m g r + 1) b) (“2:”) (2k 511+ 1) a) 23 b) 8 19. 7'5 —3(3‘5) +3 21. ('})(‘3°)/(172) = 0.3483 3) 21:0 : 13:0 b) (i) ('é)(‘5‘)/ 1 1:0 (1)6111] 00 ('§)(‘P)/ 1 1:0 (1”) 1:11] (iii) [(26) + (5)024) + (3)03) + (3)06") + (11) (1)1 / l 1:0(1')('gi1)1 AUB=[~2,4],AF)B={3} 27. 135/512i0.263672 Pr(A F) (B U C)) = Pr((A f) B) U (A F) C)) = Pr(A t) B) + Pr(A F) C) — Pr((A D 8) 0 (A f) C)). Since A, B, C are independent and (AflB)fl (AflC) = (AHA) ()(BOC) = A flBflC, P/‘(A fl(BUC)) = Pr(A)Pr(B) + Pr(A)Pr(C) — Pr(A)Pr(B)Pr(C) = Pr(A)[Pr(B) + Pr(_C) — Pr(B)Pr(C)l : Pr(A)[P1-(B) + Pr(C) — Pr(B F) C)] = Pr(A)Pr(B U C). so A and B U C are independent. 3) 0.99 b) (0.99)3 = 0.970299 33. 3/5 (§)(0.8)3(0.2)2 + (Z)(0.8)4(0.2) + (§)(0.8)5 2 0.94208 675/2048 39. a) c: = 1/50 1)) 0.82 c) 13/41 d) 2.8 e) 1.64 a) 3/(427) b) [(20) “31/(427) 0) [310(1) —31/(427) 2/[m(m —l— 1)] 45. a) Pr(X = 1) = 7/16; Pr(X = 2) = 3/8; Pr(X = 3) = 3/16 1)) 7/4 C) ox =3/4 Chapter4 Properties of the Integers: Mathematical Induction Section 4.1 —p. 208 1. b) Since 1 v 3 = (1)(2)(9)/6, the result is true for n = 1.Assume the result is true 101‘ n = k(21):1-3+214+3-5 forn=k+1:[1-3l2-4 - (k ~— 1)(k ~— 3) = [(k —l— 1)/6][k(2k k(k 2) “ k(k l 1)(2k-l—7)/6.Then consider the case WC 7- 2)1+(k + 1)(k + 3) = [We +1)(2k + 7)/61 + +7)~-6(k l 3)]— (k i 1)(2/81 1311 18)/6— (k 1)(k 2) (2k + 9)/6. Hence he result follows for 2111/1 E Z+ by the Principle 01" Mathematical Induction. c) S< ) n 1 ” n : ——— = [:1 [TU—«1) n+1 5(1) I I 1 l S(])' t —— = —— = , so 15 rue. [:1 i(i»-1) 1(2) 1+1 *' 1 Assume S(k): Z I . 1'21 1“ 1 l’ 1 1 1 Z [(1 + 1) _ 1 [:1 :1 r(r+1) ‘ (k—vl)(k+2) (1+1) + (k+1)(k+2) = Wk + 2) + 11/l(1< + 1)(k + 2)] = (k +1)/(k + 2). k —. Consider S(k + 1). ...
View Full Document

This note was uploaded on 12/27/2011 for the course CMPSC 40 taught by Professor Egiceoclu during the Fall '09 term at UCSB.

Page1 / 5

SolutionsChapter3_CS40 - Chapter 3 Set Theory Section...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online