1 Chapter 1 1–1. (a) P(A∪B ∪C) = 1−P(A∪B ∪C) = 1−0.25 = 0.75 (b) P(A∪B) = P(A)+P(B)−P(A∩B) = 0.18 1–2. (a) A B A B U U C C A∪(B∪C) = (A∪B)∪C A∩(B∩C) = (A∩B)∩C (b) A∪B A B A B U U = C A∪C C A∪(B∩C) = (A∪B)∩C(A∪C) A B A B U U = C A C C A∩(B∪C) = (A∩B)∪(A∩C) (c) (d) U U A B A B A∩B = A A∪B = B (e) A B (f) U U ABC A∩B = ∅ ⇒ A⊂B A⊂B and B⊂C ⇒ A⊂C 1–3. (a) A∩B = {5}, (b) A∪B = {1,3,4,5,6,7,8,9,10}, (c) A∩B = {2,3,4,5}, (d) A∩(B ∩C) = U = {1,2,3,4,5,6,7,8,9,10}, (e) A∩(B ∪C) = {1,2,5,6,7,8,9,10} 2 1–4. P(A) = 0.02, P(B) = 0.01, P(C) = 0.015 P(A∩B) = 0.005, P(A∩C) = 0.006 P(B ∩C) = 0.004, P(A∩B ∩C) = 0.002 P(A∪B ∪C) = 0.02+0.01+0.015−0.005−0.006−0.004+0.002 = 0.032 1–5. S = {(t ,t ): t ∈ R,t ∈ R,t ≥ 0,t ≥ 0} 1 2 1 2 1 2 t t t 2 2 2 0.15 t +t = 0.3 1 2 C B A t t t 1 0.15 1 1 A = {(t ,t ): t ∈ R,t ∈ R,0 ≤ t ≤ 0.3,0 ≤ t ≤ 0.3−t } 1 2 1 2 1 2 1 B = {(t ,t ): t ∈ R,t ∈ R,0 ≤ t ≤ 0.15,0 ≤ t ≤ 0.15} 1 2 1 2 1 2 C = {(t ,t ): t ∈ R,t ∈ R,t ≥ 0,t ≥ 0,t −0.06 ≤ t ≤ t +0.06} 1 2 1 2 1 2 1 2 1 1–6. (a) S = {(x,y): x ∈ R,y ∈ R,0 ≤ x ≤ y ≤ 24} (b) i) t2 ii) y 24 24 y – x = 1 1 23 t2 24 t1 t1 x iii) y 24 4.8 x 19.2 24 3 1–7. S = {NNNNN,NNNND,NNNDN,NNNDD,NNDNN, NNDND,NNDD,NDNNN,NDNND,NDND, NDD,DNNNN,DNNND,DNND,DND,DD} 1–8. {0,1}A = {∅,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c},{b,d}, {c,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},{a,b,c,d}} 1–9. N = Not Defective, D = Defective (a) S = {NNN,NND,NDN,NDD,DNN,DND,DDN,DDD} (b) S = {NNNN,NNND,NNDN,NDNN,DNNN} 1–10. p(cid:48) = Lot Fraction Defective 50·p(cid:48) = Lot No. of Defectives (cid:181) (cid:182)(cid:181) (cid:182) 50p(cid:48) 50(1−p(cid:48)) 0 10 P(Scrap Lot|n = 10,N = 50,p(cid:48)) = 1− (cid:181) (cid:182) 50 10 If p(cid:48) = 0.1, P(scrap lot) ∼= 0.689. She might wish to increase sample size. 1–11. 6·5 = 30 routes 1–12. 263 ·103 = 17,576,000 possible plates ⇒ scheme feasible (cid:181) (cid:182)(cid:181) (cid:182)(cid:181) (cid:182) 15 8 4 1–13. = 560,560 ways 6 2 1 (cid:181) (cid:182)(cid:181) (cid:182) 20 80 (cid:88)2 k 4−k 1–14. P(X ≤ 2) = (cid:181) (cid:182) ∼= 0.97 100 k=0 0.4 (cid:181) (cid:182)(cid:181) (cid:182) 300p(cid:48) 300(1−p(cid:48)) (cid:88)1 k 10−k 1–15. P(Accept|p(cid:48)) = (cid:181) (cid:182) 300 k=0 10 1–16. There are 512 possibilities, so the probability of randomly selecting one is 5−12. (cid:181) (cid:182) 8 1–17. = 28 comparisons 2 4 (cid:181) (cid:182) 40 1–18. = 780 tests 2 40! 1–19. P40 = = 1560 tests 2 38! (cid:181) (cid:182) 10 1–20. = 252 5 (cid:181) (cid:182)(cid:181) (cid:182) 5 5 1–21. = 25 1 1 (cid:181) (cid:182)(cid:181) (cid:182) 5 5 = 100 2 2 1–22. [1−(0.2)(0.1)(0.1)][1−(0.2)(0.1)](0.99) = 0.968 1–23. [1−(0.2)(0.1)(0.1)][1−(0.2)(0.1)](0.9) = 0.880 1–24. R = R {1−[1−(1−(1−R )(1−R ))(R )][1−R ]} S 1 2 4 5 3 1–25. S = Siberia U = Ural P(S) = 0.6,P(U) = 0.4,P(F|S) = P(F|S) = 0.5 P(F|U) = 0.7,P(F|U) = 0.3 (0.6)·(0.5) . P(S|F) = = 0.714 (0.6)(0.5)+(0.4)(0.3) 1–26. R = (0.995)(0.993)(0.994) = 0.9821 S 1–27. A: 1st ball numbered 1 B: 2nd ball numbered 2 P(B) = P(A)·P(B|A)+P(A)·P(B|A) 1 1 m−1 1 = · + · m m−1 m m m2 −m+1 = m2(m−1) 1–28. 9×9−9 = 72 possible numbers D +D even: 32 possibilities 1 2 P(D odd and D odd|D +D even) = 20. 1 2 1 2 32 5 1–29. A: over 6(cid:48) M: male F: female P(M) = 0.6,P(F) = 0.4,P(A|M) = 0.2,P(A|F) = 0.01 P(F)·P(A|F) P(F|A) = P(F)·P(A|F)+P(M)·P(A|M) (0.04)(0.01) 0.004 = = (0.4)(0.01)+(0.6)(0.2) 0.124 ∼ = 0.0323 1–30. A: defective B : production on machine i i (a) P(A) = P(B )·P(A|B )+P(B )·P(A|B )+P(B )·P(A|B ) 1 1 2 2 3 3 + P(B )·P(A|B ) 4 4 = (0.15)(0.04)+(0.30)(0.03)+(0.20)(0.05)+(0.35)(0.02) = 0.032 (0.2)(0.05) (b) P(B |A) = = 0.3125 3 0.032 1–31. r = radius (cid:179) (cid:180) r 2 π 1 2 P(closer to center) = = πr2 4 1–32. P(A∪B ∪C) = P((A∪B)∪C) (associative law) = P(A∪B)+P(C)−P((A∪B)∩C) = P(A)+P(B)−P(A∩B)+P(C)−P((A∩C)∪(B ∩C)) = P(A)+P(B)+P(C)−P(A∩B)−P(A∩C) − P(B ∩C)+P(A∩B ∩C) 1–33. For k = 2, P(A ∪A ) = P(A )+P(A )−P(A ∩A ); Thm. 1–3. 1 2 1 2 1 2 Using induction we show that if true for k −1, then true for k, i.e., 6 If (cid:88)k (cid:88) (cid:88) P(A ∪A ∪···∪A ) = P(A )− P(A ∩A )+ P(A ∩A ∩A ) 2 3 k i i j i j r i=2 2≤i<j≤k 2≤i<j<r≤k (cid:88) − P(A ∩A ∩A ∩A )+··· (Eq. 1) i j r (cid:96) 2≤i<j<r<(cid:96)≤k Then (cid:88)k (cid:88) (cid:88) P(A ∪A ∪···∪A ) = P(A )− P(A ∩A )+ P(A ∩A ∩A ) 1 2 k i i j i j r i=1 1≤i<j≤k 1≤i<j<r≤k (cid:88) − P(A ∩A ∩A ∩A )+··· (Eq. 2) i j r (cid:96) 1≤i<j<r<(cid:96)≤k By hypothesis, and letting A ∩A replace A in Eq. 1, 1 i i (cid:88)k (cid:88) P((A ∩A )∪(A ∩A )∪···∪(A ∩A )) = P(A ∩A )− P(A ∩A ∩A ) 1 2 1 3 1 k 1 i 1 i j i=2 2≤i<j≤k (cid:88) (cid:88) + P(A ∩A ∩A ∩A )− P(A ∩A ∩A ∩A ∩A )+··· (Eq. 3) 1 i j r 1 i j r (cid:96) 2≤i<j<r≤k 2≤i<j<r≤k By Thm. 1–3, P(A ∪(A ∪A ∪···∪A )) = P(A )+P(A ∪A ∪···∪A ) 1 2 3 k 1 2 3 k − P((A ∩A )∪···∪(A ∩A )) 1 2 1 k So from using Eq. 1 through 3, P(A ∪A ∪···∪A ) 1 2 k (cid:34) (cid:35) (cid:88)k (cid:88) (cid:88) = P(A )+ P(A )− P(A ∩A )+ P(A ∩A ∩A )−··· 1 i i j i j r i=2 2≤i<j≤k 2≤i<j<r≤k (cid:34) (cid:35) (cid:88)k (cid:88) (cid:88) − P(A ∩A )− P(A ∩A ∩A )+ P(A ∩A ∩A ∩A )−··· 1 i 1 i j 1 i j r i=2 2≤i<j≤k 2≤i<j<r≤k (cid:88)k (cid:88) (cid:88) = P(A )− P(A ∩A )+ P(A ∩A ∩A ) i i j i j r i=1 1≤i<j≤k 1≤i<j<r≤k +···+(−1)k−1 ·P(A ∩A ∩···∩A ) 1 2 j 7 (365)(364)···(365−n+1) 1–35. P(B) = 365n n 10 20 21 22 23 24 25 30 40 50 60 P(B) 0.117 0.411 0.444 0.476 0.507 0.538 0.569 0.706 0.891 0.970 0.994 6 2 8 1–36. P(win on 1st throw) = + = 36 36 36 P(win after 1st throw) = P(win on 4)+P(win on 5)+P(win on 6) + P(win on 8)+P(win on 9)+P(win on 10) (cid:34) (cid:35) (cid:181) (cid:182) (cid:181) (cid:182) 3 3 27 3 27 2 3 1 P(win on 4) = · + · + · +··· = 36 36 36 36 36 36 36 (cid:34) (cid:35) (cid:181) (cid:182)(cid:181) (cid:182) (cid:181) (cid:182) (cid:181) (cid:182) 4 4 26 4 26 2 4 2 P(win on 5) = · + + · +··· = 36 36 36 36 36 36 45 (cid:34) (cid:35) (cid:181) (cid:182)(cid:181) (cid:182) (cid:181) (cid:182) (cid:181) (cid:182) 5 5 25 5 25 2 5 25 P(win on 6) = + + · +··· = 36 36 36 36 36 36 396 25 P(win on 8) = P(win on 6) = 396 2 1 P(win on 9) = P(win on 5) = P(win on 10) = P(win on 4) = 45 36 (cid:183) (cid:184) 8 1 2 25 P(win) = + 2· +2· +2· = 0.4929 36 36 45 396 1–37. P8 = 8! = 40,320 8 1–38. Let B, C, D, E, X represent the events of arriving at points so labeled. 1 P(B) = P(C) = P(D) = P(E) = 4 1 2 P(X|B) = , P(X|C) = 1, P(X|D) = 1, P(X|E) = 3 5 P(X) = P(B)·P(X|B)+P(C)·P(X|C)+P(D)·P(X|D)+P(E)·P(X|E) (cid:181) (cid:182) (cid:181) (cid:182) (cid:181) (cid:182) (cid:181) (cid:182) 1 1 1 1 1 2 41 = · + ·1 + ·1 + · = 4 3 4 4 4 5 60 8 (cid:183) (cid:184) P(B )·P(A|B ) (0.5)(0.3) 3 3 1–39. P(B |A) = = 3 (cid:88)3 (0.2)(0.2)+(0.3)(0.5)+(0.5)(0.3) P(B )·P(A|B ) i i i=1 = 0.441 1–40. F: Structural Failure D : Diagnosis as Structural Failure S P(F)·P(D |F) S P(F|D ) = S P(F)·P(D |F)+P(F)·P(D |F) S S (0.25)(0.9) 0.225 = = = 0.6 (0.25)(0.9)+(0.75)(0.2) 0.225+0.150 1 Chapter 2    4 48    0 5 2–1. R = {0,1,2,3,4},P(X = 0) =   X 52   5 (cid:181) (cid:182)(cid:181) (cid:182) (cid:181) (cid:182)(cid:181) (cid:182) 4 48 4 48 1 4 2 3 P(X = 1) = (cid:181) (cid:182) , P(X = 2) = (cid:181) (cid:182) , 52 52 5 5 (cid:181) (cid:182)(cid:181) (cid:182) (cid:181) (cid:182)(cid:181) (cid:182) 4 48 4 48 3 2 4 1 P(X = 3) = (cid:181) (cid:182) , P(X = 4) = (cid:181) (cid:182) 52 52 5 5 1 1 1 1 1 1 26 2–2. µ = 0· +1· +2· +3· +4· +5· = 6 6 3 12 6 12 12 (cid:183) (cid:184) (cid:181) (cid:182) 1 1 1 1 1 1 26 2 83 σ2 = 02 · +12 · +22 · +32 · +42 · +52 · − = 6 6 3 12 6 12 12 36 (cid:90) ∞ 2–3. ce−xdx = 1 ⇒ c = 1, so 0 (cid:189) e−x if x ≥ 0 f(x) = 0 otherwise (cid:90) (cid:90) ∞ ∞ µ = xe−xdx = −xe−x|∞ + e−xdx = 1 0 0 0 (cid:90) (cid:183) (cid:90) (cid:184) ∞ ∞ σ2 = x2e−xdx−12 = −x2e−x|∞ + 2xe−xdx −1 0 0 0 (cid:183) (cid:90) (cid:184) ∞ = −2xe−x|∞ + 2e−xdx −1 0 0 = 2−1 = 1 2–4. F (t) = P (T ≤ t) = 1−e−ct;t ≥ 0 T T (cid:189) ce−ct if t ≥ 0 ∴ f (t) = F(cid:48)(t) = T T 0 otherwise 2 2–5. (a) Yes (b) No, since G (∞) (cid:54)= 1 and G (b) (cid:54)≥ G (a) if b ≥ a X X X (c) Yes 2–6. (a) f (x) = F(cid:48) (x) = e−x;0 < x < ∞ X X = 0;x ≤ 0 (c) h (x) = H(cid:48) (x) = ex;−∞ < x ≤ 0 X X = 0;x > 0 2–7. Both are since p (x) ≥ 0, all x; and Σ p (x) = 1 X allx X 2–8. The probability mass function is x p (x) X −1 1 5 0 1 10 +1 2 5 +2 3 10 ow 0 (cid:181) (cid:182) (cid:181) (cid:182) (cid:181) (cid:182) (cid:181) (cid:182) 1 1 2 3 4 E(X) = −1· + 0· + 1· + 2· = 5 10 5 10 5 (cid:181) (cid:182) (cid:181) (cid:182) (cid:181) (cid:182) (cid:181) (cid:182) (cid:181) (cid:182) 1 1 2 3 4 2 29 V(X) = (−1)2 · + 02 · + 12 · + 22 · − = 5 10 5 10 5 25