TEST BANK FOR Probability Concepts in Engineering Emphasis on Applications to Civil and Environmental Engineering 2nd Edition By Alfredo H. Ang

Exam (elaborations) TEST BANK FOR Probability Concepts in Engineering Emphasis on Applications to Civil and Environmental Engineering 2nd Edition By Alfredo H. Ang 2.1 TAB TBC CAB CBC TAC CAC 1500 1500 850 850 850 (a) Sample space of travel time from A to B = {6, 7, 9, 10, 11} Sample space of travel time from A to C = {8, 9, 10, 11, 12, 13, 14} (b) Sample space of travel cost from A to C = {850, 1500} (c) Sample space of TAC and CAC = {(8, 1500), (9, 1500), (10, 1500), (11, 850), (12, 850), (13, 850), (14, 850)} 2.2 (a) Since the possible values of settlement for Pier 1 overlap partially with those of Pier 2, it is possible that both Piers will have the same settlement. Hence, the minimum differential settlement is zero. The maximum differential settlement will happen when the settlement of Pier 2 is 10 cm and that of Pier 1 is 2 cm, which yields a differential settlement of 8 cm. Hence, the sample space of the differential settlement is zero to 8 cm. (b) If the differential settlement is assumed to be equally likely between 0 and 8 cm, the probability that it will be between 3 and 5 cm is equal to 0.25 8 2 8 0 5 3 = = − − P = 2.3 (a) (b) Wind direction E1 Wind Speed 90o 30o 60o E2 E3 E1 0 15 35 45 (c) A and B are not mutually exclusive A and C are not mutually exclusive 2.4 Possible water level (a) Inflow Outflow Inflow – Outflow + 7’ 6’ 5’ 8’ 6’ 6’ 7’ 6’ 7’ 6’ 7’ 5’ 9’ 7’ 6’ 8’ 7’ 7’ 7’ 8’ 5’ 10’ 8’ 6’ 9’ 8’ 7’ 8’ Hence possible combinations of inflow and outflow are (6’,5’), (6’,6’), (6’,7’), (7’,5’), (7’,6’), (7’,7’), (8’,5’), (8’,6’) and (8’,7’). (b) The possible water levels in the tank are 6’, 7’, 8’, 9’ and 10’. (c) Let E = at least 9 ft of water remains in the tank at the end of day. Sample points (7’, 5’), (8’, 5’) and (8’, 6’) are favourable to the event E. So P(E) = 9 3 = 3 1 . 2.5 (a) Locations of W1 Locations of W2 Load at B Load at C MA Probability E1 E2 E3 – – 0 0 0 0.15x0.2=0.03 – B 500 0 5,000 0.045 X – C 0 500 10,000 0.075 X X B – 200 0 2,000 0.05 X X B B 700 0 7,000 0.075 X X X B C ,000 0.125 X C – 0 200 4,000 0.12 X C B 500 200 9,000 0.18 X X C C 0 700 14,000 0.30 X (b) E1 and E2 are not mutually exclusive because these two events can occur together, for example, when the weight W2 is applied at C, MA is 10,000 ft-lb; hence both E1 and E2 will occur. (c) The probability of each possible value of MA is tabulated in the last column of the table above. (d) P(E1) = P(MA 5,000) = 0.075 + 0.075 + 0.125 + 0.18 + 0.30 = 0.755 P(E2) = P(1,000 ≤ MA ≤ 12,000) = 0.045 + 0.075 + 0.05 + 0.075 + 0.12 + 0.18 = 0.545 P(E3) = 0.05 + 0.075 = 0.125 P(E1 ∩ E2) = P(5,000 MA ≤ 12,000) = 0.075 + 0.075 + 0.18 = 0.33 P(E1 ∪ E2) = 1 – 0.03 = 0.97 P( 2 E ) = 0.03 + 0.125 + 0.3 = 0.455 P( 2 P(E ) = 1− P(E) = 1− 0.545 = 0.455 2.6 (a) Let A1 = Lane 1 in Route A requires major surfacing A2 = Lane 2 in Route A requires major surfacing B1 = Lane 1 in Route B requires major surfacing B2 = Lane 2 in Route B requires major surfacing P(A1) = P(A2) = 0.05 P(A2⏐A1) = 0.15 P(B1) = P(B2) = 0.15 P(B2⏐B1) = 0.45 P(Route A will require major surfacing) = P(A) = P(A1∪A2) = P(A1) + P(A2) - P(A2⏐A1) P(A1) = 0.05 + 0.05 - (0.15)(0.05) = 0.0925 P(Route B will require major surfacing) = P(B) = P(B1∪B2) = P(B1) + P(B2) - P(B2⏐B1) P(B1) = 0.15 + 0.15 - (0.45)(0.15) = 0.2325 (b) P(route between cities 1 and 3 will require major resurfacing) = P(A∪B) = P(A) + P(B) - P(A)P(B) = 0.0925 + 0.2325 – (0.0925)(0.2325) = 0.302 2.7 P(Di) = 0.1 Assume condition between welds are statistically independent (a) P( 1 D 2 D 3 D ) = P( 1 D )P( 2 D )P( 3 D ) = 0.9 x 0.9x 0.9 = 0.729 (b) P(Exactly two of the three welds are defective) = P( 1 D D2D3 ∪ D1 2 D D3 ∪ D1 D2 3 D ) = P( 1 D D2D3) + P(D1 2 D D3) + P(D1 D2 3 D ) since the three events are mutually exclusive. Hence, the probability become P = 0.9x0.1x0.1 + 0.1x0.9x0.1 + 0.1x0.1x0.9 = 0.027 (c) P(all 3 welds defective) = P(D1 D2 D3) = P3(D) = (0.1)3 = 0.001 2.8 P(E1) = 0.8; P(E2) = 0.7; P(E3) = 0.95 P(E3⏐ 2 E ) = 0.6; assume E2 and E3 are statistically independent of E1 (a) A = (E2 ∪ E3) E1 B = E E E or 2 3 1 ( ∪ ) 2 E 3 E ∪ 1 E (b) P(B) = P( 1 E ∪ 2 E 3 E ) = P( 1 E ) + P( 2 E 3 E ) – P( 1 E 2 E 3 E ) = 0.2 + P( 3 E ⏐ 2 E )P( 2 E ) – P( 1 E )P( 2 E 3 E ) = 0.2 + 0.4x0.3 – 0.2x(0.4x0.3) = 0.296 (c) P(casting ⏐concrete production not feasible at site) = P((E2 ∪ E3) E1⏐ 2 E ) = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 3 2 2 2 1 2 3 2 2 3 1 2 P E P E P E E P E P E P E E E P E P E E E E = = ∪ = 0.8 x 0.6 =0.48 2.9 E1, E2, E3 denote events tractor no. 1, 2, 3 are in good condition respectively (a) A = only tractor no. 1 is in good condition = E1 2 E 3 E B = exactly one tractor is in good condition = E1 2 E 3 E ∪ 1 E E2 3 E ∪ 1 E 2 E E3 C = at least one tractor is in good condition = E1 ∪ E2 ∪ E3 (b) Given P(E1) = P(E2) = P(E3) = 0.6 P( 2 E ⏐ 1 E ) = 0.6 or P( 2 E ⏐ 3 E ) = 0.6 P( 3 E ⏐ 1 E 2 E ) = 0.8 or P( 1 E ⏐ 2 E 3 E ) = 0.8 P(A) = P(E1 2 E 3 E ) = P(E1⏐ 2 E 3 E ) P( 2 E 3 E ) = [1 – P( 1 E ⏐ 2 E 3 E )] P( 2 E ⏐ 3 E ) P( 3 E ) = (1 - 0.8)(0.6)(0.4) = 0.048 Since E1 2 E 3 E , 1 E E2 3 E and 1 E 2 E E3 are mutually exclusive; also the probability of each of these three events is the same, P(B) = 3 x P(E1 2 E 3 E ) = 3 x 0.048 = 0.144 P(C) = P(E1 ∪ E2 ∪ E3) = 1 – P( 1 2 3 E ∪ E ∪ E ) = 1 – P( 1 E 2 E 3 E ) = 1 – P( 1 E ⏐ 2 E 3 E ) P( 2 E ⏐ 3 E ) P( 3 E ) = 1 – 0.8x0.6x0.4 = 0.808 2.10 (a) The event “both subcontractors will be available” = AB, hence since P(A∪B) = P(A) + P(B) - P(AB) ⇒ P(AB) = P(A) + P(B) - P(A∪B) = 0.6 + 0.8 - 0.9 = 0.5 (b) P(B is available⏐A is not available) = P(B⏐ A ) = P BA P A ( ) ( ) while it is clear from the following Venn diagram that P(BA) = P(B) - P(AB). A A AB B Hence P BA P A ( ( ) ) = P B P AB P A ( ) ( ) ( ) − 1− = (0.8 - 0.5)/(1 - 0.6) = 0.3/0.4 = 0.75 (c) (i) If A and B are s.i., we must have P(B⏐A) = P(B) = 0.8. However, using Bayes’ rule, P(B⏐A) = P(AB)/P(A) = 0.5/0.6 = 0.8333 So A and B are not s.i. (A’s being available boosts the chances that B will be available) (ii) From (a), P(AB) is nonzero, hence AB ≠ ∅, i.e. A and B are not m.e. (iii) Given: P(A∪B) = 0.9 ⇒ A∪B does not generate the whole sample space (otherwise the probability would be 1), i.e. A and B are not collectively exhaustive. 2.11 (a) Let L, SA, SB denote the respective events “leakage at site”, “seam of sand from X to A”, “seam of sand from X to B”. Given probabilities: P(L) = 0.01, P(SA) = 0.02, P(SB) = 0.03, P(SB⏐SA) = 0.2, Also given: independence between leakage and seams of sand, i.e. P(SB⏐L) = P(SB), P(L⏐SB) = P(L), P(SA⏐L) = P(SA), P(L ⏐SA) = P(L) The event “water in town A will be contaminated” = L∩SA, whose probability is P(L∩SA) = P(L⏐ SA) P(SA) = P(L)P(SA) = 0.01×0.02 = 0.0002. (b) The desired event is (L SA) ∪ (L SB), whose probability is P(L SA) + P(L SB) - P(L SA L SB) = P(L)P(SA) + P(L)P(SB) - P(L SA SB) = P(L) [P(SA) + P(SB) - P(SA SB)] = P(L) [P(SA) + P(SB) - P(SB⏐SA) P(SA)] = 0.01 (0.02 + 0.03 - 0.2×0.02) = 0.00046 2.12 Let A,B,C denote the respective events that the named towns are flooded. Given probabilities: P(A) = 0.2, P(B) = 0.3, P(C) = 0.1, P(B | C) = 0.6, P(A | BC) = 0.8, P( AB | C ) = 0.9, where an overbar denotes compliment of an event. (a) P(disaster year) = P(ABC) = P(A | BC)P(BC) = P(A | BC)P(B | C)P(C) = 0.8×0.6×0.1 = 0.048 (b) P(C | B) = P(BC) / P(B) = P(B | C)P(C) / P(B) = 0.6×0.1÷0.3 = 0.2 (c) The event of interest is A∪B∪C. Since this is the union of many items, we can work with its compliment instead, allowing us to apply De Morgan’s rule and rewrite as P(A∪B∪C) = 1 – P( A∪B∪C ) = 1 – P( ABC ) by De Morgan’s rule, = 1 – P( AB | C )P(C ) = 1 – 0.9×(1 – 0.1) = 1 – 0.81 = 0.19 2.13 (a) C = [(L∪M) G]] ∪[(LM)G ] (b) Since (L∪M)G is contained in G, it is mutually exclusive to (LM) G which is contained inG . Hence P(C) is simply the sum of two terms, P(C) = P[(L∪M) G]] + P[(L∩M) G ] = P(LG∪MG) + P(L)P(MG ) = P(LG) + P(MG) – P(LMG) + P(L)P(M|G )P(G ) = P(L)P(G) + P(M|G)P(G) – P(L)P(M|G)P(G)+ P(L)P(M|G )P(G ) = 0.7×0.6 + 1×0.6 – 0.7×1×0.6 + 0.7×0.5×0.4 = 0.74 (c) P( L|C) = P( LC)/P(C) = P( L{[(L∪M) G]} ∪[(LM) G ])) / P(C) = P[ L (L∪M) G]] / P(C) since ( LL)MG is an impossible event = P( LMG) / P(C) = P( L)P(MG) / P(C) = P( L)P(M|G)P(G) / P(C) = 0.3×1×0.6 / 0.74 ≅ 0.243 2.14 (a) Note that the question assumes an accident either occurs or does not occur at a given crossing each year, i.e. no more than one accident per year. There were a total of 30 + 20 + 60 + 20 = 130 accidents in 10 years, hence the yearly average is 130 / 10 = 13 accidents among 1000 crossings, hence the yearly probability of accident occurring at a given crossing is 1000 13 = 0.013 (probability per year) (b) Examining the data across the “Day” row, we see that the relative likelihood of R compared to S is 30:60, hence P(S | D) = 60/90 = 2/3 (c) Let F denote “fatal accident”. We have 50% of (30 + 20) = 0.5×50 = 25 fatal “run into train” accidents, and 80% of (60 + 20) = 0.8×80 = 64 fatal “struck by train” accidents, hence the total is P(F) = (25 + 64) / 130 ≅ 0.685 (d) (i) D and R are not mutually exclusive; they can occur together (there were 30 run-into-train accidents happened in daytime); (ii) If D and R are, we must have P(R | D) = P(R), but here P(R | D) = 30 / 90 = 1/3, while P(R) = (30 + 20) / 130 = 5/13, so D and R are not statistically independent. 2.15 F = fuel cell technology successfully marketable S = solar power technology successfully marketable F and S are statistically independent Given P(F) = 0.7; P(S) = 0.85 (i) P(energy supplied) = P(F∪S) = P(F) + P(S) – P(F)P(S) = 0.7 + 0.85 – 0.7x0.85 = 0.955 (ii) P(only one source of energy available) = P(F S ∪ F S) = P(F S ) + P( F S) = (0.7)(1-0.85) + (1-0.7)(0.85) = 0.36 2.16 a. E1 = Monday is a rainy day E2 = Tuesday is a rainy day E3 = Wednesday is a rainy day Given P(E1) = P(E2) = P(E3) = 0.3 P(E2⏐ E1) = P(E3⏐ E2) = 0.5 P(E3⏐ E1E2) = 0.2 b. P(E1E2) = P(E2⏐ E1) P(E1) = 0.5x0.3 = 0.15 c. P(E1E2 3 E ) = P( 3 E ⏐E1E2) P(E2⏐ E1) P(E1) = (1-0.2)(0.5)(0.3) = 0.12 d. P(at least one rainy day) = P(E1 ∪ E2 ∪ E3) = P(E1) + P(E2) + P(E3) - P(E1E2) - P(E2E3) - P(E1E3) + P(E1E2E3) = 0.3 + 0.3 + 0.3 - 0.3x0.5 - 0.3x0.5 – 0.1125 + 0.3x0.5x0.2 =0.52 Where P(E1E3)=P(E3|E1)P(E1)=0.375x0.3=0.1125 P(E3|E1)= P(E3|E2E1)P(E2|E1)+ P(E3| 2 E E1)P( 2 E|E1) =0.15x0.15+0.3x0.5 =0.375 2.17 Let A, B, C denote events parking lot A, B and C are available on a week day morning respectively Given: P(A) = 0.2; P(B) = 0.15; P(C) = 0.8 P(B⏐ A ) = 0.5; P(C⏐ A B ) = 0.4 (a) P(no free parking) = P( A B ) = P( A ) x P( B ⏐ A ) = 0.8 x 0.5 = 0.4 (b) P(able to park) = 1 – P(not able to park) = 1 – P( A B C ) = 1 – P( A B )xP(C ⏐ A B ) = 1 – 0.4x(1-0.4) = 1 – 0.24 =0.76 (c) P(free parking⏐able to park) P(A∪B⏐A∪B∪C) = ( ) [( )( )] P A B C P A B A B C ∪ ∪ ∪ ∪ ∪ = ( ) ( ) P A B C P A B ∪ ∪ ∪ = 0.76 1 − P(AB) = 0.76 1 − 0.4 = 0.789 2.18 C = Collapse of superstructure E = Excessive settlement Given: P(E) = 0.1; P(C) = 0.05 P(C⏐E) = 0.2 (a) P(Failure) = P(C ∪ E) = P(C) + P(E) – P(C⏐E)P(E) = 0.05 + 0.1 – 0.2x0.1 = 0.13 (b) P( EC⏐E ∪ C) = ( ) (( ) ( )) P E C P EC E C ∪ ∩ ∪ = P(EC) / 0.13 = 0.2x0.1 / 0.13 = 0.154 (c) P(EC ∪ E C) = P(EC ) + P( E C) = P(E∪C) – P(EC) = 0.13 – 0.02 =0.11 2.19 M = failure of master cylinder W = failure of wheel cylinders B = failure of brake pads Given: P(M) = 0.02; P(W) = 0.05; P(B) = 0.5 P(MW) = 0.01 B is statistically independent of M or W (a) P(WM B ) = P(WM )P( B ) = [P(W) – P(MW)] [1 – P(B)] = (0.05 – 0.01)(1 – 0.5) = 0.02 (b) P(system failure) = 1 – P(no component failure) = 1- P(M W B ) = 1- P(M W ) P( B ) = 1- {1 - P(M ∪ W)}P( B ) = 1- {1-[P(M)+P(W)-P(MW)]}P( B ) = {1 - [0.02+0.05-0.01]}(0.5) = 0.53 (c) P(only one component failed) = P(W M B) + P(W MB ) + P(WM B ) P(WM B ) = 0.02 from part (a) also P(W MB ) = P(W M) x P( B ) = [P(M)-P(MW)]P( B ) = (0.02 – 0.01)x0.5 = 0.005 also P(W M B) = P(W M ) x P( B ) = [1-P(M)-P(W)+P(MW)]P(B) = (1-0.02-0.005+0.01)x0.5 = 0.47 Since the three events W M B, W MB and WM B are all within the event of system failure P(only one component failure⏐system failure) = (0.47+0.005+0.02) / 0.53 = 0.934 2.20 E1 = Excessive snowfall in first winter E2 = Excessive snowfall in second winter E3 = Excessive snowfall in third winter (a) P(E1) = P(E2) = P(E3) = 0.1 P(E2 ⏐E1) = 0.4 = P(E3 ⏐E2) P(E3 ⏐E1 E2) = 0.2 (b) P(E1 ∪ E2) = P(E1)+P(E2)-P(E1 E2) = 0.1 + 0.1 – 0.4x0.1 = 0.16 (c) P(E1E2E3) = P(E1) P(E2 ⏐E1) P(E3 ⏐E1 E2) = 0.1x0.4x0.2 =0.008 (d) P( 2 E ⏐ 1 E ) = ? Since P( 1 E ∪ 2 E ) = P( 1 E )+P( 2 E )-P( 1 E 2 E ) = P( 1 E )+P( 2 E )-P( 2 E ⏐ 1 E )P( 1 E ) = 0.9+0.9-P( 2 E ⏐ 1 E ) 0.9 and from given relationship, P( 1 E ∪ 2 E ) = 1- P(E1E2) = 1- P(E1) P(E2 ⏐E1) = 1-0.1x0.4 = 0.96 Therefore, 1.8-0.9 P( 2 E ⏐ 1 E ) = 0.96 and P( 2 E ⏐ 1 E ) = 0.933 2.21 H1, H2, H3 denote first, second and third summer is hot respectively Given: P(H1) = P(H2) = P(H3) = 0.2 P(H2 ⏐H1) = P(H3 ⏐E2) = 0.4 (a) P(H1H2H3) = P(H1) P(H2 ⏐H1) P(H3 ⏐E2) = 0.2x0.4x0.4 = 0.032 (b) P( 2 H ⏐ 1 H ) = ? Using the hint given in part (d) of P2.20, P( 1 H ) + P( 2 H ) - P( 2 H ⏐ 1 H )P( 1 H ) = P( 1 H ∪ 2 H ) = 1 – P(H1H2) = 1 - P(H2 ⏐H1) P(H1) we have, 0.8 + 0.8 - P( 2 H ⏐ 1 H )(0.8) = 1 – 0.4x0.2 therefore, P( 2 H ⏐ 1 H ) = 0.85 (c) P(at least 1 hot summer) = 1 – P(no hot summers) = 1 – P( 1 H 2 H 3 H ) = 1 – P( 1 H )P( 2 H ⏐ 1 H ) P( 3 H ⏐ 2 H ) = 1 – 0.8x0.85x0.85 = 0.422 2.22 A = Shut down of Plant A B = Shut down of Plant B C = Shut down of Plant C Given: P(A) = 0.05, P(B) = 0.05, P(C) = 0.1 P(B⏐A) = 0.5 = P(A⏐B) C is statistically independent of A and B (a) P(complete backout⏐A) = P(BC⏐A) = P(B⏐A)P(C) = 0.5x0.1 = 0.05 (b) P(no power) = P(ABC) = P(AB)P(C) = P(B⏐A)P(A)P(C) = 0.5x0.05x0.1 =0.00025 (c) P(less than or equal to 100 MW capacity) = P (at most two plants operating) = 1 – P(all plants operating) = 1 – P( A B C ) = 1 – P(C )P( A B ) = 1 – P(C ) [1-{P(A)+P(B)- P(B⏐A)P(A)}] = 1 – 0.9[1-{0.05+0.05-0.5x0.05}] = 0.1675 2.23 P(Damage) = P(D) = 0.02 Assume damages between earth quakes are statistically independent (a) P(no damage in all three earthquakes) = P3(D) = 0.023 = 8x10-6 (b) P( 1 D D2) = P( 1 D )P(D2) = 0.98x0.02 = 0.0196 2.24 (a) Let A, B denote the event of the respective engineers spotting the error. Let E denote the event that the error is spotted, P(E) = P(A∪B) = P(A) + P(B) – P(AB) = P(A) + P(B) – P(A)P(B) = 0.8 + 0.9 – 0.8×0.9 = 0.98 (b) “Spotted by A alone” implies that B failed to spot it, hence the required probability is P(AB’|E) = P(AB’∩E)/P(E) = P(E|AB’)P(AB’) / P(E) = 1×P(A)P(B’) / P(E) = 0.8×0.1 / 0.98 ≅ 0.082 (c) With these 3 engineers checking it, the probability of not finding the error is P(C1’C2’C3’) = P(C1’)P(C2’)P(C3’) = (1 – 0.75)3 , hence P(error spotted) = 1 – (1 – 0.75)3 ≅ 0.984, which is higher than the 0.98 in (a), so the team of 3 is better. (d) The probability that the first error is detected (event D1) has been calculated in (a) to be 0.98. However, since statistical independence is given, detection of the second error (event D2) still has the same probability. Hence P(D1 D2) = P(D1)P(D2) = 0.982 ≅ 0.960 2.25 L = Failure of lattice structure A = Failure of anchorage Given: P(A) = 0.006 P(L⏐A) = 0.4 P(A⏐L) = 0.3 Hence, P(L) = P(L⏐A)P(A) / P(A⏐L) = 0.008 (a) P(antenna disk damage) = P(A∪L) = P(A) + P(L) – P(A)P(L⏐A) = 0.4 + 0.008 - 0.006x0.4 = 0.406 (b) P(only one of the two potential failure modes) = P(A L ) + P( A L) = P( L ⏐A)P(A) + P( A ⏐L)P(L)(0.008) = (1-0.4)(0.006) + (1-0.3)(0.008) = 0.0036 + 0.0056 =0.0092 (c) 0.0148 0.406 0.006 ( ) ( ) ( ) ( ) ( ( )) = = ∪ = ∪ ∪ = P A L P A P A L P AD P A A L 2.26 P(F) = 0.01 P(A⏐ F ) = 0.1 P(A⏐F) = 1 (a) FA, F A , F A, F A are set of mutually exclusive and collectively exhaustive events (b) P(FA) = P(A⏐F)P(F) = 1x0.01 = 0.01 P(F A ) = P( A ⏐F)P(F) = 0x0.01 = 0 P( F A) = P(A⏐ F )P( F ) = 0.1x0.99 = 0.099 P( F A ) = P( A ⏐ F )P( F ) = 0.9x0.99 = 0.891 (c) P(A) = P(A⏐F)P(F) + P(A⏐ F )P( F ) = 1x0.01 + 0.1x0.99 = 0.109 (d) P(F⏐A) = 0.0917 0.109 0.01 ( ) ( ) ( ) = = P A P A F P F 2.27 Given: P(D) = 0.001 P(T⏐D) = 0.85; P(T⏐ D) = 0.02 (a) DT, DT , DT, D T (b) P(DT) = P(T⏐D)P(D) = 0.85x0.001 = 0.00085 P(DT ) = P(T ⏐D)P(D) = 0.15x0.001 = 0.00015 P( DT) = P(T⏐ D)P( D) = 0.02x0.999 = 0.01998 P( D T ) = P(T ⏐ D)P( D) = 0.98x0.999 = 0.979 (c) 0.0408 0.00085 0.01998 0.00085 ( ) ( ) ( ) ( ) ( ) ( ) ( ) = + = + = P T D P D P T D P D P T D P D P DT 2.28 Given: P(RA) = P(GA) = 0.5 P(RB) = P(GB) = 0.05 P(GB⏐GA) = 0.8 P(GLT) = 0.2 Signal at C is statistically independent of those at A or B. (a) E1 = RA∪ RB E2 = GLT E3 = RA GB ∪ GA RB (b) P(stopped at least once from M to Q) = 1 – P(no stopping at all from M to Q) = 1 – P(GA GB GLT) = 1 – P(GLT)P(GA)P(GB ⏐GA) = 1 – 0.2x0.5x0.8 = 0.92 (c) P(stopped at most once from M to N) = 1 – P(stopped at both A and B) = 1 – P(RARB) = 1 - P(RA) P(RB⏐RA) From the hint given in P2.20, it can be shown that [1 0.8 0.5] 1 0.2 0.5 [1 ( ) ( )] 1 1 ( ) ( ) = 1 − − = − × − = B A A A B A P G G P G P R P R R Hence P(stopped at most once from M to N) = 1 – 0.5x0.2 = 0.9 2.29 (a) P(WSC) = P(W)P(SC) = P(W)P(S|C)P(C) = 0.1×0.3×0.05 = 0.0015 (b) P(W∪C) = P(W) + P(C) – P(WC) = P(W) + P(C) – P(W)P(C) = 0.1 + 0.05 – 0.1×0.05 = 0.145 (c) P(W∪C| S ) = P[ S (W∪C)] / P(S) = P(SW∪SC) / P(S) = [ P(SW) + P(SC) – P(SWC) ] / P(S) = [ P( S )P(W) + P( S |C)P(C) – P(W)P(SC) ] / P(S) = [0.8×0.1 + (1 – 0.3)×0.05 – 0.1×(1 – 0.3)×0.05] / (1 – 0.2) ≅ 0.139 (d) P(nice winter day) = P( S W C) = P(W )P( S C) = P(W )P( S | C)P(C) = 0.9x0.7x0.05 = 0.0315 (e) P(U) = P(U |CW)P(CW) + P(U |C W)P(C W) + P(U|CW ) P(C|W ) + P(U||C W ) P(C W ) = 1xP(W)P(C) + 0.5xP(W)P(C ) + 0.5xP(C)P(W ) + 0xP(C )P(W ) = 1x0.1x0.05 + 0.5x0.1x0.95 + 0.5x0.05x0.9 = 0.075 2.30 (a) Let L and B denote the respective events of lead and bacteria contamination, and C denote water contamination. P(C) = P(L∪B) = P(L) + P(B) – P(LB) = P(L) + P(B) – P(L)P(B)∵L and B are independent events = 0.04 + 0.02 – 0.04×0.02 = 0.0592 (b) P(L B| C) = P(CL B) / P(C) = P(C | L B)P(L B) / P(C), but P(C | L B) = 1 (∵lead alone will contaminate for sure) and P(L B) = P(L)P( B) (∵statistical independence), hence the probability = 1×P(L)P(B) / P(C) = 1×0.04×(1 – 0.02) / 0.0592 ≅ 0.662 2.31 C Water level W (x103 lb) F (x103 lb) Probability 0..5x0.2=0.1 0..5x0.4=0.2 0..5x0.4=0.2 0..5x0.2=0.1 0..5x0.4=0.2 0..5x0.4=0.2 (a) Sample space of F = {10, 20, 21, 32, 42, 64} (b) Let H be the horizontal force in 103 lb P(sliding) = P(HF) = P(F15) = 0.1 (c) P(sliding) = P(F15) P(H=15) + P(F20) P(H=20) = 0.1x0.5 + 0.1x0.1 = 0.06 2.32 Given: P(W) = 0.9 P(H) = 0.3 P(E) = 0.2 P(W⏐H) = 0.6 E is statistically independent of W or H (a) I = E(W ∪H) II = E W H (b) P[E(W ∪H)] = P(E) P(W ∪H) = P(E) x [P(W )+P(H)-P(W ⏐H)P(H)] = 0.2x(0.1+0.3-0.4x0.3) = 0.056 P( E W H) = P( E ) P(W ⏐H) P(H) = 0.8x0.4x0.3 = 0.096 (c1) Since P(W⏐H) = 0.6 ≠ 0, W and H are not mutually exclusive. Since P(W⏐H) = 0.6 ≠ 0.9 = P(W), Wand H are not statistically independent (c2) I II = E(W ∪H) ( E W H) = (E E )(W ∪H)( W H) Since E E is an empty set, I and II are mutually exclusive. From the above Venn Diagram, the union of I and II does not make up the entire sample space. Hence, I and II are not collectively exhaustive. (d) Since I and II are mutually exclusive, P(leakage) = P(I) + P(II) = 0.056 + 0.096 = 0.152 2.33 Given: P(E) = 0.15 P(G) = 0.1 P(O) = 0.2 P(E⏐O) = 2x0.15 = 0.3 G is statistically independent of E or O (a) P(shortage of all three sources) = P(EGO) = P(G) P(E⏐O) P(O) = 0.1x0.3x0.2 = 0.006 (b) P(G ∪ E) = 1-P(G E ) = 1- P(G ) P( E ) = 1-0.9x0.85 = 0.235 (c) P(GO⏐E) = P(GOE)/P(E) = 0.006/0.15 = 0.04 (d) P(at least 2 sources will be in short supply) = P(EG∪GO∪EO) = P(EG)+P(GO)+P(EO)-P(EGO)-P(EGO)-P(EGO)+P(EGO) = P(G)P(E)+P(G)P(O)+P(E⏐O)P(O)-2P(EGO) = 0.1x0.15 + 0.1x0.2 + 0.3x0.2 - 2x0.006 = 0.083 2.34 A, B, C denote failure of component A, B, C respectively N and H denote normal and ultra-high altitude respectively Given: P(A⏐N) = 0.05, P(B⏐N) = 0.03, P(C⏐N) = 0.02 P(A⏐H) = 0.07, P(B⏐H) = 0.08, P(C⏐H) = 0.03 P(N) = 0.6, P(H) = 0.4 P(B⏐A) = 2 x P(B) C is statistically independent of A or B P(system failure) = P(S) = P(S⏐N)P(N) + P(S⏐H)P(H) P(S⏐N) = P(A∪B∪C⏐N) = P(A⏐N) + P(B⏐N) + P(C⏐N) - P(AB⏐N) - P(BC⏐N) - P(AC⏐N) + P(ABC⏐N) = 0.05 + 0.03 + 0.02 – 2x0.03x0.03 – 0.03x0.02 – 0.05x0.02 + 0.02x2x0.03x0.03 = 0. Similarly, P(S⏐H) = P(A⏐H) + P(B⏐H) + P(C⏐H) - P(AB⏐H) - P(BC⏐H) - P(AC⏐H) + P(ABC⏐H) = 0.07 + 0.08 + 0.03 – 2x0.08x0.08 – 0.08x0.03 – 0.07x0.03 + 0.03x2x0.08x0.08 = 0.16308 Hence P(system failure) = 0.96636x0.6 + 0.16308x0.4 ≅ 0.123 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ( )] P S P B N P N P B H P H P S P B P S P B S P B A B C + = = ∪ ∪ = = 0.172 0.123 0.03 0.6 0.08 0.4 = × + × 2.35 Given: P(C) = 0.5 P(W) = 0.3 P(W⏐C) = 0.4 U = C ∪ W (a) Since P(W⏐C) = 0.4 ≠ 0.3=P(W) W and C are not statistically independent (b) P(U) = P(C ∪ W) = P(C) + P(W) – P(W⏐C) P(C) = 0.5 + 0.3 - 0.4x0.5 = 0.6 (c) P(CW ) = P(W ⏐C) P(C) = (1-0.4)x0.5 = 0.3 (d) 0.333 0.6 0.4 0.5 0.6 ( ) ( ) ( ) ( ) ( ) ( ) [ ( )] = × = = ∪ = ∪ ∪ ∪ = P W C P C P C W P CW P C W P CW C W P CW C W 2.36 A, B, C denote route A, B, C are congested respectively Given: P(A) = 0.6, P(B) = 0.6, P(C) = 0.4 P(A⏐B) = P(B⏐A) = 0.85 C is statistically independent of A or B P(L⏐ABC) = 0.9 P(L⏐ ABC ) = 0.3 (a) P(A B C )+ P( A BC )+ P( A B C) = P( B ⏐A)P(A)P(C )+P( A ⏐B)P(B)P(C )+P( A B )P(C) but P( A B ) = 1-P(A ∪B) = 1-P(A)-P(B)+ P(B⏐A)P(A) = 1 - 0.6 - 0.6 + 0.85x0.6 = 0.31 Hence, P(exactly one route congested) = 0.15x0.6x0.6 + 0.15x0.6x0.6 +0.31x0.4 = 0.232 (b) P(Late) = P(L⏐ABC)P(ABC)+ P(L⏐ ABC ) P( ABC ) But P(ABC) = P(AB)P(C) = P(B⏐A)P(A)P(C) = 0.85x0.6x0.4 = 0.204 Hence, P(L) = 0.9x0.204 + 0.3x(1-0.204) = 0.422 (c) If C is congested, P(ABC) = 0.85x0.6x1 = 0.51 P(L) = 0.9x0.51 + 0.3x0.49 = 0.606 2.37 (a) Let subscripts 1 and 2 denote “after first earthquake” and “after second earthquake”. Note that (i) occurrence of H1 makes H2 a certain event; (ii) H, L and N are mutually exclusive events and their union gives the whole sample space. P(heavy damage after two quakes) = P(H2 H1) + P(H2 L1) + P(H2 N1) = P(H2 | H1) P(H1) + P(H2 | L1)P(L1)+ P(H2 | N1)P(N1) = 1×0.05 + 0.5×0.2 + 0.05×(1 – 0.2 – 0.05) ≅ 0.188 (b) The required probability is [P(H2 L1) + P(H2 N1)] / P(heavy damage after two quakes) = (0.5×0.2 + 0.05×0.75) / (1×0.05 + 0.5×0.2 + 0.05×0.75) ≅ 0.733 (c) In this case, one always starts from an undamaged state, the probability of not getting heavy damage at any stage is simply (1 – 0.05) = 0.95 (not influenced by previous condition). Hence P(any heavy damage after 3 quakes) = 1 – P(no heavy damage in each of 3 quakes) = 1 – 0.953 ≅ 0.143. Alternatively, one could explicitly sum the probabilities of the three mutually exclusive events, P(H) = P(H1) + P(H2 H1’) + P(H3 H2’ H1’) = 0.05 + 0.05×0.95 + 0.05×0.95×0.95 ≅ 0.143 2.38 Given: P(L) = 0.6, P(A) = 0.3, P(H) = 0.1 S denote supply adequate, i.e. no shortage P(S⏐L) = 1, P(S⏐A) = 0.9, P(S⏐H) = 0.5 (a) P(S) = P(S⏐L)P(L)+P(S⏐A)P(A)+P(S⏐H)P(H) = 1x0.6 + 0.9x0.3 + 0.5x0.1 = 0.92 (b) 0.375 1 0.92 0.1 0.3 ( ) ( ) ( ) ( ) = − × = = P S P S A P A P A S (c) P(shortage in at least one or of the next two months) = 1 – P(no shortage in next two months) = 1 – 0.92x0.92 = 0.154 2.39 (a) P(shipment accepted on a given day)