2.5. (a) Locations Locations Load at B Load at C M A Probability E 1 E 2 E 3
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1 2.5 (a) Locations Locations of W 1 of W 2 Load at B Load at C M A Probability E 1 E 2 E x0.2=0.03 B , X C , X X B , X X B B , X X X B C , X C , X C B , X X C C , X (b) E 1 and E 2 are not mutually exclusive because these two events can occur together, for example, when the weight W 2 is applied at C, M A is 10,000 ft-lb; hence both E 1 and E 2 will occur. (c) The probability of each possible value of M A is tabulated in the last column of the table above. (d) E 1 ) = M A > 5,000) = = E 2 ) = 1,000 MA 12,000) = = E 3 ) = = E 1 E 2 ) = 5,000 < M A 12,000) = = 0.33 E 1 E 2 ) = = 0.97 E 2 ) = = PE ( 2) 1 PE ( )
2 2.10 (a) The event both subcontractors will be available = AB, hence (b) B is available A is not available) = B A ) since A B) = A) + B) - AB) AB) = A) + B) - A B) = = 0.5 = PBA ( ) PA ( ) while it is clear from the following Venn diagram that PBA ( ) = B) - AB). A AB B A Hence PBA ( ) PA ( ) = B) AB) 1 PA ( ) = ( )/(1-0.6) = 0.3/0.4 = 0.75 (c) (i) If A and B are s.i., we must have B A) = B) = 0.8. However, using Bayes rule, B A) = AB)/A) = 0.5/0.6 = So A and B are not s.i. (A s being available boosts the chances that B will be available) (ii) From (a), AB) is nonzero, hence AB, i.e. A and B are not m.e. (iii) Given: 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.
3 2.17 Let A, B, C denote events parking lot A, B and C are available on a week day morning respectively Given: A) = 0.2; B) = 0.15; C) = 0.8 B A ) = 0.5; C A B ) = 0.4 (a) no free parking) = A B ) = A ) x B A ) = 0.8 x 0.5 = 0.4 (b) able to park) = 1 not able to park) = 1 A B C ) = 1 A B )xc A B ) = 1 0.4x(1-0.4) = =0.76 (c) free parking able to park) A B A B C) P[( A B)( A B C)] = A B C) A B) = A B C) = = 1 AB) =
4 2.1 T AB T BC C AB C BC T AC C AC (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 T AC and C AC = {(8, 1500), (9, 1500), (10, 1500), (11, 850), (12, 850), (13, 850), (14, 850)}
5 2.26 F) = 0.01 A F ) = 0.1 A F) = 1 (a) FA, F A, F A, F A are set of mutually exclusive and collectively exhaustive events (b) FA) = A F)F) = 1x0.01 = 0.01 F A ) = A F)F) = 0x0.01 = 0 F A) = A F ) F ) = 0.1x0.99 = F A ) = A F ) F ) = 0.9x0.99 = (c) A) = A F)F) + A F ) F ) = 1x x0.99 = A F) F) 0.01 (d) F A) = A) 0.109
6 2.29 (a) WSC) = W)SC) = W)S C)C) = = (b) W C) = W) + C) WC) = W) + C) W)C) = = (c) W C S ) = P[ S (W C)] / S ) = S W S C) / S ) = [ S W) + S C) S WC) ] / S ) = [ S )W) + S C)C) W) S C) ] / S ) = [ (1 0.3) (1 0.3) 0.05] / (1 0.2) (d) nice winter day) = S W C) = W ) S C) = W ) S C)C) = 0.9x0.7x0.05 = (e) U) = U CW)CW) + U C W)C W) + U CW ) C W ) + U C W ) C W ) = 1xW)C) + 0.5xW)C ) + 0.5xC)W ) + 0xC )W ) = 1x0.1x x0.1x x0.05x0.9 = 0.075
7 2.50 Let H and S denote Hard and Soft ground, respectively, and let L denote a successful landing. Given probabilities: L H) = 0.9; L S) = 0.5; H) = 3S), but since ground is either hard or soft H) = 0.75, S) = 0.25 (a) Using theorem of total probability, L) = L H)H) + L S)S) = = 0.8 (b) Let E denote penetration. Given: E S) = 0.9; E H) = 0.2 (i) The updated probability of hard ground, P (H) H E) = E H)H) / [E H)H) + E S)S)] by Bayes theorem = / ( ) = 0.4 (ii) Using the updated probabilities P (H) = 0.4, P (S) = = 0.6, the updated probability of a successful landing now becomes P (L) = L H)P (H) + L S)P (S) = = 0.66
8 2.56 (a) Let A = presence of anomaly and D = detection of anomaly by geophysical techniques. We are given D A) = 0.5 D A) = = 0.5, and also D A ) = 0 D A ) = 1. We need A D ) = D A)A) / D ) by Bayes theorem, in which D ) = D A)A) + D A )A ) = = 0.85, hence A D ) = / (b) (i) With the updated anomaly probability P*(A) = 0.15/0.85 (thus P*(A ) = 0.7/0.85), and also a better detection probability D A) = 0.8 (thus D A) = 0.2), the probability of having no anomaly, given no detection, is now P**(A ) = A D ) = D A )P*(A ) / D ) = 1 (0.7/0.85) / [1 (0.7/0.85) (0.15/0.85)] (ii) Total probability of a safe foundation = safe A)P**(A) + safe A )P**(A ) = 0.80 ( ) (iii) A failure probability of p means that out of a large number (N) of similar systems, p N of them are expected to fail. The total failure cost would be p N $ , which when divided into N gives the average or expected cost per system = p $ Hence we have the formula Expected loss = (probability of failure) (failure loss) = ( ) $ $8300 If the system is known for sure to be anomaly free, however, it has only = chance of failure, in which case the expected loss would be $ = $100. Hence, ($ $100) = $8200 in expected loss is saved when the site can be verified to be anomaly free
9 2.32 Given: W) = 0.9 H) = 0.3 E) = 0.2 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)] = E) W H) = E) x [W )+H)-W H)H)] = 0.2x( x0.3) = E W H) = E ) W H) H) = 0.8x0.4x0.3 = (c1) Since W H) = 0.6 0, W and H are not mutually exclusive. Since W H) = = W), W and 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.
10 (d) Since I and II are mutually exclusive, leakage) = I) + II) = = 0.152
11 2.41 Let L, N, H denote the event of low, normal and high demand respectively; also O and G denote oil and gas supply is low respectively. (a) Given normal energy demand, probability of energy crisis = E N) = O G) = O)+G)-OG) = O)+G)-G O)O) = x0.2 = 0.5 (b) E) = E L) L) + E N) N) + E H) H) = OG)x x0.6 +1x0.1 = G O)O)x = 0.43
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