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Not all problems can be solved by analysis of data Set Theory Sample space : collection of all possibilities Sample point : each possibility Event : subset of sample space Probability Theory. Formal analysis of uncertainties and probability. Constructions job needs 3 bulldozers
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Not all problems can be solved by analysis of data Set Theory Sample space: collection of all possibilities Sample point: each possibility Event: subset of sample space Probability Theory Formal analysis of uncertainties and probability
Constructions job needs 3 bulldozers • Each bulldozer could be operative or not operative after 6 months • We are interested in the operative status of the 3 bulldozers
E – at least 2 bulldozers operative E1 – operator #1 no job
1,2 3,4,5 1 2 9 Travel Time Total travel time T: {4,5,6,7} Suppose 1,2 are equally likely Suppose 3,4,5 are equally likely P (T=5) =?
2 9 2 1 Assume statistical independence
P (1) = 2/3 P (2) = 1/3 P (3) = ¼ P (4) = ¼ P (5) = 1/2 P (T=5) = 2/12+1/12 = 1/4
1 Km Number of crack on a highway S = {0,1,2,……….∞} discrete
100 ton B A RA RB Continuous S E RA : 0-100 RA 0 90 100
Union: either E1 or E2 occur E1∪E2 Intersection: both E1 and E2 occurE1∩ E2 or E1 E2
E1 = road 1 closed E2 = road 2 closed 1 2 1 E3 = road 3 closed A C B 3 2 Examples B A No communication between A and B = E1E2 No communication between A and B = E3∪E1E2
E1 = river 1 floods E2 = river 2 floods 1 2 Town Flood in town = E1∪E2
E1 = 1 settles Ē1 = 1 does not settle E2 = 2 settles Ē2 = 2 does not settle 2 1 Example - pair of footings Settlement occurs = E1∪E2 Tilting occurs = E1Ē2∪Ē1E2
Mutually exclusive m.e. E1 and E2 are m.e. if occurrence of one precludes the occurrence of the other S E1 E2
Examples of m.e. events • Zero and 1 quakes in a year • E1 and Ē1 • E1 and E2 in the footing example?! • E1 Ē2 and Ē1E2 in the footing examples?!
Collectively exhaustive c.e. A set of events are c.e. if the union makes up the sample space
E 2.7 Contractors A and B bidding for jobs (1) bidding for different jobs (2) bidding for same job A B A B
A B (3) A, B are the only bidders (4) B is a subcontractor of A only – also there are other bidders beside A A B
Operating rules AB = BA A∪B = B∪A (A∪B)C = AC ∪ BC
B (A∪B)(B∪C) = AB∪AC∪BB∪BC = AB∪AC∪B = AB∪B∪AC = B∪AC B C B Algebra: (a+b) (b+c) = ab+ac+b2+bc ≠b+ac