de Morgan’s rule E 1 = pipe 1 breaks E 2 = pipe 2 breaks 1 2 Water Supply E = failure in water supply = E 1 ∪ E 2 no failure in water supply = Ē = E 1 ∪ E 2

Alternatively pipe 1 does not break pipe 2 does not break No failure in water supply

Event of “no failure” Extension to n events

de Morgan #2

Basis of Probability Estimation a) Subjective assumption e.g. P(Q) = 1/2 b) Relative frequency e.g. P(Q)=502/1000 c) Bayesian (a)+(b) judgment + limited observation

Axioms of probability theory if A and B are m.e sample space

S 1 Example:

Probability of Union in general E1E1 E2E2

P(E 1 ∪ E 2 ∪ E 3 ) = P(E 1 )+P(E 2 ∪ E 3 ) – P(E 1 (E 2 ∪ E 3 )) P(E 2 ) + P(E 3 ) – P(E 2 E 3 ) P(E 1 E 2 ∪ E 1 E 3 ) P(E 1 E 2 ) + P(E 1 E 3 ) – P(E 1 E 2 E 1 E 3 ) – P(E 1 E 2 E 1 E 3 )

=P(E 1 ) + P(E 2 ) + P(E 3 ) – P(E 1 E 2 ) – P(E 1 E 3 ) – P (E 2 E 3 ) + P(E 1 E 2 E 3 )  7 Terms P(E 1 ∪ E 2 ∪ E 3 ∪ E 4 )

Using de Morgan’s rule P (intersection) conditional probability

Conditional probability highway accident 100 Km XX B A P(A) = 30/100 ; P(B) = 40/100 ; P(AB) = 10/100

Define conditional probability P(A|B) = P (accident in region A given accident reported region B) = 10/40 = 10/40 Similarly P(B|A) = 10/30

Observe Hence or

New sample space for given B 40 Km 