MEGN 537 – Probabilistic Biomechanics Ch.2 – Mathematics of Probability

1. MEGN 537 – Probabilistic BiomechanicsCh.2 – Mathematics of Probability Anthony J Petrella, PhD

3. Probability Relations • Experimental outcomes can be represented by set theory relationships • Union: A1A3, elements belong to A1 or A3 or both • P(A1A3) = P(A1) + P(A3) - P(A1A3) = A1+A3-A2 • Intersection: A1A3, elements belong to A1 and A3 • P(A1A3) = P(A3|A1) * P(A1) = A2 (multiplication rule) • Complement: A’, elements that do not belong to A • P(A’) = 1 – P(A) S

4. Important Concepts (review) • Conditional Probability P(AB) = P(A|B) * P(B) = P(B|A) * P(A) • Statistical independence P(AB) = P(A) * P(B) = P(B) * P(A) • Mutual Exclusivity P(AB) = 0

5. Conditional Probability • The likelihood that event B will occur if event A has already occurred • P(AB) = P(B|A) * P(A) • P(B|A) = P(AB) / P(A) • Requires that P(A) ≠ 0 • Remember • P(AB) = P(A|B) * P(B) = P(B|A) * P(A) S A B

6. Theorem of Total Probability • For collectively exhaustive and mutually exclusive events (E1, E2, …, En): P(A) = P(A|E1)*P(E1)+P(A|E2)*P(E2) + … + P(A|En)*P(En) E3 S E4 E1 A En E2

7. Building on the idea of total probability, what is the probability that the event A was “caused” by the event Ek(inverse probability) If Ei are mutually exclusive events Bayes’ Theorem E3 S Theorem of Total Probability E4 E1 A En E2 • For a specific • event Ek: Conditional Probability For collectively exhaustive and mutually exclusive events

8. Example • Given a truss structure: • If the likelihood of failure of any member is 10-5, what is the likelihood of truss failure. • The truss is statically determinant – failure of one member implies failure of the truss. 1 4 3 2 5 w

9. Answer • Application of Total Probability • Let E = member failure F = truss failure P(Ei) = 10-5 P(F|Ei) = 1 • P(F) = P(F|E1)P(E1)+…P(F|E5)P(E5) = 1*(10-5)+…+1*(10-5) = 5*(1*(10-5)) = 0.00005

10. Answer • Application of DeMorgan’s Rule • Let E = member failure; F = truss failure P(Ei) = 10-5 P(Ei’) = 1 - 10-5 • P(F) = P(E1E2E3E4E5) = 1 - P((E1E2E3E4E5)’) = 1 - P(E1’E2’E3’E4’E5’) [statistically independent] = 1 - P(E1’) P(E2’) P(E3’) P(E4’) P(E5’) = 1 - (1 - 10-5)5 = 0.00005

11. Example • A firm that manufactures syringes tested 1000 samples and finds that 10% are defective. • The company concludes that a randomly selected shipment from the production process has a probability of 0.1 of being defective, P(D)=0.1. • The firm also finds that the inspectors typically had a 5% error rate on both defective and non-defective parts. If “A” is the event of accepting the syringe as good P(A|D)=0.05, P(A|D’)=0.95 • What is the probability of shipping a defective syringe at the current operating conditions? • What is the probability of shipping a non-defective syringe at the current operating conditions? • What is the probability of shipping a defective syringe if the company reduces its defect production rate to 0.01?

12. Answer: Using Bayes’ Theorem • P(D) = 0.1 • P(D) = 0.01 P(D|A) = P(A|D)*P(D) P(A|D)*P(D)+P(A|D’)*P(D’) P(D|A) = (0.05*0.1) = 0.0058 0.05*0.1+0.95*0.9 P(D’|A) = P(A|D’)*P(D’) = 0.9942 P(A|D)*P(D)+P(A|D’)*P(D’) P(D|A) = (0.05*0.01) = 0.000531 0.05*0.01+0.95*0.99