MEGN 537 – Probabilistic Biomechanics Ch.2 – Mathematics of Probability Anthony J Petrella, PhD.

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MEGN 537 – Probabilistic Biomechanics Ch.2 – Mathematics of Probability Anthony J Petrella, PhD

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

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

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 AB

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) A S E1 E2 E3 E4 En

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

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. w 1 2 3 4 5

Answer Application of Total Probability Let E = member failure F = truss failure P(E i ) = 10 -5 P(F|E i ) = 1 P(F) = P(F|E 1 )P(E 1 )+…P(F|E 5 )P(E 5 ) = 1*(10 -5 )+…+1*(10 -5 ) = 5*(1*(10 -5 )) = 0.00005

Answer Application of DeMorgan’s Rule Let E = member failure; F = truss failure P(E i ) = 10 -5 P(E i ’) = 1 - 10 -5 P(F)= P(E 1  E 2  E 3  E 4  E 5 ) = 1 - P((E 1  E 2  E 3  E 4  E 5 )’) = 1 - P(E 1 ’E 2 ’E 3 ’E 4 ’E 5 ’) [statistically independent] = 1 - P(E 1 ’) P(E 2 ’) P(E 3 ’) P(E 4 ’) P(E 5 ’) = 1 - (1 - 10 -5 ) 5 = 0.00005

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 1.What is the probability of shipping a defective syringe at the current operating conditions? 2.What is the probability of shipping a non-defective syringe at the current operating conditions? 3.What is the probability of shipping a defective syringe if the company reduces its defect production rate to 0.01?

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) = (0.05*0.01) = 0.000531 0.05*0.01+0.95*0.99 P(D’|A) = P(A|D’)*P(D’) = 0.9942 P(A|D)*P(D)+P(A|D’)*P(D’)

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