1. MEGN 537 – Probabilistic Biomechanics Ch
MEGN 537 – Probabilistic Biomechanics Ch.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. Bayes’ Theorem
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 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)) =

10. Answer Application of DeMorgan’s Rule
(AB)’ = A’B’ Complement of union = intersection of complements
(A  B)’ = A’  B’ Complement of intersection = union of complements
Application of DeMorgan’s Rule
Let E = member failure; F = truss failure P(Ei) = 10-5 P(Ei’) =
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 - ( )5 =

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) = * *0.9
P(D’|A) = P(A|D’)*P(D’) = P(A|D)*P(D)+P(A|D’)*P(D’)
P(D|A) = (0.05*0.01) = * *0.99