MEGN 537 – Probabilistic Biomechanics Ch. 1 – Introduction Ch MEGN 537 – Probabilistic Biomechanics Ch.1 – Introduction Ch.2 – Mathematics of Probability Anthony J Petrella, PhD

Ch.1 - Introduction

Uncertainty Uncertainty present in physical systems Repeated measurement yields variability Dimensional tolerances, respiration rate, tissue material properties, joint loading, etc. What impact does this uncertainty have on performance?

Strength-Based Reliability Safety factor shows acceptable design Some percentage of the time, stress may exceed strength

Reliability Definitions Probability of Failure POF = 0.001, 0.0001 Probability of Survival or Reliability Reliability = 0.999 (three 9s), 0.9999 (four 9s) POF + POS = 1

Reliability-based Design Design for Six Sigma Concept developed by Bill Smith in 1993 Motorola owns six sigma trademark Six sigma corresponds to 3.4 failures per 1,000,000 POF = 0.000,003,4 or Reliability = 0.999,997,6 Design Excellence or BlackBelt programs Many companies have implemented their versions GE and Honeywell boast 100s of millions of dollars saved

Uncertainty Sources of uncertainty Inherent / repeated measurement Statistical uncertainty – limited availability of sampling size means actual distribution unknown Modeling uncertainty – how good is the model? Cognitive or qualitative sources – intellectual abstraction of reality, human factors

Course Objectives Ability to understand and apply probability theory and probabilistic analysis methods To assess impact of uncertainty in parameters (inputs) on performance (outcomes) Determine the appropriate distribution to represent a dataset To apply this knowledge to real biomechanical systems

Ch.2 - Mathematics of Probability

Definitions Probability: The likelihood of an event occurring Event: Represents the outcome of a single experiment (or single simulation) Experiment: An occurrence that has an uncertain outcome (die toss , coin toss, tensile test) – usually based on a physical model Simulation: An occurrence that has an uncertain outcome – usually based on an analytical or computational model

Example – Coin Toss OR = add, AND = multiply If you flip a coin two times, what is the probability of: seeing “heads” one time? seeing “heads” two times?

Example – Coin Toss OR = add, AND = multiply If you flip a coin two times, what is the probability of : seeing “heads” one time? option 1: heads (0.5) AND tails (0.5) = 0.25 option 2: tails (0.5) AND heads (0.5) = 0.25 option 1 OR option 2 = 0.25 + 0.25 = 0.5 seeing “heads” two times? option 1: heads (0.5) AND heads (0.5) = 0.25

Example – TKR Casting A knee implant casting process is known to produce a defective part 5% of the time If 10 castings were tested, find the probability of: a) no defective parts b) exactly one defective part c) at least one defective part d) no more than one defective part

Permutations & Combinations Number of permutations of r objects from a set of n distinct objects (ordered sequence) Number of combinations in which r objects can be selected from a set of n distinct objects n objects taken r at a time Independent of order

Example – Answers Must consider combinations for each # of defects Probability Combinations 1 2 3 4 5 6 7 8 9 10 0.5987369 0.95 0.3151247 0.05 0.0746348 45 0.0104751 120 0.0009648 210 0.0000609 252 0.0000027 8.03789E-08 1.58643E-09 1.85547E-11 9.76563E-14   Sum 1.00

Example - Answers no defective parts b) exactly one defective part P(0 defects) = P(part 1 no defect)*P(part 2 no defect)… = (1-0.05)^10 = 0.598 P(1 defect) = P(part 1 defect)*P(part 2 no defect)… = (0.05)*(0.95)^9 *10 = 0.315

Example - Answers c) at least one defective part d) no more than one defective part P(≥ 1 defect) = P(1defect) + P(2 defects) + P(3 defects)… = 1- P(0 defects) = 1-0.598 = 0.402 P(≤ 1 defect) = P(0 defects) + P(1 defect) = 0.598 + 0.315 = 0.913

Definitions Sample Space (S): The set of all basic outcomes of an experiment Mutually Exclusive: Events that preclude occurrence of one another Collectively Exhaustive: No other events are possible S A B

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

Special Cases If the events are statistically independent P(AB) = P(B|A) * P(A) = P(B) * P(A) If the events are mutually exclusive P(AB) = 0 P(AB) = P(A) + P(B) - P(AB) = P(A) + P(B) S A B

Example For a randomly chosen automobile: Let A={car has 4 cylinders} B={car has 6 cylinders}. Since events are mutually exclusive, if B occurs, then A cannot occur. So P(A|B) = 0 ≠ P(A). If 2 events are mutually exclusive, they cannot be independent…when A & B are mutually exclusive, the information that A occurred says something about B (it cannot have occurred), so independence is precluded

Rules of Set Theory Commutative: AB = BA, AB = BA Associative: (AB)C = A(BC) Distributive: (AB)C = (AC)(BC) Complementary: P(A) + P(A’) = 1 de Morgan’s Rule: (AB)’ = A’B’ Complement of union = intersection of complements (A  B)’ = A’  B’ Complement of intersection = union of complements

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 Multiplication Rule: P(AB) = P(A|B) * P(B) = P(B|A) * P(A) S A B

Example Common knee injuries include: PCL tear (A), MCL sprain (B), meniscus tear (C) Injury statistics as reported by epidemiology literature: a) What does the Venn Diagram look like? Injury A B C A  B A  C B  C A  B  C Probability 0.14 0.23 0.37 0.08 0.09 0.13 0.05

Example Injury A B C A  B A  C B  C A  B  C Probability 0.14 0.23 0.37 0.08 0.09 0.13 0.05 A B C S 0.02 0.03 0.04 0.07 0.05 0.08 0.20 0.51

Example A B C S 0.02 0.03 0.04 0.07 0.05 0.08 0.20 0.51 b) What is the probability that a patient with an MCL sprain (B) will later sustain a PCL tear (A)? P(A|B) = P(A  B) = 0.08 = 0.348 P(B) 0.23

Example A B C S 0.02 0.03 0.04 0.07 0.05 0.08 0.20 0.51 c) If a patient has sustained either an MCL sprain (B) or a meniscus tear (C) or both, what is the probability of a later PCL tear (A)? P(A| BC) = P(A  (BC) = 0.03+0.04+0.05 = 0.26 P(BC) 0.47

Example d) If a patient has sustained at least one knee injury in the past, what is the probability he will later tear his PCL? P(A|at least one) = P(A | ABC) = P(A  (ABC) P(ABC) P(A  (ABC) = P(A) = 0.14 = 0.28 P(ABC) P(ABC) 0.49