1 / 28

MEGN 537 – Probabilistic Biomechanics Ch.1 – Introduction Ch.2 – Mathematics of Probability

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

kali
Download Presentation

MEGN 537 – Probabilistic Biomechanics Ch.1 – Introduction Ch.2 – Mathematics of Probability

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

  2. Ch.1 - Introduction

  3. 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?

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

  5. 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

  6. 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

  7. 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

  8. 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

  9. Ch.2 - Mathematics of Probability

  10. 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

  11. 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?

  12. 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) = 0.25option 2: tails (0.5) AND heads (0.5) = 0.25option 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

  13. Example – TKR Casting • A knee implant casting process is known to produce a defective part 5% of the timeIf 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

  14. 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 Permutations & Combinations

  15. Must consider combinations for each # of defects Example – Answers

  16. 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

  17. 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

  18. 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

  19. 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

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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?

  25. Example S A B 0.02 0.07 0.03 0.05 0.04 0.08 0.20 0.51 C

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

  27. Example 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)? S A B 0.02 0.07 0.03 0.05 0.04 0.08 0.20 P(A| BC) = P(A  (BC) = 0.03+0.04+0.05 = 0.26 P(BC) 0.47 0.51 C

  28. 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

More Related