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

samples of#failures of#

POF

samples of#survivals of#

POS

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:

a)seeing “heads” one time?

b)seeing “heads” two times?

Example – Coin Toss

• OR = add, AND = multiply

• If you flip a coin two times, what is the probability of :

a)seeing “heads” one time?option 1: heads (0.5) AND tails (0.5) = 0.25

option 2: tails (0.5) AND heads (0.5) = 0.25option 1 OR option 2 = 0.25 + 0.25 = 0.5

b)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 partsb) exactly one defective partc) at least one defective partd) no more than one defective part

Permutations & Combinations

!rn!n

Prn

• 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

!rn!r!n

r

nCrn

Example – Answers

# of Defects Probability Combinations 1 2 3 4 5 6 7 8 9 10

0 0.5987369 1 0.95 0.95 0.95 0.95 0.95 0.95 0.95 0.95 0.95 0.95

1 0.3151247 10 0.05 0.95 0.95 0.95 0.95 0.95 0.95 0.95 0.95 0.95

2 0.0746348 45 0.05 0.05 0.95 0.95 0.95 0.95 0.95 0.95 0.95 0.95

3 0.0104751 120 0.05 0.05 0.05 0.95 0.95 0.95 0.95 0.95 0.95 0.95

4 0.0009648 210 0.05 0.05 0.05 0.05 0.95 0.95 0.95 0.95 0.95 0.95

5 0.0000609 252 0.05 0.05 0.05 0.05 0.05 0.95 0.95 0.95 0.95 0.95

6 0.0000027 210 0.05 0.05 0.05 0.05 0.05 0.05 0.95 0.95 0.95 0.95

7 8.03789E-08 120 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.95 0.95 0.95

8 1.58643E-09 45 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.95 0.95

9 1.85547E-11 10 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.95

10 9.76563E-14 1 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05

                         

Sum 1.00                      

• Must consider combinations for each # of defects

Example - Answers

a) 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

• Experimental outcomes can be represented by set theory relationships

• Union: A1A3, elements belong to A1 or A3 or both• P(A1A3) = P(A1) + P(A3) - P(A1A3) = A1+A3-A2

• Intersection: A1A3, elements belong to A1 and A3• P(A1A3) = P(A3|A1) * P(A1) = A2 (multiplication rule)

• Complement: A’, elements that do not belong to A• P(A’) = 1 – P(A)

Probability Relations

S

Special Cases

• If the events are statistically independent• P(AB) = P(B|A) * P(A) = P(B) * P(A)

• If the events are mutually exclusive• P(AB) = 0• P(AB) = P(A) + P(B) - P(AB)

= 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: AB = BA, AB = BA • Associative: (AB)C = A(BC)• Distributive: (AB)C = (AC)(BC)• Complementary: P(A) + P(A’) = 1• de Morgan’s Rule:

• (AB)’ = 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(AB) = P(B|A) * P(A)• P(B|A) = P(AB) / P(A)• Requires that P(A) ≠ 0

• Multiplication Rule:• P(AB) = P(A|B) * P(B) = P(B|A) * P(A)

S

A B

• 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?

InjuryA 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

InjuryA 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

A B

C

S

0.02 0.03

0.04

0.07

0.050.08

0.20

0.51

b)What is the probability that a patient with an MCL sprain (B) will later sustain a PCL tear (A)?

Example

P(A|B) = P(A B) = 0.08 = 0.348 P(B) 0.23

A B

C

S

0.02 0.03

0.04

0.07

0.050.08

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)?

Example

P(A| BC) = P(A (BC) = 0.03+0.04+0.05 = 0.26 P(BC) 0.47

A B

C

S

0.02 0.03

0.04

0.07

0.050.08

0.20

0.51

d)If a patient has sustained at least one knee injury in the past, what is the probability he will later tear his PCL?

Example

P(A (ABC) = P(A) = 0.14 = 0.28 P(ABC) P(ABC) 0.49

P(A|at least one) = P(A | ABC) = P(A (ABC) P(ABC)

A B

C

S

0.02 0.03

0.04

0.07

0.050.08

0.20

0.51