Transcript MEGN 537 – Probabilistic Biomechanics Ch.1 – Introduction

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
# of failures
POF 
# of samples
• POF = 0.001, 0.0001
• Probability of Survival or Reliability
# of survivals
POS 
# of samples
• 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:
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) = 0.25
option 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 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)
n!
nPr 
n  r !
• 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
n
n!
n Cr    
 r  r! n  r !
Example – Answers
• Must consider combinations for each # of defects
# 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
Example - Answers
a) no defective parts
P(0 defects)
= P(part 1 no defect)*P(part 2 no defect)…
= (1-0.05)^10 = 0.598
b) exactly one defective part
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
P(≥ 1 defect)
= P(1defect) + P(2 defects) + P(3 defects)…
= 1- P(0 defects) = 1-0.598 = 0.402
d) no more than one defective part
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)
A
S
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
S
A
• Multiplication Rule:
• P(AB) = P(A|B) * P(B) = P(B|A) * P(A)
B
Example
• Common knee injuries include: PCL tear (A),
MCL sprain (B), meniscus tear (C)
• Injury statistics as reported by epidemiology
literature:
Injury
Probability
A
B
C
AB
AC
BC
ABC
0.14
0.23
0.37
0.08
0.09
0.13
0.05
a) What does the Venn Diagram look like?
Example
Injury
Probability
A
B
C
AB
AC
BC
ABC
0.14
0.23
0.37
0.08
0.09
0.13
0.05
S
A
B
0.02
0.03
0.07
0.05
0.04
0.08
0.20
0.51
C
Example
S
A
B
0.02
0.03
0.07
0.05
0.04
0.08
0.20
0.51
C
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)
P(B)
= 0.08 =
0.23
0.348
Example
S
A
B
0.02
0.03
0.07
0.05
0.04
0.08
0.20
0.51
C
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)
P(BC)
= 0.03+0.04+0.05 = 0.26
0.47
Example
S
A
B
0.02
0.03
0.07
0.05
0.04
0.08
0.20
0.51
C
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