Interpretations of Probability in Uncertainty Analysis
Review of Probability
Jake Blanchard
Spring 2010
Uncertainty Analysis for Engineers
1
Introduction
Interpretations of Probability
◦
Classical – If an event can occur in N equally
likely and different ways, and if n of these have
an attribute A, then the probability of the
occurrence of A, denoted Pr(A), is defined as
n/N
◦
Example: the probability of drawing an Ace
from a full deck of cards is 1/13 (4/52)
Uncertainty Analysis for Engineers
2
Introduction
Interpretations of Probability
◦
Frequency (empirical) – If an experiment is
conducted N times, and a particular attribute A
occurs n times, then the limit of n/N as N
becomes large is defined as the probability of A
◦
Example: If, in the past, 73 cars out of 10,000 are
defective (coming from a particular factory), then
the probability that a car will be defective is
0.0073
◦
This interpretation is the most common among
statisticians
Uncertainty Analysis for Engineers
3
Introduction
Interpretations of Probability
◦
Subjective – Pr(A) is a measure of the degree
of belief one holds in a specified proposition A
◦
Broader than other interpretations
◦
Probability is directly related to the odds one
would wager on a specific proposition
Uncertainty Analysis for Engineers
4
Set Theory
Set=collection of distinct objects
Union: C
1
=A
B
Intersection: C
2
=A
B=AB
Complement (“not”): C
3
=A
=null set
I=entire set
m(A)=number of elements in set A
Uncertainty Analysis for Engineers
5
Identities
A
=A
A
I=I
A
=
A
=
A
I=A
A
A=A
A
A=A
=I
A
A=
A
A=I
Uncertainty Analysis for Engineers
6
Example
The EZ Company employs 10 non-
professional employees: 3 assemblers, 5
machinists, and 2 clerks
m(A)=3; m(M)=5; m(C)=2; m(I)=10
Q=set of workers who are both a
machinist and an assembler
Q=AM=Z; m(Q)=0
F=all factory workers; F=A
M
m(F)=m(A
M)=8
Uncertainty Analysis for Engineers
7
Example (continued)
Suppose the company employs 8
engineers, 3 supervisors, and 2 employees
who are both an engineer and a
supervisor
m(E
S)
=m(E)+m(S)-m(ES)
=10+5-2=13
Uncertainty Analysis for Engineers
8
Probability
Rolling dice
m(I)=6
A=rolling a 2
Pr(A)=m(A)/m(I)=1/6
Uncertainty Analysis for Engineers
9
Definition
Two events are
independent
if the
occurrence of one does not change the
probability of occurrence of the other
Uncertainty Analysis for Engineers
10
Probability Laws
Pr(A)=1-Pr(A)
If A and B are independent, then
◦
Pr(A and B)=Pr(AB)=Pr(A)Pr(B)
If A and B are mutually exclusive
[m(AB)=0] then
◦
Pr(A or B)=Pr(A
B)=Pr(A)+Pr(B)
In general
◦
Pr(A and/or B)
◦
=Pr(A
B)
◦
=Pr(A)+Pr(B)-Pr(AB)
Uncertainty Analysis for Engineers
11
One More Probability Law
Pr(A and/or B and/or C)
=Pr(A
B
C)
=Pr(A)+Pr(B)+Pr(C)-Pr(AB)-Pr(AC)-
Pr(BC)+Pr(ABC)
Uncertainty Analysis for Engineers
12
Example
Consider a 3-stage process, as diagrammed below
Our goal is to find the probability of success of the
entire operation, assuming all individual probabilities
are independent
Branches represent parallel redundancy, so success
in a stage requires success of, for example, A or B
Uncertainty Analysis for Engineers
13
A
B
C
F
E
D
Pr(C)=0.
95
Pr(B)=0.8
Pr(A)=0.
9
Pr(D)=0.
9
Pr(F)=0.5
Pr(E)=0.9
Solution
Pr(S)=Pr(I)Pr(II)Pr(III)
where I, II, and III represent the three stages
Pr(I)=Pr(A)+Pr(B)-Pr(AB)
=0.9+0.8-0.9*0.8=0.98 (success requires A
or B to succeed)
Pr(II)=Pr(C)=0.95
Pr(III)=Pr(D)+Pr(E)+Pr(F)-Pr(DE)-Pr(EF)-
Pr(DF)+Pr(DEF)
=0.9+0.9+0.5-0.9*0.9-0.9*0.5-
0.9*0.5+0.9*0.9*0.5=0.995
So, Pr(S)=0.98*0.95*0.995=0.926
Uncertainty Analysis for Engineers
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Another Example
What if 2 events are not independent?
Consider the system below, where event G
is in both stages
Uncertainty Analysis for Engineers
15
G
H
J
G
Conditional Probability
Conditional probability [Pr(B|A)] of an
event B with respect to some other event
A is the probability that B will occur, given
that A has taken place
For our example, Pr(II|I) represents the
probability of successful operation of stage
II, given successful operation of stage I
Once A has occurred, then A replaces I as
the sample space of interest, so the size of
AB relative to the new set is given by
m(AB)/m(A)
Uncertainty Analysis for Engineers
16
Cond. Probability (cont.)
Pr(B|A)=[m(AB)/m(I)]/[m(A)/m(I)]
=Pr(AB)/Pr(A)
Or
Pr(A and B)=Pr(AB)=Pr(A)Pr(B|A)
Extending…
Pr(A and B and C)=Pr(ABC)=
Pr(A)Pr(B|A)Pr(C|AB)
Pr(C|AB) is probability of C, given that A and
B have occurred
Uncertainty Analysis for Engineers
17
Example
75% of transistors come from vendor 1
and 25% from vendor 2
99% of supply from vendor 1 and 90% of
supply from vendor 2 are acceptable
If we randomly pick a transistor, what is
the probability that it came from vendor 1
and is defective?
Also, what is the probability that the
transistor is defective, irrespective of the
vendor
Uncertainty Analysis for Engineers
18
Solution
A
1
=transistor from vendor 1
A
2
=transistor from vendor 2
B
1
=good transistor
B
2
=bad transistor
Pr(A
1
)=0.75; Pr(A
2
)=0.25
Pr(B
1
|A
1
)=0.99; Pr(B
2
|A
1
)=0.01
Pr(B
1
|A
2
)=0.90; Pr(B
2
|A
2
)=0.10
Uncertainty Analysis for Engineers
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Solution (cont.)
Pr(A
1
B
2
)=Pr(A
1
)
Pr(B
2
|A
1
)=0.75*0.01=0.0075
Pr(A
2
B
2
)=Pr(A
2
)
Pr(B
2
|A
2
)=0.25*0.1=0.025
Pr(B
2
)=0.0075+0.025=0.0325
Uncertainty Analysis for Engineers
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A Generalization
If B depends on a series of previous
events (A
i
) then
Uncertainty Analysis for Engineers
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Example (2.25)
Hurricanes: C1=Category 1,
C2=Category 2, etc.
P(C1)=.35, P(C2)=.25, P(C3)=.14,
P(C4)=.05, P(C5)=.01
D=Damage; P(D|C1)=.05, P(D|C2)=.1,
P(D|C3)=.25, P(D|C4)=.6, P(D|C5)=1.0
What is probability of damage?
P(D)=P(D|C1)P(C1)+ P(D|C2)P(C2)+
P(D|C3)P(C3)+ P(D|C4)P(C4)+
P(D|C5)P(C5)=0.1175
Uncertainty Analysis for Engineers
22
This review delves into classical, frequency, and subjective interpretations of probability, discussing set theory and identities within uncertainty analysis. Examples and illustrations are provided to aid in understanding the concepts presented.
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Presentation Transcript
Review of Probability Jake Blanchard Spring 2010 Uncertainty Analysis for Engineers 1
Introduction Interpretations of Probability Classical If an event can occur in N equally likely and different ways, and if n of these have an attribute A, then the probability of the occurrence of A, denoted Pr(A), is defined as n/N Example: the probability of drawing an Ace from a full deck of cards is 1/13 (4/52) Uncertainty Analysis for Engineers 2
Introduction Interpretations of Probability Frequency (empirical) If an experiment is conducted N times, and a particular attribute A occurs n times, then the limit of n/N as N becomes large is defined as the probability of A Example: If, in the past, 73 cars out of 10,000 are defective (coming from a particular factory), then the probability that a car will be defective is 0.0073 This interpretation is the most common among statisticians Uncertainty Analysis for Engineers 3
Introduction Interpretations of Probability Subjective Pr(A) is a measure of the degree of belief one holds in a specified proposition A Broader than other interpretations Probability is directly related to the odds one would wager on a specific proposition Uncertainty Analysis for Engineers 4
Set Theory Set=collection of distinct objects Union: C1=A B Intersection: C2=A B=AB Complement ( not ): C3=A =null set I=entire set m(A)=number of elements in set A Uncertainty Analysis for Engineers 5
Identities A =A A I=I A = A = A I=A A A=A A A=A =I A A= A A=I Uncertainty Analysis for Engineers 6
Example The EZ Company employs 10 non- professional employees: 3 assemblers, 5 machinists, and 2 clerks m(A)=3; m(M)=5; m(C)=2; m(I)=10 Q=set of workers who are both a machinist and an assembler Q=AM=Z; m(Q)=0 F=all factory workers; F=A M m(F)=m(A M)=8 Uncertainty Analysis for Engineers 7
Example (continued) Suppose the company employs 8 engineers, 3 supervisors, and 2 employees who are both an engineer and a supervisor m(E S) =m(E)+m(S)-m(ES) =10+5-2=13 Uncertainty Analysis for Engineers 8
Probability Rolling dice m(I)=6 A=rolling a 2 Pr(A)=m(A)/m(I)=1/6 Uncertainty Analysis for Engineers 9
Definition Two events are independent if the occurrence of one does not change the probability of occurrence of the other Uncertainty Analysis for Engineers 10
Probability Laws Pr(A)=1-Pr(A) If A and B are independent, then Pr(A and B)=Pr(AB)=Pr(A)Pr(B) If A and B are mutually exclusive [m(AB)=0] then Pr(A or B)=Pr(A B)=Pr(A)+Pr(B) In general Pr(A and/or B) =Pr(A B) =Pr(A)+Pr(B)-Pr(AB) Uncertainty Analysis for Engineers 11
One More Probability Law Pr(A and/or B and/or C) =Pr(A B C) =Pr(A)+Pr(B)+Pr(C)-Pr(AB)-Pr(AC)- Pr(BC)+Pr(ABC) Uncertainty Analysis for Engineers 12
Example Consider a 3-stage process, as diagrammed below Our goal is to find the probability of success of the entire operation, assuming all individual probabilities are independent Branches represent parallel redundancy, so success in a stage requires success of, for example, A or B Pr(A)=0. 9 Pr(D)=0. 9 D A C E Pr(E)=0.9 B Pr(C)=0. 95 F Pr(B)=0.8 Pr(F)=0.5 Uncertainty Analysis for Engineers 13
Solution Pr(S)=Pr(I)Pr(II)Pr(III) where I, II, and III represent the three stages Pr(I)=Pr(A)+Pr(B)-Pr(AB) =0.9+0.8-0.9*0.8=0.98 (success requires A or B to succeed) Pr(II)=Pr(C)=0.95 Pr(III)=Pr(D)+Pr(E)+Pr(F)-Pr(DE)-Pr(EF)- Pr(DF)+Pr(DEF) =0.9+0.9+0.5-0.9*0.9-0.9*0.5- 0.9*0.5+0.9*0.9*0.5=0.995 So, Pr(S)=0.98*0.95*0.995=0.926 Uncertainty Analysis for Engineers 14
Another Example What if 2 events are not independent? Consider the system below, where event G is in both stages G G H J Uncertainty Analysis for Engineers 15
Conditional Probability Conditional probability [Pr(B|A)] of an event B with respect to some other event A is the probability that B will occur, given that A has taken place For our example, Pr(II|I) represents the probability of successful operation of stage II, given successful operation of stage I Once A has occurred, then A replaces I as the sample space of interest, so the size of AB relative to the new set is given by m(AB)/m(A) Uncertainty Analysis for Engineers 16
Cond. Probability (cont.) Pr(B|A)=[m(AB)/m(I)]/[m(A)/m(I)] =Pr(AB)/Pr(A) Or Pr(A and B)=Pr(AB)=Pr(A)Pr(B|A) Extending Pr(A and B and C)=Pr(ABC)= Pr(A)Pr(B|A)Pr(C|AB) Pr(C|AB) is probability of C, given that A and B have occurred Uncertainty Analysis for Engineers 17
Example 75% of transistors come from vendor 1 and 25% from vendor 2 99% of supply from vendor 1 and 90% of supply from vendor 2 are acceptable If we randomly pick a transistor, what is the probability that it came from vendor 1 and is defective? Also, what is the probability that the transistor is defective, irrespective of the vendor Uncertainty Analysis for Engineers 18
Solution A1=transistor from vendor 1 A2=transistor from vendor 2 B1=good transistor B2=bad transistor Pr(A1)=0.75; Pr(A2)=0.25 Pr(B1|A1)=0.99; Pr(B2|A1)=0.01 Pr(B1|A2)=0.90; Pr(B2|A2)=0.10 Uncertainty Analysis for Engineers 19
Solution (cont.) Pr(A1B2)=Pr(A1) Pr(B2|A1)=0.75*0.01=0.0075 Pr(A2B2)=Pr(A2) Pr(B2|A2)=0.25*0.1=0.025 Pr(B2)=0.0075+0.025=0.0325 Uncertainty Analysis for Engineers 20
A Generalization If B depends on a series of previous events (Ai) then n ( ) ( ) Pr 1 = = Pr( ) Pr | B B A A i i i Uncertainty Analysis for Engineers 21
Example (2.25) Hurricanes: C1=Category 1, C2=Category 2, etc. P(C1)=.35, P(C2)=.25, P(C3)=.14, P(C4)=.05, P(C5)=.01 D=Damage; P(D|C1)=.05, P(D|C2)=.1, P(D|C3)=.25, P(D|C4)=.6, P(D|C5)=1.0 What is probability of damage? P(D)=P(D|C1)P(C1)+ P(D|C2)P(C2)+ P(D|C3)P(C3)+ P(D|C4)P(C4)+ P(D|C5)P(C5)=0.1175 Uncertainty Analysis for Engineers 22
