### Review of Probability

```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
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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
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Introduction

Interpretations of Probability
◦ Subjective – Pr(A) is a measure of the degree
of belief one holds in a specified proposition A
◦ Probability is directly related to the odds one
would wager on a specific proposition
Uncertainty Analysis for Engineers
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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
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Identities
A   =A
 A  I=I
 A   = A = 
 A I=A
 A  A=A
 A A=A
  =I
 A A= 
 A  A=I

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Example
The EZ Company employs 10 nonprofessional 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

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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
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Probability
Rolling dice
 m(I)=6
 A=rolling a 2
 Pr(A)=m(A)/m(I)=1/6

Uncertainty Analysis for Engineers
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Definition

Two events are independent if the
occurrence of one does not change the
probability of occurrence of the other
Uncertainty Analysis for Engineers
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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
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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
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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
A
D
C
B
Pr(B)=0.8
Pr(C)=0.
95
E
Pr(D)=0.
9
Pr(E)=0.9
F
Pr(F)=0.5
Uncertainty Analysis for Engineers
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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*0.50.9*0.5+0.9*0.9*0.5=0.995
 So, Pr(S)=0.98*0.95*0.995=0.926

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Another Example
What if 2 events are not independent?
 Consider the system below, where event G
is in both stages

G
H
Uncertainty Analysis for Engineers
G
J
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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
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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
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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
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Solution
A1=transistor from vendor 1
 A2=transistor from vendor 2
 B1=good 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

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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

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A Generalization

If B depends on a series of previous
events (Ai) then
n
Pr(B)   PrB | Ai  Pr Ai 
i 1
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

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