Transcript No Slide Title

ENM 207
Lecture 7
Example:
There are three columns entitled “Art” (A) ,“Books” (B) and
“Cinema” (C) in a new magazine. Reading habits of a
randomly selected reader with respect to choose columns are:
read regularly
probability
A
B
C
A B
.14 .23 .37 .08
AC B C
.09
.13
A B C
.05
(The probability of reading only A column is 14%)

What is the probability of column A given that they red column B?
P A  B 0.08
P A \ B 

 0.348
PB
0.23
 What is the probability of reading A given that they
are reading B or C columns?
P A \ B C 
P  A   B  C 
PB C

P  A  B    A  C 
P  B   P C   P  B  C 
0.08  0.09  0.05 0.12


 0.255
0.23  0.37  0.13 0.47
 What is the probability of reading column A given
that they are reading at least one column?
P A \ readsat leastone  P A \ A  B  C 
P A   A  B  C 
P A
0.14



 0.286
P A  B  C 
P A  B  C  0.49
A
B
C
The intersection of A with union of ABC is equal A itself.

What is the probability of reading A or B columns given that they read C
columns?
P A  B \ C  
P A  B   C  P A  C   PB  C 

PC 
PC 
P  A  C   P B  C   P  A  C  B  C 
PC 
P  A  C   P B  C   P  A  B  C 

PC 
0.09  0.13  0.05 0.17


0.37
0.37

Example: Mendenhall 3.63
solution
• There are five suppliers and a company will choose at least 2
suppliers, therefore, we must find the number of ways to
select 2, 3, 4 and 5 suppliers from the 5.
• At least 2 means that , a company can choose 2 or 3 or 4 or 5
suppliers from 5.
• The total number of obtions are
 5   5  5   5
5!
5!
5!
5!



 26
     
 2   3   4   5  2! 3! 3! 2! 4! 1! 5! 0!
Example:3.65
• Define the following events :
(A) Part is supplied by company A
( B) Part is supplied by company B
( C) Part is defective
From the problem, P(A)=0.8, P(B)=0.2,
P(C\A)= 0.05 , P(C\B)= 0.03
We know the given part is defective and we want to find
The probability it come from company A
And the probability it come from company B
P(A\C)=? , P(B\C)=?
P A  C   P A PC \ A  0.8  0.05  0.04
C is the defective part. It can come from company A or company B.
That is
We need P(BC)
PB  C   PB  PC \ B   0.2  0.03  0.006
PB \ C  
PB  C  0.006

 0.13
PC 
0.046
As a result we say that the company A is more likely to have supplied the
defective part.
Total Probability And Bayes Theorem
Partitions Total Probability And Bayes Theorem
A partition of the sample space may be defined like this.
S
B1
B2
A B 1
A B 2
A B 3
B3
B4
Definition:
If B1,B2,,,,Bk are disjoint subsets of S ( mutually exclusive events ) , and if
B1 B2 B3,,,,,,, Bk=S then these subsets are said to form a partition of S.
If k events , BI ( I=1,2,,,k) , form a partition and A is an arbitrary event with
respect to S, then we may write
Total Probability Law
A   A  B1    A  B2   ....   A  Bk 
So that
P A  P A  B1   P A  B2   ....  P A  Bk 
Since the events (ABi) are pairwise mutually exclusive events
P  A 
k
 P Bi   P  A \ Bi 
i 1
is called “total probability law”
Bayes Theorem
• Another important result of the total probability law is known as
“Bayes Theorem”:
If B1,B2,,,,Bk constitute a partition of the sample space S and A is an arbitrary
event on S , then for r = 1,2, ,k


P B r   P  A \ B r  

P B r \ A 
 BAYES THEOREM
k

PBi  P A \ Bi  

i 1


PBr \ A 
PBr  A
P B r   P  A \ B r 

k
P A
PBi  P A \ Bi 

i 1
The numerator is a result of multiplication rule and the denumerator is a result of
total probability law.
Ex: 2.35 page 56 (Montgomery)
A manufacturer of telemetry equipment microprocessors from 3 different
facilities. According to the historical data which are recorded by manufacturer company :
Supplying facility
Fraction defective
Fraction supplied by
1
0,02
0,15
2
0,01
0,80
3
0,03
0,05
B1
B2
A
B3
The director of manufacturing randomly selects a microprocessor , after test
procedure , finds that it is defective . compute the probability that this defective prod
came from facility 3.
Let A be the event that an item is defective
B1 be the event that the item comes from facility 1
B2 be the event that the item comes from facility 2
B3 be the event that the item comes from facility 3
We can find this probability using BAYES THEOREM.
P B3 \ A 
P B3   P  A \ B3 
P B1  P  A \ B1   P B2  P  A \ B2   P B3   P  A \ B3 
P B3  A
0.05  0.03
3


P  A
0.15  0.02  0.80  0.01  0.05  0.03 25
Example:
• Three machines A,B,C produce respectively 50%, 30%, 20% of
the total number of items of a factory .
• The percentages of defective output of these machines are 3% ,
4% , 5%,respectively. If an item is selected at random, find the
probability that the item is defective.
• Let X be the event that an item is defective.
Which theorem or which rule is used to solve this problem?
Total probability law.
P X   P A P X \ A  PB  P X \ B   PC  P X \ C 
 0.50 0.03  0.30 0.04  0.20 0.05
 0.037
Another way to solve this problem: Tree Diagram
0.03
the probability of
any part
produced by
machine A is
%50
0.50
D defective of
machine A
A
R reliable
0.04
0.30
D
B
R
0.20
C
0.05
D
R
Example:
• Consider the factory in the preceeding example. Suppose an
item is selected at random and is found to be defective. Find the
probability that the item was produced by machine A;
• What is the P(A\X)?
• Which theorem or which law can be used to solve this problem?
P A \ X  
P  A  P  X \ A
P A P X \ A  PB  P X \ B   PC  P X \ C 
0.50 0.03
15


0.50 0.03  0.30 0.04  0.20 0.05 37
Ex: 4.11 p 62/schaums outline series
In a certain college, 25% of the students failed math, 15% of the students failed
chemistry, 10% of the students failed both math and chemistry. A student is
selected at random.
a) If he failed chemistry what is the probability that he failed math?
b) If he failed math, what is the probability that he failed chemistry?
c) What is the probability that he failed math or chemistry?
Let M  students who failed math
C  students who failed chemistry
PM   0.25
1
2
3
PC   0.15
PM  C   0.10
PM  C  0.10 10
PM \ C  


P C 
0.15 15
PC  M  0.10 10
PC \ M  


PM 
0.25 15
PM  C   PM   PC   PM  C   0.25  0.15  0.10  0.30
SYSTEM RELIABILITY
System- electronic, mechanical or a combination of both –are
composed of components.
A component of a system is represented by a capital letter.
Two system each composed of tree components A; B; C are shown
below.
Systems according to their components connections can be classified
in two groups. Such as series and parallel.
A
A
B
C
B
A)series system
C
B)parallel system
Definition:
• If the system fails when any of the components fails, it is called a
series system.
A
B
C
•If the system fails only when all of its components fail, it is called
a parallel system.
A
B
C
D
A
B
C
E
This system is composed of five components A, B, C, D and E as shown above.
Components D and E are from a two-component parallel system. This subsystem
is connected in series with A, B and C.
subsystem
subsystem
A
B
D
E
F
G
C
Two parallel subsystems are connected in series. The first parallel
subsystem contains three componenets such as A, B and C.
The second contains two series subsystems: the first composed of D and E,
the second composed of F and G.
The reliability of a series system is
P(system functions ) = P(all componenets function)
The multiplicative rule can be applied because the components operate
independently of each other in a series system. If there is k components in a series
system
P(series system functions) = P( A functions) . P(B functions). . P(K functions)
PA  PB  PC .... PK
where pI is the probability that ith component functions, I=A,B,,,K
The reliability of a parallel system containing k components can be
calculated in a similar manner. Since a parallel system will fail only if all
components fail,
P parallel system fails  1  PA  1  PB ....1  PC 
Where pI is the probability that the ith component functions,
i=A,B,,,,,K.
Example:
A
Given that PA=0.90 PB=0.95
B
C
PC=0.90 find the reliability of the series system
shown in this figure.
Psystem functions  0.90 0.95 0.90  0.7695
Summary of system reliability
1. If the system is in series form that all components are
connected as a series.

P seriessystem functions
n
 Peach component functions
i
If there are k components like A1,A2,,,Ak

P seriessystem functions
k
 P Ai 
i
Because a series system will fail if any one of its components fail.
2.
If the system is parallel and also if it contains k components,
 k


P parallel system fails  
(1  P Ai ) 



 i


Because as you know a parallel system will fail only if all components will fail.
From this equation
k


  1 
P parallel system functions

i



1  P Ai 


Note: These formulations can be used to calculate the reliabilities of seriesm
systems , parallel systems , or any combinations of them. These systems must
satisfy the assumption that the components operate independently.
Example: 17.22 : First subcircuits involves comp. A,B,C in parallel
A
P(first subcircuits)=1-(1-PA)(1-PB)(1-PC)
P(A)=0.95, P(B)=0.95 , P(C)=0.90, P(D)=0.90, P(E)=0.98
B
C
P(second subcircuits)=1-P(D)P(E)
D
E
.
If the subcircuits are connected in series
subsystem
subsystem
A
B
D
E
F
G
C
P(two subcircuits are connected in series)=P(first subcircuits)P(second subcircuits)
 If two subcircuits are connected in parallel
A
B
C
D
E
two subcircuits are connected in parallel)=1-(1-P(first scir.))(1-P(sec.scir.))