SNS College of Engineering
Department of Computer Science and
Engineering
Bayes’ Theorem
Presented By,
S.Yamuna
AP/CSE
7/28/2017
Artificial Intelligence
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Bayes’ Theorem
 An insurance company divides its clients into two
categories: those who are accident prone and those
who are not. Statistics show there is a 40% chance an
accident prone person will have an accident within 1
year whereas there is a 20% chance non-accident
prone people will have an accident within the first
year.
 If 30% of the population is accident prone, what is the
probability that a new policyholder has an accident
within 1 year?
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Bayes’ Theorem
 Let A be the event a person is accident prone
 Let F be the event a person has an accident
within 1 year
P A  F C 
A
AC
P AC  F C 
F
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P  A  F Artificial Intelligence
P AC  F 
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Bayes’ Theorem
 Notice we’ve divided up or partitioned the
sample space along accident prone and nonaccident prone
P A  F C 
A
AC
P AC  F C 
F
P A  F 
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P AC  F 
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Bayes’ Theorem
 Notice that A  F  andAC  F  are mutually
exclusive events and that
A  F   AC  F   F
 Therefore
PF   PA  F   P AC  F 
 We need to find P  A  F  and P AC  F 
 How?
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Bayes’ Theorem
 Recall from conditional probability
P E  F 
 P E |F  
P F 
P E |F   P F   P E  F 
P F  E 
 P F |E  
P E 
P F |E   P E   P E  F   P F  E 
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Bayes’ Theorem
 Thus:


P  A  F   P F | A  P A
P AC  F   P F | AC  P AC 
P(F|AC) = 0.2 since
non-accident prone
people have a 20%
chance of having an
accident within 1 year
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P(A) = 0.30 since
30% of population
is accident prone
P(F|A) = 0.40 since if
a person is accident
prone, then his chance
of having an accident
within 1 year is 40%
P(AC) = 1- P(A) = 0.70
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Bayes’ Theorem
 Updating our Venn Diagram
P A  F C 
A
P AC  F C 
AC
F
P A  F   P F | AP  A
P AC  F   P F | AC P AC 

 Notice again that PF   PA  F   P AC  F
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
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Bayes’ Theorem
 So the probability of having an accident within
1 year is:
C





 P F  P AF  P A F
 P F | AP A  P F | AC P AC 
 0.400.30   0.200.70   0.26
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Bayes’ Theorem
 Using Tree Diagrams:
Accident
w/in 1 year
P(F|A)=0.40
Accident Prone
P(A) = 0.30
No Accident
w/in 1 year
P(FC|A)=0.60
Not Accident Prone
P(AC) = 0.70
Accident w/in
1 year
P(F|AC)=0.20
No Accident
w/in 1 year
P(FC|AC)=0.80
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P A  F 
P A  F C 
P AC  F 
P AC  F C 
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Bayes’ Theorem
 Notice you can have an accident within 1 year
by following branch A until F is reached
 The probability that F is reached via branch A is
given by P F | A  P  A
 In other words, the probability of being accident
prone and having one within 1 year is
P  A  F   P F | A  P A
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Bayes’ Theorem
 You can also have an accident within 1 year by
following branch AC until F is reached
 The probability that F is reached via branch AC is
given by P F | AC  P AC 
 In other words, the probability of NOT being
accident prone and having one within 1 year is
P AC  F   P F | AC  P AC 
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Bayes’ Theorem
 What would happen if we had partitioned our
sample space over more events, say A1 , A2 ,, An,
all them mutually exclusive?
 Venn Diagram
A1
A2
......
An-1
An
F
(etc.)
P A1  F  P A2  F  Artificial Intelligence
P A  F 
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n1
P An  F 
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Bayes’ Theorem

P F   P A1  F   P A2  F     P An  F 
 For each P Ai  F   P F | Ai P Ai 
P F   P A1  F   P A2  F     P An  F 
 P F | A1 P A1   P F | A2 P A2     P F | An P An 
n
  P F | Ai P Ai 
i 1
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Bayes’ Theorem
 Tree Diagram
P  A1 
P  A2 

P An1 
P  An 
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P F |A1 
P F C |A1 
P F |A2 
P F C |A2 
P F | An1 
P F C | An1 
P F | An 
P F C | An 
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Bayes’ Theorem
 Notice that F can be reached via A1 , A2 ,, An
branches
 Multiplying across each branch tells us the
probability of the intersection
 Adding up all these products gives:
n
P F    P F | Ai P Ai 
i 1
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Bayes’ Theorem
 Ex: 2 (text tractor example) Suppose there are 3
assembly lines: Red, White, and Blue. Chances of
a tractor not starting for each line are 6%, 11%,
and 8%. We know 48% are red and 31% are blue.
The rest are white. What % don’t start?
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Bayes’ Theorem
 Soln.
R: red
W: white
B: blue
N: not starting
P(R) = 0.48
P(W) = 0.21
P(B) = 0.31
P(N | R) = 0.06
P(N | W) = 0.11
P(N | B) = 0.08
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Bayes’ Theorem
 Soln.
P N   P N |R   P R   P N |W   P W   P N |B   P B 
 0.06   0.48   0.11   0.21   0.08   0.31 
 0.0767
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Bayes’ Theorem
 Main theorem:
Suppose we know P E |F  . We would like to
use this information to find P F |E  if possible.
Discovered by Reverend Thomas Bayes
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Bayes’ Theorem
 Main theorem:
 Ex. Suppose B1 and B2 partition a space and A is
some event.
 Use PB1 , PB2 , PA|B1 , and P  A|B2  to
determine P B1 | A.
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Bayes’ Theorem
P B1  A
 Recall the formulas: P B1 | A 
P  A
P B1  A  P A  B1   P A|B1   P B1 
P A  P B1   P A|B1   P B2   P A|B2 
P B1  A
 So, P B1 | A 
P  A
P A|B1   P B1 

P B1   P A|B1   P B2   P A|B2 
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Bayes’ Theorem
 Bayes’ Theorem:
P Bk | A 
P A|Bk   P Bk 
n
 P A|B  P B 
i 1
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i
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i
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Bayes’ Theorem
 Ex. 4 (text tractor example) 3 assembly lines:
Red, White, and Blue. Some tractors don’t start
(see Ex. 2). Find prob. of each line producing a
non-starting tractor.
P(R) = 0.48
P(W) = 0.21
P(B) = 0.31
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P(N | R) = 0.06
P(N | W) = 0.11
P(N | B) = 0.08
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Bayes’ Theorem
 Soln.
Find P(R | N), P(W | N), and P(B | N)
P(R) = 0.48
P(N | R) = 0.06
P(W) = 0.21
P(N | W) = 0.11
P(B) = 0.31
P(N | B) = 0.08
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Bayes’ Theorem
 Soln.
P R  N 
P R |N  
P N 
P N |R   P R 

P N |R   P R   P N |W   P W   P N |B   P B 
0.06   0.48 

0.06   0.48   0.11   0.21   0.08   0.31 
 0.3755
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Bayes’ Theorem
 Soln.
P W |N  
P N |W   P W 
P N |R   P R   P N |W   P W   P N |B   P B 
0.11   0.21 

0.06   0.48   0.11   0.21   0.08   0.31 
 0.3012
P B |N   0.3233
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Bayes’ Theorem
 Focus on the Project:
We want to find the following probabilities:
P S |Y  T  C  and P F |Y  T  C  .
To get these, use Bayes’ Theorem
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Bayes’ Theorem
 Focus on the Project:
P S |Y  T  C  
P Y  T  C | S   P S 
P Y  T  C | S   P S   P Y  T  C |F   P F 
P F |Y  T  C  
P Y  T  C |F   P F 
P Y  T  C | S   P S   P Y  T  C |F   P F 
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Bayes’ Theorem
 Focus on the Project:
P S |Y  T  C  
P Y  T  C | S   P S 
P Y  T  C | S   P S   P Y  T  C |F   P F 
In Excel, we find the probability to be approx.
0.4774
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Bayes’ Theorem
 Focus on the Project:
P Y  T  C |F   P F 
P F |Y  T  C  
P Y  T  C | S   P S   P Y  T  C |F   P F 
In Excel,we find the probability to be approx.
0.5226
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Bayes’ Theorem
 Focus on the Project:
Let Z be the value of a loan work out for a
borrower with 7 years, Bachelor’s, Normal…
E Z   Success  Prob. Success  Failure  Prob. Failure
 4,000 ,000  0.4774   250 ,000 0.5226 
 $2,040 ,000
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Bayes’ Theorem
 Focus on the Project:
Since foreclosure value is $2,100,000 and on
average we would receive $2,040,000 from a
borrower with John Sanders characteristics, we
should foreclose.
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Bayes’ Theorem
 Focus on the Project:
However, there were only 239 records
containing 7 years experience.
Look at range of value 6, 7, and 8 (1 year more
and less)
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Bayes’ Theorem
 Focus on the Project:
Use DCOUNT function with an extra “Years in
Business” heading
Years In
Former Bank
Business
BR
>=6
Education Level
State Of
Economy
Loan Paid
Years In
Back?
Business
yes
<=8
Same for “no”
Added a new column
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Bayes’ Theorem
 Focus on the Project:
From this you get 349 successful and 323 failed
records
Let Y  be a borrower with 6, 7, or 8 years
experience
349
PY | S   1470
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and
323
PY |F   1779
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Bayes’ Theorem
 Focus on the Project:
P Y   T  C | S   P Y | S   P T | S   P C | S 
 0.2374   0.5301   0.5823 
 0.0733
P Y   T  C |F   P Y |F   P T |F   P C |F 
 0.1816   0.5314   0.5222 
 0.0504
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Bayes’ Theorem
 Focus on the Project:
Use Bayes’ Theorem to get new probabilities
P S |Y   T  C   0.5575
P F |Y   T  C   0.4425
Z  : 6, 7, or 8 years, Bachelor’s, Normal
(indicates work out)
E Z   $2,341 ,000
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Bayes’ Theorem
 Focus on the Project:
We can look at a large range of years.
Look at range of value 5, 6, 7, 8, and 9 (2 years
more and less)
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Bayes’ Theorem
 Focus on the Project:
Use DCOUNT function with an extra “Years in
Business” heading
Years In
Former Bank Business
BR
>=5
Education Level
State Of
Economy
Loan Paid Years In
Back?
Business
yes
<=9
Same for “no”
Added a new column
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Bayes’ Theorem
 Focus on the Project:
From this you get 566 successful and 564 failed
records
Let Y  be a borrower with 5, 6, 7, 8, or 9 years
exper.
566
PY "| S   1470
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and
564
PY "|F   1779
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Bayes’ Theorem
 Focus on the Project:
P Y   T  C | S   P Y | S   P T | S   P C | S 
 0.3850   0.5301   0.5823 
 0.1189
P Y   T  C |F   P Y |F   P T |F   P C |F 
 0.3170   0.5314   0.5222 
 0.0880
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Bayes’ Theorem
 Focus on the Project:
Use Bayes’ Theorem to get new probabilities
P S |Y   T  C   0.5392
P F |Y   T  C   0.4608
Z  : 5, 6, 7, 8, or 9 years, Bachelor’s, Normal
E Z   $2,272,000 (indicates work out)
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Bayes’ Theorem
 Focus on the Project:
Since E Z  and E Z  both indicated a work out
while onlyE Z 
indicated a foreclosure, we
will work out a new payment schedule.
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