1. Bayes’Theorem
Special Type of Conditional Probability

2. Recall- Conditional Probability
 P(Y  T  C|S) will be used to calculate
P(S|Y  T  C)
 P(Y  T  C|F) will be used to calculate
P(F|Y  T  C)
 HOW?????
 We will learn in the next lesson?
 BAYES THEOREM

3. Definition of Partition
Let the events B1, B2, , Bn be non-empty subsets
of a sample space S for an experiment. The Bi’s
are a partition of S if the intersection of any two of
them is empty, and if their union is S. This may be
stated symbolically in the following way.
1. Bi  Bj = , unless i = j.
2. B1  B2    Bn = S.

4. Partition Example
S
B1 B2 B3

5. Example 1
Your retail business is considering holding
a sidewalk sale promotion next Saturday.
Past experience indicates that the
probability of a successful sale is 60%, if it
does not rain. This drops to 30% if it does
rain on Saturday. A phone call to the
weather bureau finds an estimated
probability of 20% for rain. What is the
probability that you have a successful
sale?

6. Example 1
Events
R- rains next Saturday
N -does not rain next Saturday.
A -sale is successful
U- sale is unsuccessful.
Given
P(A|N) = 0.6 and P(A|R) = 0.3.
P(R) = 0.2.
In addition we know R and N are complementary events
P(N)=1-P(R)=0.8
Our goal is to compute P(A).
)
R
N
( c


7. Using Venn diagram –Method1
Event A is the
disjoint union of
event R  A
&
event N  A
S=RN
R N
A
P(A) = P(R  A) + P(N  A)

8. P(A)- Probability that you have a
Successful Sale
We need P(R  A) and P(N  A)
Recall from conditional probability
P(R  A)= P(R )* P(A|R)=0.2*0.3=0.06
Similarly
P(N  A)= P(N )* P(A|N)=0.8*0.6=0.48
Using P(A) = P(R  A) + P(N  A)
=0.06+0.48=0.54

9. Let us examine P(A|R)
 Consider P(A|R)
 The conditional
probability that sale is
successful given that it
rains
 Using conditional
probability formula
)
R
(
P
)
A
R
(
P
)
R
|
A
(
P


S=RN
R N
A

10. Tree Diagram-Method 2
Bayes’, Partitions
Saturday
R
N
A R  A 0.20.3 = 0.06
A N  A 0.80.6 = 0.48
U R  U 0.20.7 = 0.14
U N  U 0.80.4 = 0.32
0.2
0.8
0.7
0.3
0.6
0.4
Probability
Conditional
Probability
Probability
Event
*Each Branch of the tree represents the intersection of two events
*The four branches represent Mutually Exclusive events
P(R ) P(A|R)
P(N ) P(A|N)

11. Method 2-Tree Diagram
Using P(A) = P(R  A) + P(N  A)
=0.06+0.48=0.54

12. Extension of Example1
Consider P(R|A)
The conditional probability that it rains given
that sale is successful
the How do we calculate?
Using conditional probability formula
)
N
(
P
)
N
|
A
(
P
)
R
(
P
)
R
|
A
(
P
)
R
(
P
)
R
|
A
(
P
)
A
(
P
)
A
R
(
P
)
A
|
R
(
P







8
0
6
0
2
0
3
0
2
0
3
0
.
.
.
.
.
.




=
= 0.1111
*show slide 7

13. Example 2
 In a recent New York Times article, it was
reported that light trucks, which include
SUV’s, pick-up trucks and minivans,
accounted for 40% of all personal vehicles on
the road in 2002. Assume the rest are cars.
Of every 100,000 car accidents, 20 involve a
fatality; of every 100,000 light truck accidents,
25 involve a fatality. If a fatal accident is
chosen at random, what is the probability the
accident involved a light truck?

14. Example 2
Events
C- Cars
T –Light truck
F –Fatal Accident
N- Not a Fatal Accident
Given
P(F|C) = 20/10000 and P(F|T) = 25/100000
P(T) = 0.4
In addition we know C and T are complementary events
P(C)=1-P(T)=0.6
Our goal is to compute the conditional probability of a Light truck
accident given that it is fatal P(T|F).
)
T
C
( c


15. Goal P(T|F)
Consider P(T|F)
Conditional probability
of a Light truck accident
given that it is fatal
Using conditional
probability formula
)
F
(
P
)
F
T
(
P
)
F
|
T
(
P


S=CT
C T
F

16. P(T|F)-Method1
Consider P(T|F)
Conditional probability of a Light truck
accident given that it is fatal
How do we calculate?
Using conditional probability formula
)
C
(
P
)
C
|
F
(
P
)
T
(
P
)
T
|
F
(
P
)
T
(
P
)
T
|
F
(
P
)
F
(
P
)
F
T
(
P
)
F
|
T
(
P







)
.
)(
.
(
)
.
)(
.
(
)
.
)(
.
(
6
0
0002
0
4
0
00025
0
4
0
00025
0

=
= 0.4545

17. Tree Diagram- Method2
Vehicle
C
T
F C  F 0.6 0.0002 = .00012
F T  F 0.40.00025= 0.0001
N C  N 0.6 0.9998 = 0.59988
N T N 0.40.99975= .3999
0.6
0.4
0.9998
0.0002
0.00025
0.99975
Probability
Conditional
Probability
Probability
Event

18. Tree Diagram- Method2
)
F
C
(
P
)
F
T
(
P
)
F
T
(
P
)
F
(
P
)
F
T
(
P
)
F
|
T
(
P







)
.
)(
.
(
)
.
)(
.
(
)
.
)(
.
(
6
0
0002
0
4
0
00025
0
4
0
00025
0

=
= 0.4545

19. Partition
S
B1 B2 B3
A
)
(
)
|
(
)
(
)
|
(
)
(
)
|
(
)
( 3
3
2
2
1
1 B
P
B
A
P
B
P
B
A
P
B
P
B
A
P
A
P 






20. Law of Total Probability
))
(
)
(
)
((
))
(
(
)
(
)
(
2
1
2
1
n
n
B
A
B
A
B
A
P
B
B
B
A
P
S
A
P
A
P
















Let the events B1, B2, , Bn partition the finite discrete sample
space S for an experiment and let A be an event defined on S.

21. Law of Total Probability
























n
i
i
i
n
n
n
n
B
P
B
A
P
B
P
B
A
P
B
P
B
A
P
B
P
B
A
P
B
A
P
B
A
P
B
A
P
B
A
B
A
B
A
P
1
2
2
1
1
2
1
2
1
)
(
)
|
(
)
(
)
|
(
)
(
)
|
(
)
(
)
|
(
)
(
)
(
)
(
))
(
)
(
)
((



.
)
(
)
|
(
)
(
1




n
i
i
i B
P
B
A
P
A
P

22. Bayes’ Theorem
 Suppose that the events B1, B2, B3, . . . , Bn
partition the sample space S for some
experiment and that A is an event defined on
S. For any integer, k, such that
we have
n
k 

1
     
   


 n
j
j
j
k
k
k
B
P
B
A
P
B
P
B
A
P
A
B
P
1
|
|
|

23. Focus on the Project
Recall
 P(Y  T  C|S) will be used to calculate
P(S|Y  T  C)
 P(Y  T  C|F) will be used to calculate
P(F|Y  T  C)

24. How can Bayes’ Theorem help us with the
decision on whether or not to attempt a loan work
out?
Partitions
1. Event S
2. Event F
Given
P(Y  T  C|S)
P(Y  T  C|F)
Need
P(S|Y  T  C)
P(F|Y  T  C)

25. Using Bayes Theorem
P(S|Y  T  C)  0.477
 
)
536
.
0
(
)
021
.
0
(
)
464
.
0
(
)
022
.
0
(
)
464
.
0
(
)
022
.
0
(
)
(
)
|
(
)
(
)
|
(
)
(
)
|
(
|


















F
P
F
C
T
Y
P
S
P
S
C
T
Y
P
S
P
S
C
T
Y
P
C
T
Y
S
P
 
.
)
536
.
0
(
)
021
.
0
(
)
464
.
0
(
)
022
.
0
(
)
536
.
0
(
)
021
.
0
(
)
(
)
|
(
)
(
)
|
(
)
(
)
|
(
|


















F
P
F
C
T
Y
P
S
P
S
C
T
Y
P
F
P
F
C
T
Y
P
C
T
Y
F
P
LOAN FOCUS EXCEL-BAYES
P(F|Y  T  C)  0.523

26. RECALL
 Z is the random variable giving the amount of money,
in dollars, that Acadia Bank receives from a future
loan work out attempt to borrowers with the same
characteristics as Mr. Sanders, in normal times.
)
523
.
0
(
000
,
250
$
)
477
.
0
(
000
,
000
,
4
$
)
|
(
000
,
250
$
)
|
(
000
,
000
,
4
$
)
000
,
250
$
(
000
,
250
$
)
000
,
000
,
4
$
(
000
,
000
,
4
$
)
(


















C
T
Y
F
P
C
T
Y
S
P
Z
P
Z
P
Z
E
E(Z)  $2,040,000.

27. Decision
EXPECTED VALUE OF A WORKOUT=E(Z)  $2,040,000
FORECLOSURE VALUE- $2,100,000
RECALL
FORECLOSURE VALUE> EXPECTED VALUE OF A WORKOUT
DECISION
FORECLOSURE

28. Further Investigation I
 let Y  be the event that a borrower has 6, 7, or 8 years of
experience in the business.
Using the range
Let Z be the random variable giving the amount of
money, in dollars, that Acadia Bank receives from a
future loan work out attempt to borrowers with Y  and a
Bachelor’s Degree, in normal times. When all of the
calculations are redone, with Y  replacing Y, we find that P(Y 
 T  C|S)  0.073 and P(Y   T  C|F)  0.050.
Former Bank
Years In
Business
Years In
Business
Education
Level
State Of
Economy
Loan Paid
Back
BR >=6 <=8 yes

29. Calculations
P(Y   T  C|S)  0.073
P(Y   T  C|F)  0.050
P(S|Y   T  C)  0.558
P(F|Y   T  C)  0.442
The expected value of Z is E(Z )  $2,341,000.
Since this is above the foreclosure value of
$2,100,000, a loan work out attempt is
indicated.

30. Further Investigation II
 Let Y" be the event that a borrower has 5, 6,
7, 8, or 9 years of experience in the business
 Let Z" be the random variable giving the
amount of money, in dollars, that Acadia Bank
receives from a future loan work out attempt
to borrowers with 5, 6, 7, 8, or 9 years
experience and a Bachelor's Degree, in
normal times. Redoing our work yields the
follow results.

31. Similarly can calculate E(Z  )
 Make at a decision- Foreclose vs. Workout
 Data indicates Loan work out

32. Close call for Acadia Bank loan officers
Based upon all of our calculations, we
recommend that Acadia Bank enter into a
work out arrangement with Mr. Sanders.