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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
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.
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?
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).
)RN( c
Using Venn diagram –Method1
Event A is the
disjoint union of
event R A
&
event N A
S=RN
R N
A
P(A) = P(R A) + P(N A)
P(A)- Probability that you have a Successful SaleWe 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
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
)AR(P)R|A(P
S=RN
R N
A
Tree Diagram-Method 2Bayes’, Partitions
Saturday
R
N
A R A 0.20.3 = 0.06
A N A 0.80.6 = 0.48
U R U 0.20.7 = 0.14
U N U 0.80.4 = 0.32
0.2
0.8
0.7
0.3
0.6
0.4
ProbabilityConditionalProbability
ProbabilityEvent
*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)
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
)AR(P)A|R(P
80602030
2030
....
..
=
= 0.1111*show slide 7
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?
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).
)TC( c
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
)FT(P)F|T(P
S=CT
C T
F
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
)FT(P)F|T(P
).)(.().)(.(
).)(.(
600002040000250
40000250
=
= 0.4545
Tree Diagram- Method2
Vehicle
C
T
F C F 0.6 0.0002 = .00012
F T F 0.40.00025= 0.0001
N C N 0.6 0.9998 = 0.59988
N T N 0.40.99975= .3999
0.6
0.4
0.9998
0.0002
0.00025
0.99975
ProbabilityConditionalProbability
ProbabilityEvent
Tree Diagram- Method2
)FC(P)FT(P
)FT(P
)F(P
)FT(P)F|T(P
).)(.().)(.(
).)(.(
600002040000250
40000250
=
= 0.4545
Law of Total Probability
))()()((
))((
)()(
21
21
n
n
BABABAP
BBBAP
SAPAP
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.
Law of Total Probability
n
iii
nn
n
n
BPBAP
BPBAPBPBAPBPBAP
BAPBAPBAP
BABABAP
1
2211
21
21
)()|(
)()|()()|()()|(
)()()(
))()()((
.)()|()(1
n
iii BPBAPAP
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
nk 1
n
jjj
kkk
BPBAP
BPBAPABP
1
|
||
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)
How can Bayes’ Theorem help us with thedecision 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)
Using Bayes Theorem
P(S|Y T C) 0.477
)536.0()021.0()464.0()022.0()464.0()022.0(
)()|()()|()()|(
|
FPFCTYPSPSCTYPSPSCTYP
CTYSP
.)536.0()021.0()464.0()022.0(
)536.0()021.0(
)()|()()|()()|(
|
FPFCTYPSPSCTYPFPFCTYP
CTYFP
LOAN FOCUS EXCEL-BAYES
P(F|Y T C) 0.523
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$(000,000,4$)(
CTYFPCTYSP
ZPZPZE
E(Z) $2,040,000.
Decision
EXPECTED VALUE OF A WORKOUT=E(Z) $2,040,000
FORECLOSURE VALUE- $2,100,000
RECALL FORECLOSURE VALUE> EXPECTED VALUE OF A WORKOUT
DECISIONFORECLOSURE
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 BankYears In Business
Years In Business
Education Level
State Of Economy
Loan Paid Back
BR >=6 <=8 yes
Calculations
P(Y T C|S) 0.073 P(Y T C|F) 0.050P(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.
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.
Similarly can calculate E(Z ) Make at a decision- Foreclose vs. Workout Data indicates Loan work out