Conditional Probability
example:
 Toss a balanced die once and record the
number on the top face.
 Let E be the event that a 1 shows on the top
face.
 Let F be the event that the number on the
top face is odd.
– What is P(E)?
– What is the Probability of the event E if we are told
that the number on the top face is odd, that is,
we know that the event F has occurred?
Conditional Probability
Key idea: The original sample space no
longer applies.
 The new or reduced sample space is
S={1, 3, 5}
 Notice that the new sample space
consists only of the outcomes in F.
 P(E occurs given that F occurs) = 1/3
 Notation: P(E|F) = 1/3

Conditional Probability

Def. The conditional probability of E
given F is the probability that an event,
E, will occur given that another event,
F, has occurred
P( E  F )
P( E | F ) 
P( F )
if
P( F )  0
Conditional Probability
P( A  B)
P( A B) 
P( B)
A
S
B

If the outcomes of an experiment are
equally likely, then
number of outcomes in E  F
P( E | F ) 
number of outcomes in F

Example:
Earned degrees in the United States in
recent year
Female
Male
Total
B
616
529
1145
529
P( Male | B) 
 0.4620
1145
770
P( Male ) 
 0.4735
1626
M
194
171
365
P
30
44
74
D
16
26
42
Total
856
770
1626
P( E  F )
P( E | F ) 
P( F )
Conditional Probability can be rewritten as follows
P( E  F )  P( E | F ) * P( F )
Example:
E: dollar falls in value against the yen
F: supplier demands renegotiation of contract
P ( E )  0.40
P ( F | E )  0 .8
Find P( E  F )
P( E  F )  0.8 * 0.4  0.32
Independent Eventssec2
If the probability of the occurrence of event A
is the same regardless of whether or not an
outcome B occurs, then the outcomes A and
B are said to be independent of one another.
Symbolically, if
P( A | B)  P( A)
then A and B are independent events.
Independent Events
P( A  B)  P( A | B) P( B)
then we can also state the following
relationship for independent events:
P( A  B)  P( A) P( B)
if and only if
A and B are independent events.
Example
A coin is tossed and a single 6-sided die is
rolled. Find the probability of getting a
head on the coin and a 3 on the die.
 Probabilities:
P(head) = 1/2
P(3) = 1/6
P(head and 3) = 1/2 * 1/6 = 1/12

Independence Formula –3 events

Example:
If E, F, and G are independent, then
P( E  F  G)  P( E ) * P( F ) * P(G)
The Notion of Independence applied
to Conditional Probability

If E, F, and G are independent given that
an event H has occurred, then
P( E  F  G | H )  P( E | H ) * P( F | H ) * P(G | H )
Important


Independent Events vs. Mutually Exclusive
Events (Disjoint Events)
If two events are Independent,
P( A | B)  P( A)
P( A  B)  P( A) P( B)

If two events are Mutually Exclusive Events
then they do not share common outcomes
Focus on the Project
How can conditional probability help us
with the decision on whether or not to
attempt a loan work out?
How might our information about John
Sanders change this probability?
Focus on the Project
Recall:
Events
S- An attempted workout is a Success
F- An attempted workout is a Failure
P(S)=.464
P(F)=.536
How might our information about John
Sanders change this probability?
CalculationsExpected Values
More Events
Y- 7 years of experience
T- Bachelor’s Degree
C- Normal times
Conditional Probabilities
P(S|Y)=?
P(F|Y)=?
P(S|T)=?
P(F|T)=?
P(S|C)=?
P(F|C)=?
Each team will
have their client
data
Each team will
have to calculate
complementary formula
P(F|Y)=1- P(S|Y)
Indicates that the event occurred at the
given bank
Assumption
Similar clients
P( S | Y )  P( S BR | YBR ) 
P( S BR  YBR ) 
number in S BR and YBR
number of BR records
P( S BR | YBR ) 
P( S BR  YBR )
.
P(YBR )
P(YBR ) 
number in YBR
number of BR records
number in S BR and YBR
.
number in YBR
Recall-BR Bank
Range1
Ranges
Former Bank
Years In
Business
Education
Level
State Of
Economy
BR
Loan Paid
Back?
yes
Range2
Former Bank
BR
Years In
Business
Education
Level
State Of
Economy
7
Loan Paid
Back?
yes
Range3
Former Bank
Years In
Business
Education
Level
State Of
Economy
BR
Loan Paid
Back?
no
Range4
Former Bank
BR
Years In
Business
7
Education
Level
State Of
Economy
Loan Paid
Back?
no
Using DCOUNT
Counting
Number
Successful
1,470
Range1
Number
Successful
With Y
105
Range2
Number
Failed
Number
Failed Number
With Y With Y
1,779
Range3
134
239
Range4
105+134
P(S|Y) & P(F|Y) –BR Bank
BR Bank
Number with Number with Estimated
Estimated
S BR and Y BR
Y BR
P (S BR |Y BR ) P (F BR |Y BR )
105
239
Acadia Bank
Estimated
P (S |Y )
0.439
Estimated
P (F |Y )
0.561
0.439
0.561
ZY -The Money bank receives from loan work out attempt to a
borrower with 7 years experience
expected value of ZY.
E ( ZY )  $4,000,000  P( Z  $4,000,000)  $250,000  P( Z  $250,000)
 $4,000,000  P( S | Y )  $250,000  P( F | Y )
=4,000,000* .439 +250,000*.561
=$1,897,490
Analysis of E(Zy)?
Foreclosure value - $ 2,100,000
 E(Zy)=$ 1,897,490
 This piece of information E(Zy)
indicates FORECLOSURE

Decision?
Recall
 Bank Forecloses a loan if
Benefits of Foreclosure > Benefits of Workout

Bank enters a Loan Workout if
Expected Value Workout > Expected Value Foreclose
Similarly
You can calculate
E(Zt), E(Zc) for the Team Project1
 Do a Similar analysis using E(Zt), E(Zc)
RANDOM VARIABLES
Zt -The Money bank receives from loan work out attempt to a

borrower with Bachelor’s Degree
Zc -The Money bank receives from loan work out attempt to a
borrower during normal economy
Calculations
Conditional Probability
Recall Events
Y- 7 years of experience
T- Bachelor’s Degree
C- Normal times
Conditional Probabilities
P(Y|S)=?
P(Y|F)=?
P(T|S)=?
P(T|F)=?
P(C|S)=?
P(C|F)=?
Each team will
have their client
data
Each team will
have to calculate
Important – Here we cannot use the complementary formula
P(Y|S) & P(Y|F) –BR Bank
BR Bank
Number with Number with Estimated Number with Number with Estimated
Y BR and S BR
S BR
P (Y BR |S BR ) Y BR and F BR
F BR
P (Y BR |F BR )
105
1,470
Estimated
P(Y|S)
0.071
0.071
Estimated
P(Y|F)
0.075
134
1,779
0.075
P(Y|S) –BR Bank
P(YBR  S BR ) 
number in YBR and S BR
number of BR records
P( S BR ) 
number in S BR
number of BR records
number in YBR and S BR
P(YBR | S BR ) 
.
number in S BR
P(YBR|SBR)  105/1,470  0.071.
P(Y|S)  P(YBR|SBR)  0.071.
*Slide 20
Next step
P(Y|S)  0.071 (BR)
Similarly
Recall-The Notion of Independence
applied to Conditional Probability
P(T|S)  0.530 (Cajun)
P(C|S)  0.582 (Dupont)
We know Y, T, and C are independent events, even when they
are conditioned upon S or F. Hence,
P(Y  T  C|S) = P(Y|S)P(T|S)P(C|S)
 (0.071)(0.530)(0.582)  0.022
Similarly, can calculate
P(Y  T  C|F) = P(Y|F)P(T|F)P(C|F)
Next step
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?

Summary
Conditional Probabability Formula
P( E | F ) 
P( E  F )
P( F )
if
P( F )  0
If two events are Independent,
P ( A | B )  P ( A)
P ( A  B )  P ( A) * P ( B )
Independence Formula –3 events
P( E  F  G)  P( E ) * P( F ) * P(G)
The Notion of Independence applied to Conditional Probability
P( E  F  G | H )  P( E | H ) * P( F | H ) * P(G | H )
Summary
Calculations –Expected Value
 Calculations – Conditional Probability
