Conditional Probability
Conditional Probability
 A newspaper editor has 120 letters
from irate readers about the firing of
a high school basketball coach.
 The letters are divided among parents
and students, in support of or against
the coach
 They have space to print only one of
these letters.
Conditional Probability
 The break down of the letters:
Written by
students
Written by
parents
Total
Support coach
16
44
60
Against coach
8
52
60
Total
24
96
120
 What are the chances that a student letter
supporting the coach will be chosen?
Conditional Probability
 Let’s look at a Venn Diagram:
 Let C: event the letter is from a student
 Let T: event the letter favors the coach
PT  C C 
C
8
120
PC  T
C

T
16
120
44
120
52
120
PC  T 

P T  C 
C

Conditional Probability
 From the Venn Diagram:
 Slim chance a student letter supporting the
coach will be printed: P C  T   16  0.133
120
 Could be unfair: student letters support the
coach by a ratio of 2 : 1
PT  C  0.133 2


 This fact is evident since
PC 
0.20 3
Conditional Probability
 What does
PT  C  0.133 2




PC
0.20 3
tell us?
 Given the letter came from a student,
the chance it supports the coach is twothirds
 In other words: 20% of the letters came
from students. Of those, two-thirds
were in favor of the coach
Conditional Probability
 Notice previous Venn Diagram probabilities
were all relative to sample space:
 For example: P T  C  

16
120
P T  C 
P C 
looks at probability a letter supports
a teacher based on a reduced sample
 16 
space, student letters only


P T  C   120  16 2



P C 
 24  24 3


120


Conditional Probability
 What does this mean?
 Knowing some info beforehand can change a
probability
 Ex: Probability of rolling a 12 with 2 dice is
1/36, but if you know the first die is a 4, the
probability is 0. If the first die is a 6, the
probability is 1/6
 Determining a probability after some information
is known is called conditional probability
Conditional Probability
This is a conditional
probability
 Notation
 PE | F  means the probability of E
happening given that F has already
occurred
 Definition
P E  F 
, where PF   0
 P E | F  
P F 
Conditional Probability
P E  F 
 The formula P E | F  
P F 
 PE  F   PF   PE | F 
 PE  F   PE   PF | E 
Note:
PE | F   PF | E 
implies:
Notice the
reversal of
the events
E and F
Very Important!
These are two
different things.
They aren’t always
equal.
Conditional Probability
 Ex: Suppose 22% of Math 115A
students plan to major in accounting
(A) and 67% on Math 115A students
are male (M). The probability of being
a male or an accounting major in Math
115A is 75%. Find P A | M and PM | A.
Conditional Probability
 Sol:
P A  M 
P A | M  
PM 
First find P A  M 
P A  M   P A  PM   P A  M 
 0.22  0.67  0.75
 0.14
Conditional Probability
 Sol:
P A  M 
P A | M  
P M 
0.14

0.67
 0.2090
Conditional Probability
 Sol:
PM  A
P M | A 
P  A
0.14

0.22
 0.6364
Conditional Probability
 Sometimes one event has no effect
on another
 Example: flipping a coin twice
 Such events are called independent
events
 Definition: Two events E and F are
independent if PE | F   PE  or PF | E   PF 
Conditional Probability
 Implications:

P E | F   P E 
P E  F 
 P E 
P F 
P E  F   P E   P F 
So, two events
E and F are
independent if
this is true.
Conditional Probability
 The property of independence can be
extended to more than two events:
PE1  E2    En   PE1   PE2     PEn 
assuming that E1 , E2 , , En are all
independent.
Conditional Probabilities
 INDEPENDENT EVENTS AND
MUTUALLY EXCLUSIVE EVENTS ARE
NOT THE SAME
 Mutually exclusive: PE  F   0
 Independence: P E | F   P E 
P E  F   P E   P F 
Conditional Probability
 Ex: Suppose we roll toss a fair coin 4
times. Let A be the event that the
first toss is heads and let B be the
event that there are exactly three
heads. Are events A and B
independent?
 HHHH , HHHT , HHTH , HHTT ,
 HTHH , HTHT , HTTH , HTTT , 


S 

THHH , THHT , THTH , THTT , 
TTHH , TTHT , TTTH , TTTT

Conditional Probability
 Soln:
For A and B to be independent,
P A  B  P A  PB
P A  168 
1
2
and PB  
P A  B   163  0.1875
 HHHH , HHHT , HHTH , HHTT ,
 HTHH , HTHT , HTTH , HTTT , 


S 

THHH
,
THHT
,
THTH
,
THTT
,


TTHH , TTHT , TTTH , TTTT

4
1
16
4
P A  PB   12  14  18  0.125

Different, so
dependent
Conditional Probability
 Ex: Suppose you apply to two
graduate schools: University of Arizona
and Stanford University. Let A be the
event that you are accepted at Arizona
and S be the event of being accepted
at Stanford. If P A  0.7 and PS   0.2 ,
and your acceptance at the schools is
independent, find the probability of
being accepted at either school.
Conditional Probability
 Soln: Find P A  S  .
P A  S   P A  PS   P A  S 
Since A and S are independent,
P A  S   P A  PS 
 0.7  0.2
 0.14
Conditional Probability
 Soln:
P A  S   P A  PS   P A  S 
 0.7  0.2  0.14
 0.76
There is a 76% chance of being
accepted by a graduate school.
Conditional Probability
 Independence holds for complements
as well.
 Ex: Using previous example, find the
probability of being accepted by
Arizona and not by Stanford.
Conditional Probability


 Soln: Find P A  S C .


 
P A  S C  P  A  P S C
 0.7   1  0.2
 0.7  0.8
 0.56
Conditional Probability
 Ex: Using previous example, find the
probability of being accepted by
exactly one school.
 Sol: Find probability of Arizona and not
Stanford or Stanford and not Arizona.

 
P A  S C  S  AC

Conditional Probability
 Sol: (continued)
Since Arizona and Stanford are
mutually exclusive (you can’t attend
both universities)

P A S
C
  S  A   PA  S   PS  A 
C
C
 
C
 
 P A  P S C  PS   P AC
(using independence)
Conditional Probability
 Soln: (continued)

 


 

 P A  PS   PS   PA 
P A  S C  S  AC  P A  S C  P S  AC
C
 0.7   0.8  0.2  0.3
 0.56  0.06
 0.62
C
Conditional Probability
 Independence holds across conditional
probabilities as well.
 If E, F, and G are three events with E
and F independent, then
PE  F | G  PE | G  PF | G
Conditional Probability
 Focus on the Project:
Recall: PS   0.464 and PF   0.536
However, this is for a general borrower
Want to find probability of success for
our borrower
Conditional Probability
 Focus on the Project:
Start by finding PS | Y  and PF | Y 
We can find expected value of a loan
work out for a borrower with 7 years
of experience.
Conditional Probability
 Focus on the Project:
To find PS | Y  we use the info from the
DCOUNT function
PS  Y 
PS | Y  
PY 
This can be approximated by counting the
number of successful 7 year records divided
by total number of 7 year records
Conditional Probability
 Focus on the Project:
Technically, we have the following:
PS BR  YBR 
PS | Y   PS BR | YBR  
PYBR 
So, PS | Y  
105
239
 0.4393
Why “technically”? Because
we’re assuming that the loan
workouts BR bank made were
made for similar types of
borrowers for the other three.
So we’re extrapolating a
probability from one bank and
using it for all the banks.
Conditional Probability
 Focus on the Project:
Similarly,
P F  Y 
P F | Y  
PY 
This can be approximated by counting
the number of failed 7 year records
divided by total number of 7 year
records
Conditional Probability
 Focus on the Project:
Technically, we have the following:
PFBR  YBR 
PF | Y   PFBR | YBR  
PYBR 
So, PF | Y  
134
239
 0.5607
Conditional Probability
 Focus on the Project:
Let Z Y be the variable giving the value of
a loan work out for a borrower with 7
years experience
Find EZY 
Conditional Probability
 Focus on the Project:
E Z Y   Success  Prob. Success  Failure  Prob. Failure
 4,000,000  0.4393  250,0000.5607 
 $1,897,000
This indicates that looking at only the
years of experience, we should
foreclose (guaranteed $2.1 million)
Conditional Probability
 Focus on the Project:
Of course, we haven’t accounted for the
other two factors (education and
economy)
Using similar calculations, find the
following:
PS | T , PF | T , PS | C , and PF | C 
Conditional Probability
 Focus on the Project:
510
PS | T   1154
 0.4419
644
PF | T   1154
 0.5581
807
PS | C   1547
 0.5217
740
PF | C   1547
 0.4783
Conditional Probability
 Focus on the Project:
Let Z T represent value of a loan work
out for a borrower with a Bachelor’s
Degree
Let Z C represent value of a loan work
out for a borrower with a loan during a
Normal economy
Conditional Probability
 Focus on the Project:
Find EZT  and E Z C 
E Z T   Success  Prob. Success  Failure  Prob. Failure
 4,000,000  0.4419  250,0000.5581
 $1,907,000
E Z C   Success  Prob. Success  Failure  Prob. Failure
 4,000,000  0.5217   250,0000.4783
 $2,206,000
Conditional Probability
 Focus on the Project:
 So, two of the three individual
expected values indicates a
foreclosure:
EZ Y   $1,897,000
EZT   $1,907,000
E Z C   $2,206,000
Conditional Probability
 Focus on the Project:
Can’t use these expected values for the
final decision
None has all 3 characteristics
combined:
EZY for example has all education
levels and all economic conditions
included
Conditional Probability
 Focus on the Project:
Now perform some calculations to be
used later
PY | S , PT | S , and PC | S 
We will use the given bank data:
That is PY | S  is really PYBR | S BR 
and so on…
Conditional Probability
 Focus on the Project:
We can find PY  T  C | S 
PY  T  C | S   PY | S   PT | S   PC | S 
since Y, T, and C are independent
Also
PY  T  C | F   PY | F   PT | F   PC | F 
Conditional Probability
 Focus on the Project:
PY | S   PYBR | S BR 
number in YBR and S BR

number in S BR
105

1470
 0.0714
Similarly:
510
PT | S  
 0.5301
962
807
PC | S  
 0.5823
1386
Conditional Probability
 Focus on the Project:
PY  T  C | S   PY | S   PT | S   PC | S 
 0.0714  0.5301  0.5823
 0.0220
Conditional Probability
 Focus on the Project:
134
PY | F  
 0.0753
1779
644
PT | F  
 0.5314
1212
740
PC | F  
 0.5222
1417
Conditional Probability
 Focus on the Project:
PY  T  C | F   PY | F   PT | F   PC | F 
 0.0753  0.5314  0.5222
 0.0209
Conditional Probability
 Focus on the Project:
Now that we have found PY  T  C | S 
and PY  T  C | F  we will use these
values to find PS | Y  T  C  and
PF | Y  T  C 