conditional probability. a newspaper editor has 120 letters from irate readers about the firing of...
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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:
What are the chances that a student letter supporting the coach will be chosen?
Written by students
Written by parents
Total
Support coach 16 44 60
Against coach 8 52 60
Total 24 96 120
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
C T
120
16
120
8
120
44
120
52 CTCP
CCTP
TCP
CCTP
Conditional Probability From the Venn Diagram:
Slim chance a student letter supporting the coach will be printed:
Could be unfair: student letters support the coach by a ratio of 2 : 1
This fact is evident since
133.0120
16TCP
3
2
20.0
133.0
CP
CTP
Conditional Probability
What does tell us?
Given the letter came from a student, the chance it supports the coach is two-thirds
In other words: 20% of the letters came from students. Of those, two-thirds were in favor of the coach
3
2
20.0
133.0
CP
CTP
Conditional Probability Notice previous Venn Diagram probabilities
were all relative to sample space:
For example:
looks at probability a letter supports a teacher based on a reduced sample space, student letters only
CPCTP
120
16CTP
3
2
24
16
12024
12016
CP
CTP
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
Notation means the probability of E
happening given that F has already occurred
Definition
FEP |
0 where, |
FPFP
FEPFEP
This is a conditional probability
Conditional Probability
The formula implies:
|FP
FEPFEP
FEPFPFEP |
EFPEPFEP |
Notice the reversal of the events E and F
Note: EFPFEP || 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 and .
MAP | AMP |
Conditional Probability
Sol:
First find
MPMAP
MAP
|
MAP
14.0
75.067.022.0
MAPMPAPMAP
Conditional Probability
Sol:
2090.067.0
14.0
|
MP
MAPMAP
Conditional Probability
Sol:
6364.022.0
14.0
|
AP
AMPAMP
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 or EPFEP | FPEFP |
Conditional Probability
Implications:
FPEPFEP
EPFP
FEP
EPFEP
|
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:
assuming that are all independent.
nn EPEPEPEEEP 2121
nEEE ,,, 21
Conditional Probabilities
INDEPENDENT EVENTS AND MUTUALLY EXCLUSIVE EVENTS ARE NOT THE SAME
Mutually exclusive:
Independence:
0FEP
FPEPFEP
EPFEP
|
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?
TTTTTTTHTTHTTTHH
THTTTHTHTHHTTHHH
HTTTHTTHHTHTHTHH
HHTTHHTHHHHTHHHH
S
,,,
,,,,
,,,,
,,,,
Conditional Probability
Soln:For A and B to be independent,
and
Different, sodependent
TTTTTTTHTTHTTTHH
THTTTHTHTHHTTHHH
HTTTHTTHHTHTHTHH
HHTTHHTHHHHTHHHH
S
,,,
,,,,
,,,,
,,,,
21
168 AP 4
1164 BP
1875.0163 BAP
BPAPBAP
125.081
41
21 BPAP
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 and , and your acceptance at the schools is independent, find the probability of being accepted at either school.
7.0AP 2.0SP
Conditional Probability
Soln: Find .
Since A and S are independent,
SAP
SAPSPAPSAP
14.0
2.07.0
SPAPSAP
Conditional Probability
Soln:
There is a 76% chance of being accepted by a graduate school.
76.0
14.02.07.0
SAPSPAPSAP
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 .
56.0
8.07.0
2.017.0
CC SPAPSAP
CSAP
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.
CC ASSAP
Conditional Probability
Sol: (continued)Since Arizona and Stanford are mutually exclusive (you can’t attend both universities)
(using independence)
CCCC ASPSAPASSAP
CC APSPSPAP
Conditional Probability
Soln: (continued)
62.0
06.056.0
3.02.08.07.0
CC
CCCC
APSPSPAP
ASPSAPASSAP
Conditional Probability
Independence holds across conditional probabilities as well.
If E, F, and G are three events with E and F independent, then
GFPGEPGFEP |||
Conditional Probability
Focus on the Project: Recall: and
However, this is for a general borrower
Want to find probability of success for our borrower
464.0SP 536.0FP
Conditional Probability
Focus on the Project: Start by finding and
We can find expected value of a loan work out for a borrower with 7 years of experience.
YSP | YFP |
Conditional Probability Focus on the Project:
To find we use the info from the DCOUNT function
This can be approximated by counting the number of successful 7 year records divided by total number of 7 year records
YSP |
YPYSP
YSP
|
Conditional Probability
Focus on the Project: Technically, we have the following:
So,
BR
BRBRBRBR YP
YSPYSPYSP
||
4393.0| 239105 YSP
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,
This can be approximated by counting the number of failed 7 year records divided by total number of 7 year records
YPYFP
YFP
|
Conditional Probability
Focus on the Project: Technically, we have the following:
So,
BR
BRBRBRBR YP
YFPYFPYFP
||
5607.0| 239134 YFP
Conditional Probability
Focus on the Project: Let be the variable giving the value of a loan work out for a borrower with 7 years experience
Find
YZ
YZE
Conditional Probability
Focus on the Project:
This indicates that looking at only the years of experience, we should foreclose (guaranteed $2.1 million)
000,897,1$
5607.0000,2504393.0000,000,4
Failure Prob. Failure Success Prob. Success
YZE
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:
CFPCSPTFPTSP | and,|,|,|
Conditional Probability
Focus on the Project:
5581.0| 1154644 TFP
4419.0| 1154510 TSP
5217.0| 1547807 CSP
4783.0| 1547740 CFP
Conditional Probability
Focus on the Project: Let represent value of a loan work out for a borrower with a Bachelor’s Degree
Let represent value of a loan work out for a borrower with a loan during a Normal economy
TZ
CZ
Conditional Probability
Focus on the Project: Find and TZE CZE
000,907,1$
5581.0000,2504419.0000,000,4
Failure Prob. Failure Success Prob. Success
TZE
000,206,2$
4783.0000,2505217.0000,000,4
Failure Prob. Failure Success Prob. Success
CZE
Conditional Probability
Focus on the Project: So, two of the three individual
expected values indicates a foreclosure:
000,897,1$YZE
000,206,2$CZE
000,907,1$TZE
Conditional Probability Focus on the Project:
Can’t use these expected values for the final decision
None has all 3 characteristics combined:
for example has all education levels and all economic conditions included
YZE
Conditional Probability Focus on the Project:
Now perform some calculations to be used later
We will use the given bank data:That is is reallyand so on…
SCPSTPSYP | and,|,|
SYP | BRBR SYP |
Conditional Probability
Focus on the Project: We can find
since Y, T, and C are independent
Also
SCPSTPSYPSCTYP |||| SCTYP |
FCPFTPFYPFCTYP ||||
Conditional Probability
Focus on the Project:
Similarly:
0714.01470
105
in number
and in number
||
BR
BRBR
BRBR
S
SY
SYPSYP
5823.01386
807| SCP
5301.0962
510| STP
Conditional Probability
Focus on the Project:
0220.0
5823.05301.00714.0
||||
SCPSTPSYPSCTYP
Conditional Probability
Focus on the Project:
0753.01779
134| FYP
5222.01417
740| FCP
5314.01212
644| FTP
Conditional Probability
Focus on the Project:
0209.0
5222.05314.00753.0
||||
FCPFTPFYPFCTYP
Conditional Probability
Focus on the Project:
Now that we have found and we will use these values to find and
SCTYP | FCTYP |
CTYSP | CTYFP |