conditional probability. a newspaper editor has 120 letters from irate readers about the firing of...

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

Support coach 16 44 60

Against coach 8 52 60

Total 24 96 120

Let’s look at a Venn Diagram: Let C: event the letter is from a student Let T: event the letter favors the coach

52 CTCP

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

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

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

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

Notation means the probability of E

happening given that F has already occurred

Definition

0 where, |

FEPFEP

This is a conditional probability

The formula implies:

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.

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 |

First find

75.067.022.0

MAPMPAPMAP

2090.067.0

MAPMAP

6364.022.0

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 |

Implications:

FPEPFEP

So, two events E and F are independent if this is true.

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:

FPEPFEP

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

Soln:For A and B to be independent,

Different, sodependent

TTTTTTTHTTHTTTHH

THTTTHTHTHHTTHHH

HTTTHTTHHTHTHTHH

HHTTHHTHHHHTHHHH

168 AP 4

1164 BP

1875.0163 BAP

BPAPBAP

125.081

21 BPAP

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

Soln: Find .

Since A and S are independent,

SAPSPAPSAP

2.07.0

SPAPSAP

There is a 76% chance of being accepted by a graduate school.

14.02.07.0

SAPSPAPSAP

Independence holds for complements as well.

Ex: Using previous example, find the probability of being accepted by Arizona and not by Stanford.

Soln: Find .

8.07.0

2.017.0

CC SPAPSAP

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

Sol: (continued)Since Arizona and Stanford are mutually exclusive (you can’t attend both universities)

(using independence)

CCCC ASPSAPASSAP

CC APSPSPAP

Soln: (continued)

06.056.0

3.02.08.07.0

APSPSPAP

ASPSAPASSAP

Independence holds across conditional probabilities as well.

If E, F, and G are three events with E and F independent, then

GFPGEPGFEP |||

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

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

Focus on the Project: Technically, we have the following:

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.

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

Focus on the Project: Technically, we have the following:

BRBRBRBR YP

YFPYFPYFP

5607.0| 239134 YFP

Focus on the Project: Let be the variable giving the value of a loan work out for a borrower with 7 years experience

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

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,|,|,|

5581.0| 1154644 TFP

4419.0| 1154510 TSP

5217.0| 1547807 CSP

4783.0| 1547740 CFP

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

Focus on the Project: Find and TZE CZE

000,907,1$

5581.0000,2504419.0000,000,4

000,206,2$

4783.0000,2505217.0000,000,4

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

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

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 |

Focus on the Project: We can find

since Y, T, and C are independent

SCPSTPSYPSCTYP |||| SCTYP |

FCPFTPFYPFCTYP ||||

Similarly:

0714.01470

in number

and in number

SYPSYP

5823.01386

807| SCP

5301.0962

510| STP

0220.0

5823.05301.00714.0

SCPSTPSYPSCTYP

0753.01779

134| FYP

5222.01417

740| FCP

5314.01212

644| FTP

0209.0

5222.05314.00753.0

FCPFTPFYPFCTYP

Now that we have found and we will use these values to find and

SCTYP | FCTYP |

CTYSP | CTYFP |

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