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bp Bayes Theorem lecture – a revised conditional probability 9/22/

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P (A) is an example of a simple (marginal) probability. A simple probability is the
summation of all possible joint probabilities. P (A) = P (A B) + P (A C) + P (A D)
+ ...+ P (A Z). That is to say the probability of A is the summation of all the ways A
can occur with all the other things which may occur when A occurs.
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For example, the probability of picking a queen from a deck of cards P (Q) = 4/
because we know there are 52 cards in a deck of cards and of the 52 cards 4 are queens.
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However, we could break this simple probability into two joint probabilities, i.e. P (Q) =
P (Q R) + P (Q B) = 2/52+2/52 = 4/52. Here we explain the probability of picking a
queen as the sum of the probability of picking a red queen plus the probability of picking
a black queen.
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We could break it down further and say the probability of picking a queen is equal to the
probability of picking the queen of hearts plus the probability of picking the queen of
diamonds plus the probability of picking the queen of clubs plus the probability picking
the queen of spades; P (Q) = P (Q H) + P (Q D) + P (Q C) + P (Q S) = 1/
+1/52 +1/52 +1/52 = 4/52.
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There purpose of going through this exercise is to make Bayes formula (p156 of the 16th
ed) a little easier to understand.
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Remember
P (A B) referred to as the probability of A and B, or the probability of A
intersection B, is an example of a joint probability. (multiplication rule applies).
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P (A|B) referred to as the probability of A given is an example of a conditional
probability
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The big distinction between a conditional probability and joint probability is that with a
conditional probability you know something has occurred which alters the likelihood of
the outcome you are searching for. In expressing the probability as a ratio, the
denominator will be the same for a simple (marginal) and a joint probability, but be
different for the conditional probability.
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If you have 30 cups of coffee with the lids on and you want to determine the probability
of randomly selecting a cup with cream P(C) (which is a simple or marginal
probability), the probability (expressed as a ratio) will be the number of coffees with
cream (in the numerator) divided by 30 (in the denominator) which is the total number of
coffees you are selecting from.
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If you want to determine the probability of randomly selecting a cup with cream and
sugar P(C S) (which is a joint probability), the probability (expressed as a ratio) will
be the number of coffees with cream and sugar (in the numerator) divided by 30 (in the
denominator) which is the total number of coffees you are selecting from.
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However, if you select a coffee and take the lid off the cup and you see it has cream in it and you now what to determine the probability it has sugar in it (which is a conditional probability the event is conditioned by the fact that you now know you have cream in the coffee) the probability (expressed as a ratio) will be the number of coffees with cream and sugar (in the numerator) divided by the number of coffees which have cream in them (in the denominator). The denominator will not be 30 as it was for the simple or joint events, but rather some subset of the total 30 coffees, i.e. the number of coffees with cream.

The Conditional Probability formula involves dividing a joint probability by a simple probability.

The Joint Probability formula involves multiplying a conditional probability by a simple probability.

P (A B) = P (A|B) P (B) or P (B|A) P (A)

A Conditional Probability takes into account information about the occurrence of one event to predict the probability of another event.

In solving for the conditional probability P (A|B), the probability of the event to the right of the slash line (i.e.B) always appears in the denominator in the conditional probability formula. P (A|B) = P (A B) / P (B) P (B|A) = P (A B) / P (A)

Bayes Theorem is a revised conditional probability. The formula on page 156 ( 16 th ed.) assumes that B can happen in conjunction with two other things happening simultaneously; A 1 and A 2. The formula skips a few steps but can be derived if we use our basic probability rules.

- P (A 1 |B) = P (A 1 B) / P (B)

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Where P (B) = P (B A 1 ) + P (B A 2 ), thus
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- P (A 1 |B) = P (A 1 B) / [P (B A 1 ) + P (B A 2 )]

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Where P (A 1 B) and P (B A 1 ) = P (A 1 ) P (B|A 1 ); and
Where P (B A 2 ) = P (A 2 ) P (B|A 2 )
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- Therefore P (A 1 |B) = [P (A 1 ) P (B|A 1 )] / [P (A 1 ) P (B|A 1 ) + P (A 2 ) P (B|A 2 )]

An example: An electronics company has an engineering position open. The probability that an applicant is capable is .70. Each applicant is given a written and oral exam. A capable applicant passes with the probability of .90, while an incapable applicant passes

with a probability of .40. Let C represent capable; NC represent not capable; P represent passing test; and NP represent not passing the test.

Each of the numbers provided in the problem must be represented using one of our four basic probability formulas; P (A); P (A U B); P (A B); or P (A|B)

The information provided is: P (C) =. P (NC) = .30 found by: 1-.70, assuming an applicant is either capable or not capable. P (P|C) =.90 i.e. A capable applicant passes with the probability of .90 is analogous to saying the probability an applicant passes if the applicant is capable is .9 0 P (P|NC) = .40 i.e. An incapable (non capable) applicant passes with the probability of .40 is analogous to saying the probability an applicant passes if the applicant is incapable is .40.

The Bayes question is Given the applicant passes the test, what is the probability that he/she is capable? i.e. P (C|P) = ??

P (C|P) = P (C P) / P (P)

- Solving for the denominator: P (P) = P (P C) + P (P NC). i.e. the probability of passing the test is really the sum of the probability that you are capable and you pass plus the probability that you are not capable and you still pass.

P(P) = P (P C) + P (P NC) Where P (P C) = P (C) P (P|C) and where P (P NC) = P (NC) P (P|NC) Thus P (P) = P (C) P (P|C) + P (NC) P (P|NC) which = (.70) (.90) + (.30)(.40) =.

- Solving for the numerator: P (C P) is actually already done because the denominator also involved solving P (C P); P (P C) = P (C) P (P|C)= (.70)(.90) = .
- Solving for the answer: .63/.75 =. Given the applicant passes the exam the probability he is capable is.