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Reversal of conditioning

Students in a freshman mathematics class come from three different high schools. Their mathematical preparation varies. In order to group them appropriately in classsections, they are given a diagnostic test. Let H i be the event that a student tested is from high school i , 1 i 3 . Let F be the event the student fails the test. Suppose data indicate

P ( H 1 ) = 0 . 2 , P ( H 2 ) = 0 . 5 , P ( H 3 ) = 0 . 3 , P ( F | H 1 ) = 0 . 10 , P ( F | H 2 ) = 0 . 02 , P ( F | H 3 ) = 0 . 06

A student passes the exam. Determine for each i the conditional probability P ( H i | F c ) that the student is from high school i .

SOLUTION

P ( F c ) = P ( F c | H 1 ) P ( H 1 ) + P ( F c | H 2 ) P ( H 2 ) + P ( F c | H 3 ) P ( H 3 ) = 0 . 90 0 . 2 + 0 . 98 0 . 5 + 0 . 94 0 . 3 = 0 . 952

Then

P ( H 1 | F c ) = P ( F c H 1 ) P ( F c ) = P ( F c | H 1 ) P ( H 1 ) P ( F c ) = 180 952 = 0 . 1891

Similarly,

P ( H 2 | F c ) = P ( F c | H 2 ) P ( H 2 ) P ( F c ) = 590 952 = 0 . 5147 and P ( H 3 | F c ) = P ( F c | H 3 ) P ( H 3 ) P ( F c ) = 282 952 = 0 . 2962
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The basic pattern utilized in the reversal is the following.

(CP3) Bayes' rule If E i = 1 n A i (as in the law of total probability), then

P ( A i | E ) = P ( A i E ) P ( E ) = P ( E | A i ) P ( A i ) P ( E ) 1 i n The law of total probability yields P ( E )

Such reversals are desirable in a variety of practical situations.

A compound selection and reversal

Begin with items in two lots:

  1. Three items, one defective.
  2. Four items, one defective.

One item is selected from lot 1 (on an equally likely basis); this item is added to lot 2; a selection is then made from lot 2 (also on an equally likely basis). This second itemis good. What is the probability the item selected from lot 1 was good?

SOLUTION

Let G 1 be the event the first item (from lot 1) was good, and G 2 be the event the second item (from the augmented lot 2) is good. We want to determine P ( G 1 | G 2 ) . Now the data are interpreted as

P ( G 1 ) = 2 / 3 , P ( G 2 | G 1 ) = 4 / 5 , P ( G 2 | G 1 c ) = 3 / 5

By the law of total probability (CP2) ,

P ( G 2 ) = P ( G 1 ) P ( G 2 | G 1 ) + P ( G 1 c ) P ( G 2 | G 1 c ) = 2 3 4 5 + 1 3 3 5 = 11 15

By Bayes' rule (CP3) ,

P ( G 1 | G 2 ) = P ( G 2 | G 1 ) P ( G 1 ) P ( G 2 ) = 4 / 5 × 2 / 3 11 / 15 = 8 11 0 . 73
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Additional problems requiring reversals

  • Medical tests . Suppose D is the event a patient has a certain disease and T is the event a test for the disease is positive. Data are usually of the form: prior probability P ( D ) (or prior odds P ( D ) / P ( D c ) ), probability P ( T | D c ) of a false positive , and probability P ( T c | D ) of a false negative. The desired probabilities are P ( D | T ) and P ( D c | T c ) .
  • Safety alarm . If D is the event a dangerous condition exists (say a steam pressure is too high) and T is the event the safety alarm operates, then data are usually of the form P ( D ) , P ( T | D c ) , and P ( T c | D ) , or equivalently (e.g., P ( T c | D c ) and P ( T | D ) ). Again, the desired probabilities are that the safety alarms signals correctly, P ( D | T ) and P ( D c | T c ) .
  • Job success . If H is the event of success on a job, and E is the event that an individual interviewed has certain desirable characteristics, thedata are usually prior P ( H ) and reliability of the characteristics as predictors in the form P ( E | H ) and P ( E | H c ) . The desired probability is P ( H | E ) .
  • Presence of oil . If H is the event of the presence of oil at a proposed well site, and E is the event of certain geological structure (salt dome or fault), the data are usually P ( H ) (or the odds), P ( E | H ) , and P ( E | H c ) . The desired probability is P ( H | E ) .
  • Market condition . Before launching a new product on the national market, a firm usually examines the condition of a test market as anindicator of the national market. If H is the event the national market is favorable and E is the event the test market is favorable, data are a prior estimate P ( H ) of the likelihood the national market is sound, and data P ( E | H ) and P ( E | H c ) indicating the reliability of the test market. What is desired is P ( H | E ), the likelihood the national market is favorable, given the test market is favorable.
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Questions & Answers

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Mostly, they use nano carbon for electronics and for materials to be strengthened.
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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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