Difference between revisions of "Bayes Theorem"

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(Created page with "Bayes Theorem proposes to look at the end of a sequence of events, so we know the outcome, and we wish to find the probability of a certain event in the sequence (posterior pr...")
 
 
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The general formulation of Bayes Theorem is:
 
The general formulation of Bayes Theorem is:
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<math>\huge {P(A_i|X)} =\frac</math>{{P}(X|A_i) P(A_i)}{\sum_{i=1}^{i=n}{P(X | A_i) P(A_i)}}<math></math>
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<math>{P(A_i|X)} =\frac</math>{{P}(X|A_i) P(A_i)}{\sum_{i=1}^{i=n}{P(X | A_i) P(A_i)}}<math></math>
  
 
Applied to the example <math>P(X|A_i)</math> represents the percentage of production that goes to machine  <math>A_i</math>.  <math>P(A_i)</math> represent the chance that this machine delivers a defective item. The multiplication of the two represent the overall probability that a specific machine delivered a defective item. <math>{\sum_{i=1}^{i=n}{P(X | A_i) P(A_i)}}</math> represents the total overal chance of getting a defective item in the full population (0.024 + 0.005 + 0.010 = 0.039)
 
Applied to the example <math>P(X|A_i)</math> represents the percentage of production that goes to machine  <math>A_i</math>.  <math>P(A_i)</math> represent the chance that this machine delivers a defective item. The multiplication of the two represent the overall probability that a specific machine delivered a defective item. <math>{\sum_{i=1}^{i=n}{P(X | A_i) P(A_i)}}</math> represents the total overal chance of getting a defective item in the full population (0.024 + 0.005 + 0.010 = 0.039)

Latest revision as of 19:53, 27 August 2020

Bayes Theorem proposes to look at the end of a sequence of events, so we know the outcome, and we wish to find the probability of a certain event in the sequence (posterior probability). As an example say a production line has 3 machines A, B and C.

A B C
40% 50% 10%

If we see defective items it is not certain from which machine they came. With Bayes theorem it is possible to identify the most likely machine and check that one first. Suppose the waste rates are:

A B C
6% 1% 10%

Then the probabilities per machine are:

  • A 0.4
    • 0.94 acceptable: 0.376
    • 0.06 defective: 0.024
  • B 0.5
    • 0.99 acceptable: 0.495
    • 0.01 defective: 0.005
  • C 0.1
    • 0.90 acceptable: 0.090
    • 0.10 defective: 0.010

The overall probability of a defective item is: {{{ P(defective) = 0.024 + 0.005 + 0.010 = 0.039 }}} This can also be written as A or B or C)

To look at a particular outcome, for example for Machine A and a defective item, we could write the probability as {{{ P(A and defective) = P(defective/defective from any machine) }}} Which is: {{{ P(A and defective) = P(0.024 / 0.039) = 0.615 }}}

The general formulation of Bayes Theorem is: Failed to parse (syntax error): {\displaystyle {P(A_i|X)} =\frac} {{P}(X|A_i) P(A_i)}{\sum_{i=1}^{i=n}{P(X | A_i) P(A_i)}}

Applied to the example represents the percentage of production that goes to machine . represent the chance that this machine delivers a defective item. The multiplication of the two represent the overall probability that a specific machine delivered a defective item. represents the total overal chance of getting a defective item in the full population (0.024 + 0.005 + 0.010 = 0.039)

Part of: M7-Quantitative Decision Methods-S2 - Probability and decision trees See also: http://en.wikipedia.org/wiki/Bayes%27_theorem