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Bayesian or not Bayesian,this is my problem

  1. Apr 24, 2008 #1
    I have a set of 10000 elements to buy; I want to do some quality assurance activity in order to avoid to buy invalid elements. I have an automatic test that provides the probability of correctness of an element; moreover I know that the test gives a 20% of false positive and a 30 % of false negative.
    1)Suppose that I have an element of which the test estimate a probability of correctness of 75%; which is the posterior probability?
    I’m afraid I’m unable to apply the Bayesian theorem because I do not know the number of correct elements. My instinct tells me to count the frequency in the past in which an element was really correct when the estimated probability by the tool was less or equal then 75%; suppose that such frequency is 50%. Consequently, my instinct will say that an element of which the test estimate a probability of correctness of 75% has a posterior probability to be correct of 50%. Am I right???

    2) Suppose that I cannot check all the elements by eye but I need to use the automatic test to prioritize the elements to check; suppose that I run the automatic test on all the elements and then I order the elements according to the probability of correctness provided by the test; suppose that I’ve analyzed all the elements with a probability of correctness lower 70%; how to calculate the probability that there is an incorrect elements in the remaining ones (i.e. the ones with a probability of correctness higher than 70%)???

    Please help me!
  2. jcsd
  3. Nov 9, 2008 #2
    Can you tell me if this is a proper question picked up from somewhere or is any data missing here? If all correct, tell me if the answer for 1. is 15/16?
  4. Nov 10, 2008 #3
    I don't understand your question.
  5. Nov 11, 2008 #4
    A way to check your calculation is to count the overall defective percentage p[0]. Suppose p[0] = p. Then the probability of having a defective part conditional on a "pass score" p[0|1] can be found as:

    p[0|1]p[1] = p[0,1] = p[1|0]p[0]
    p[0|1] = p[1|0]p[0]/p[1]
    p[0|1] = 0.2 p[0]/p[1]
    p[0|1] = 0.2 p[0]/(1-p[0])
    p[0|1] = 0.2 p/(1-p).

    I would definitely have logged in as EnumaElish had PF administration awarded that account the privilege of posting replies, after I reset my e-mail address Tuesday, October 28, 2008.
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