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Homework Assigment - Quality Control Problem: We know that the probability of an item being defective, p, can only be .05,.1 or .2. However, we have no additional information on which on if these three is more likely If we test items until a defective item has been found, what is the probability that p is .05, .1 or .2 given you had to test 1, 2, 3, 4 or 5 items (I am asking for 135 posterior probabilities)? Setup a probability table in Microsoft Excel to perform the calculations above and draw conclusions from the calculated (posterior) probabilities

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Answer #1

Here we have to use Bayes's Theorem.

e女 be conducted and P = o.2 be the three 1 1,2,3 By Bayes theme ee th melax ﹀ 丘 (J$15(1-J$15)^($116-1))/((SJ$15(1-SJ$15)시$116-1))t(SK$15(1-SK$15)시$116-1))+(SL$15(1-SL$15)^($116-1))) Probability 0.05 0.143 0.160 0.178 0.196 0.216 0.2 0.571 0.538 0.504 0.469 0.435 No. of test 0.286 0.303 0.319 0.334 0.349 4If there was i no. of test then (i-1) were not defective and the i th was defective as we were testing until first defective.

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