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# 2.10.9 - statement of the problem the conditional...

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- 1 - © John J Currano, 12/12/2008 Example from Section 2.10, Slide 9 Example : If a water specimen contains nitrates and is tested using a colormetric test, it will turn red 95% of the time. When the test is used on specimens that do not contain nitrates, the water turns red 10% of the time. From past experience, we are confident that 30% of the water specimens tested at a particular lab contain nitrates. (a) If a water specimen is randomly chosen from those sent to the lab for testing, what is the probability that it will turn red when tested? Let B 1 be the event that the specimen contains nitrates, B 2 = be the event that the specimen does not contain nitrates, and A be the event that the water turns red. We are told so by the Law of Total Probability , (b) If a water specimen is randomly selected and turns red when tested, what is the probability that it actually contains nitrates? Here we are asked for a conditional probability, . This value was not given in the

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Unformatted text preview: statement of the problem; the conditional probabilities “the other way around” were given, , as were the unconditional probabilities of the partitioning sets B 1 and B 2 . The Law of Total Probability and the Multiplicative Law of Probability , together with the definition of conditional probability, allow us to calculate the desired probability in situations like this, and this method is called Bayes’ Rule . by the definition of conditional probability, so by the Law of Total Probability and the Multiplicative Law of Probability, - 2 - © John J Currano, 12/12/2008 . This is what is known as Bayes’ Rule or Bayes’ Theorem . Since we have already calculated P ( A ) in part (a), and also calculated the numerator of the fraction in the process of doing that, we see that The following tree, obtained using the text’s applet, illustrates these computations. In the tree, B = B 1 , ~ B = = B 2 , and ~ A = ....
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2.10.9 - statement of the problem the conditional...

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