BAYES THEOREM in .NET

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2-7 BAYES THEOREM
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chosen every several minutes. Assume that the samples are independent. (a) What is the probability that ve successive samples were all produced in cavity one of the mold (b) What is the probability that ve successive samples were all produced in the same cavity of the mold (c) What is the probability that four out of ve successive samples were produced in cavity one of the mold 2-90. The following circuit operates if and only if there is a path of functional devices from left to right. The probability that each device functions is as shown. Assume that the probability that a device is functional does not depend on whether or not other devices are functional. What is the probability that the circuit operates
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other devices are functional. What is the probability that the circuit operates
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2-91. The following circuit operates if and only if there is a path of functional devices from left to right. The probability each device functions is as shown. Assume that the probability that a device functions does not depend on whether or not
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2-92. An optical storage device uses an error recovery procedure that requires an immediate satisfactory readback of any written data. If the readback is not successful after three writing operations, that sector of the disk is eliminated as unacceptable for data storage. On an acceptable portion of the disk, the probability of a satisfactory readback is 0.98. Assume the readbacks are independent. What is the probability that an acceptable portion of the disk is eliminated as unacceptable for data storage 2-93. A batch of 500 containers for frozen orange juice contains 5 that are defective. Two are selected, at random, without replacement, from the batch. Let A and B denote the events that the rst and second container selected is defective, respectively. (a) Are A and B independent events (b) If the sampling were done with replacement, would A and B be independent
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BAYES THEOREM
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In some examples, we do not have a complete table of information such as the parts in Table 2-3. We might know one conditional probability but would like to calculate a different one. In the semiconductor contamination problem in Example 2-22, we might ask the following: If the semiconductor chip in the product fails, what is the probability that the chip was exposed to high levels of contamination From the de nition of conditional probability, P1A B2 P1A B2P1B2 P1B A2 P1B A2P1A2
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Now considering the second and last terms in the expression above, we can write
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P1A B2
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P1B A2P1A2 P1B2
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(2-11)
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This is a useful result that enables us to solve for P1A B2 in terms of P1B A2.
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CHAPTER 2 PROBABILITY
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EXAMPLE 2-29
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We can answer the question posed at the start of this section as follows: The probability requested can be expressed as P1H 0 F2. Then, P1H F2 P1F H2P1H2 P1F2 0.1010.202 0.0235 0.85
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The value of P(F) in the denominator of our solution was found in Example 2-20. In general, if P(B) in the denominator of Equation 2-11 is written using the Total Probability Rule in Equation 2-8, we obtain the following general result, which is known as Bayes Theorem.
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Bayes Theorem
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If E1, E2, p , Ek are k mutually exclusive and exhaustive events and B is any event, P1E1 B2 P1B E1 2P1E1 2 P1B E1 2P1E1 2 p P1B E2 2P1E2 2 P1B Ek 2P1Ek 2 for P1B2 0 (2-12)
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EXAMPLE 2-30
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Because a new medical procedure has been shown to be effective in the early detection of an illness, a medical screening of the population is proposed. The probability that the test correctly identi es someone with the illness as positive is 0.99, and the probability that the test correctly identi es someone without the illness as negative is 0.95. The incidence of the illness in the general population is 0.0001. You take the test, and the result is positive. What is the probability that you have the illness Let D denote the event that you have the illness, and let S denote the event that the test signals positive. The probability requested can be denoted as P1D S2 . The probability that the test correctly signals someone without the illness as negative is 0.95. Consequently, the probability of a positive test without the illness is P1S D 2 From Bayes Theorem, P1D S2 0.9910.00012 30.9910.00012 1 506 0.002 P1S D2P1D2 3P1S D2P1D2 P1S D 2P1D 2 4 0.0511 0.00012 4 0.05
Surprisingly, even though the test is effective, in the sense that P1S 0 D2 is high and P1S D 2 is low, because the incidence of the illness in the general population is low, the chances are quite small that you actually have the disease even if the test is positive.