A new analytical method to detect pollutants in water is being tested. This new

Question

A new analytical method to detect pollutants in water is being tested. This new method of chemical analysis is important because if adopted, it could be used to detect three different contaminates organic pollutants, volatile solvents, and chlorinated compounds instead of having to use a single test for each pollutant. The makers of the test claim that it can detect high levels of organic pollutants with 99.4% accuracy, volatile solvents with 99.92% accuracy, and chlorinated compounds with 89.5% accuracy If a pollutant is not present, the test does not signal. Samples are prepared for the calibration of the test and 60% of them are contaminated with organic pollutants, 27% with volatile solvents, and 13% with traces of chlorinated compounds. A test sample is selected randomly (a) What is the probability that the test will signal? (b) If the test signals, what is the probability that the chlorinated compounds are present? Carry out all calculations exactly, then round the final answers to four decimal places (e.g. 98.7654)

Explanation / Answer

Let the event that the sample is contaminated with

(i) organic pollutant be denoted by A

(ii) volatile solvent be denoted by B

(iii) traces of chlorinated compounds by C

And let X denote the event that the test will signal.

Thus, from the data

P(A) = 0.6 , P(X|A) = 0.994

P(B) = 0.27, P(X|B) = 0.992

P(C) = 0.13, P(X|C) = 0.895

(a) Probability that the test will signal, P(X) = P(A)*P(X|A) + P(B)*P(X|B) + P(C)*P(X|C) = 0.6*0.994 + 0.27*0.992 + 0.13*0.895 = 0.9806

(b) P(C|X) = P(C)*P(X|C)/P(X) = 0.13*0.895/0.9806 = 0.1187

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