2.8.3 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 99.2% accuracy, volatile solvents with 99.94% 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)

Answer:a) 98.1388% b) 11.8556%Step-by-step explanation:Let A, B and C be the following events A=The sample is contaminated with organic pollutants AND is detected by the test. B=The sample is contaminated with volatile solvents AND is detected by the test. C=The sample is contaminated with chlorinated compounds AND is detected by the test. a) A, B and C are disjoints sets and the probability that the test will signal is  P(A∪B∪C) = P(A) + P(B) + P(C) On the other hand, since the probability of taking a sample does not depend on previous choices, the events are independent, so P(A) = P(The sample is contaminated with organic pollutants)*P(The sample is detected by the test) = 0.6*0.992 = 0.5952 Similarly, we find that P(B) = 0.27*0.9994 = 0.269838 P(C) = 0.13*0.895 = 0.11635 Hence, the likelihood the test will signal is 0.5952 + 0.269838 + 0.11635 = 0.981388 or 98.1388% In order not to make heavy the notation, let's all P(Chl) = Probability the sample is contaminated with chlorinated compounds = 0.13 P(V) = Probability the sample is contaminated with volatile solvents = 0.27 P(O) = Probability the sample is contaminated with organic pollutants = 0.6 P(S | Chl) = the probability that the sample signals GIVEN THAT is contaminated with chlorinated compounds = 0.895 P(S | V) = the probability that the sample signals GIVEN THAT is contaminated with volatile solvents = 0.9994 P(S | O) = the probability that the sample signals GIVEN THAT is contaminated with  organic pollutants = 0.992 b) If the test signals, what is the probability that the chlorinated compounds are present? We want P(Chl | S)  By the BAYE'S THEOREM  [tex]\large P(Chl|S)=\frac{P(Chl)P(S|Chl)}{P(Chl)P(S|Chl)+P(V)P(S|V)+P(O)P(S|O)}=\\=\frac{0.13*0.895}{0.13*0.895+0.27*0.9994+0.6*0.992}=0.1185556575\approx 11.8556\%[/tex] So, the probability wanted is 11.8556%