Question

On any given day, the probability that a large local river is polluted by carbon tetrachloride is 0.12. Each day, a test is conducted to determine whether the river is polluted by carbon tetrachloride. This test has proved correct 84 percent of the time.1. What proportion of days does the test indicate carbon tetrachloride pollution? 2. Suppose that on a particular day the test indicated carbon tetrachloride pollution What is the probability that such pollution actually exists? 3. What percentage of days where the test is negative will the river actually be polluted?

Answer 1

a) 0.2416

b) 0.4172

c) 0.0253

Explanation:

Since the result of the test should be independent of the time , then the that the test number of times that test proves correct is independent of the days the river is correct .

denoting event a A=the test proves correct and B=the river is polluted

a) the test indicates pollution when

- the river is polluted and the test is correct

- the river is not polluted and the test fails

then

P(test indicates pollution)= P(A)*P(B)+ (1-P(A))*(1-P(B)) = 0.12*0.84+0.88*0.16 = 0.2416

b) according to Bayes

P(A∩B)= P(A/B)*P(B) → P(A/B)=P(A∩B)/P(B)

then

P(pollution exists/test indicates pollution)=P(A∩B)/P(B) = 0.84*0.12 / 0.2416 = 0.4172

c) since

P(test indicates no pollution)= P(A)*(1-P(B))+ (1-P(A))*P(B) = 0.84*0.88+ 0.16*0.12 = 0.7584

the rate of false positives is

P(river is polluted/test indicates no pollution) = 0.12*0.16 / 0.7584 = 0.0253