Question

Pavement requires aggregate of specific size. Aggregates for a highway pavement are extracted from a gravel pit. Based on experience with this area, it is known that the probability of good-quality aggregate is only 80%. In order to weed out poor-quality aggregate, engineers on site use a quick test, which is not entirely reliable. The probability that a good-quality aggregate passes is 90%, whereas the probability that a poor-quality aggregate will pass is 20%.
⦁   What is the probability that a random sample of gravel from the pit passes the test?
⦁   What is the probability that a passing sample is poor-quality?
⦁   What is the overall accuracy of this test?

Answer 1

Let R shows the event that gravel is of good quality aggregate and R' shows the event that it is not good quality aggregate. So we have

P(R) = 0.80, P(R') = 1 - P(R) = 0.20

Let P shows the event that gravel from the pit pass the test and F shows the event that gravel fail the test. So we have

P(P | R) = 0.90, P(F | R) = 1 - P(P|R) = 0.10

P(P | R') = 0.20, P(F |R') =1 - P(P|R') =0.80

By the law of total probability, the probability that a random sample of gravel from the pit passes the test is

P(P) = P(P|R)P(R) + P(P|R')P(R') = 0.90 * 0.80 + 0.20 * 0.20 = 0.72 + 0.04 = 0.76

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The probability that a passing sample is poor-quality is

P(R' | P) = [ P(P | R')P(R') ]/ P(P) = [ 0.20 * 0.20] / 0.76 = 0.0526

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The accuracy of the test:

Sensitivity:

P(P | R) =0.90

Specificity:

P(F | R') = 0.80

Positive predictive value: is the probability that person has disease given that test gives positive result so

P(R | P) = [P(P | R)P(R) ] / P(P) = [ 0.90 *0.80] / 0.76 = 0.9474

Negative predictive value:

P(R' | F) = [P(F | R')P(R') ] / [1- P(P)] = [ 0.80 *0.20] / 0.24 = 0.6667

Accuracy = P(P|R)*P(R) + P(R|R')P(R') = 0.90 * 0.80 + 0.80 * (1- 0.80) = 0.88

Answer: 0.88

Note:

Accuracy= Sensitivity × Prevalence + Specificity × (1 − Prevalence)