A new medical test has been designed to detect the presence of a genetic disease

Question

A new medical test has been designed to detect the presence of a genetic disease. Among those who have the disease, the probability that the disease will be detected by the new test (true positive) is 0.9 and the probability that the disease will not be detected (false negative) is 0.1.

The probability that the test will not detect the disease in those who do not have it (true negative) is 0.85 and the probability that the test will erroneously detect the presence of the disease in those who do not actually have it (false positive) is 0.15.

Suppose 1000 people take the test and assume that 0.03% of the population who take this test have the disease.

If the is test administered to an individual is positive, what is the probability that the person has the disease? Note your answer is the same for any population size -- not just 1000.

P(Have Disease | +) =

Explanation / Answer

P (have disease) = 0.03% = 0.0003

So, P(don't have disease) = 1-0.0003=0.9997

P (have disease | +) * P (+) = P (have disease and +)

P (+) = P (have disease and +) + P (don't have disease and +)

= P ( + | have disease) * P(have disease) + P (+ | don't have disease) * P (don't have disease) = 0.9*0.0003 + 0.15*0.9997

So, rearranging,

P (have disease |+) = (0.9*0.0003) / (0.9*0.0003+0.15*0.9997) = (0.00027) / (0.00027+0.149955) = 0.00027 / 0.150225 = 0.00179

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