A laboratory analyzed N biological specimens using two diagnostic tests, Test 1 and Test 2, to detect the presence of a…

Choice B is correct. The proportion of all N specimens in which the protein was detected by both tests is given by a/N. We are given three conditions based on the table:

1. Among the specimens where Test 1 detected the protein (a + b), the proportion where Test 2 also detected it is p, so a/(a + b) = p, which means a + b = a/p.

2. Among the specimens where Test 2 detected the protein (a + c), the proportion where Test 1 also detected it is q, so a/(a + c) = q, which means a + c = a/q.

3. The protein was detected by at least one of the two tests in a fraction k of all N specimens. The number of specimens where at least one test detected the protein is a + b + c. Thus, (a + b + c)/N = k, or a + b + c = kN.

Notice that (a + b) + (a + c) = a + b + c + a, so:

a + b + c = (a + b) + (a + c) - a

Substituting the expressions for (a + b) and (a + c) into this equation yields:

kN = a/p + a/q - a

kN = a(1/p + 1/q - 1)

kN = a((q + p - pq)/(pq))

Dividing both sides by N and isolating a/N gives:

a/N = (kpq)/(p + q - pq)

Choice A is incorrect because it fails to subtract the overlap a when expressing the union of the two test results, treating a + b + c as (a + b) + (a + c), which yields a/N = (kpq)/(p + q).

Choice C is incorrect because it inverts the ratio in the conditional probability, setting a + b = pa and a + c = qa instead of a = p(a + b) and a = q(a + c).

Choice D is incorrect because it mistakenly assumes the joint proportion can be found by simply multiplying the union rate by the conditional rates, as if the events were independent.