A test for a certain rare disease is assumed to be correct 95% of the time. Let

Question

A test for a certain rare disease is assumed to be correct 95% of the time. Let S denote the event that a person has the disease, and T denote the event that the test result is positive. We have Pr[T|S] = 0.95 and Pr[ T| S] = 0.95. Assume that Pr[S] = 0.001, i.e., the probability that a randomly drawn person has the disease is 0.001. Compute Pr[S|T], the probability that one has the disease if one is tested positive. Assume that one improves the test so that Pr[T|S] = 0.998, i.e., if one has the disease, the test will come out positive with probability 0.998. Other things remain unchanged. Compute Pr[S|T]. Assume that another improvement on the test improves Pr[ T| S] = 0.998, but have Pr[T|S] remain at 0.95. Compute Pr[S|T].

Explanation / Answer

Solution

Back-up Theory

P(A/B) = P(AB)/P(B)

P(A/B) = {P(B/A).P(A)}/P(B)

P(A) = {P(A/B).P(B)} + {P(A/BC).P(BC)}

Now, to work out the solution,

Given T= event test is positive; S = event a person has the disease

P(T/S) = P(TC/SC) = 0.95 and P(S) = 0.001

By implication, P(T/SC) = 0.05 and P(SC) = 0.999

Part (a)

P(S/T) = {P(T/S).P(S)}/P(T) = (0.95x0.001)/0.0509 = 0.0187 ANSWER

[Because, P(T) = {P(T/S).P(S)} + {P(T/SC).P(SC)} = (0.95x0.001) + (0.05x0.999) = 0.00095 + 0.04995 = 0.0509]

Part (b)

We are given P(T/S) = 0.998. Since nothing is said about P(TC/SC), we will take it as 0.95 and hence P(T/SC) = 0.05

P(S/T) = {P(T/S).P(S)}/P(T) = (0.998x0.001)/ 0.050948 = 0.0196 ANSWER

[Because, P(T) = {P(T/S).P(S)} + {P(T/SC).P(SC)} = (0.998x0.001) + (0.05x0.999) = 0.000998 + 0.04995 = 0.050948]

Part (c)

We are given P(T/S) = 0.95 and P(TC/SC) = 0.998 implying P(T/SC) = 0.002

P(S/T) = {P(T/S).P(S)}/P(T) = (0.95x0.001)/0.002948 = 0.3223 ANSWER

Because, P(T) = {P(T/S).P(S)} + {P(T/SC).P(SC)} = (0.95x0.001) + (0.002x0.999) = 0.00095 + 0.001998 = 0.002948]

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