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[not T|not S] = 0.95. Assume that Pr[S] = 0.001, i.e., the probability that a randomly drawn person has the disease is 0.001. (a) Compute Pr[5|T], the probability that one has the disease if one is tested positive. (b) 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]. (c) Assume that another improvement on the test improves Pr[not T|not S] = 0.998, but have Pr[T|S] remain at 0.95. Compute Pr[S|T].

Explanation / Answer

(a) First we calculate P(T) as P(S)=0.001 thus

here, S=having the disease, T=positive test, therefore we have p(T|S)=0.95, P(T|S`)=0.05 and P(S)=0.001

so P(T)=P(S)P(T|S)+P(T|S`)P(S`) = (0.001)(0.95) + (0.999)(0.05) =0.0509

Now P(S|T)= P(T|S)P(S)/P(T) = (0.95)(0.001)/(0.0509)
P(S|T) = 0.018664

(b) Since now P(T|S)=0.998 we recalculate P(T) as above

by solving we get P(T)= P(S)P(T|S)+P(T|S`)P(S`) = (0.001)(0.998)+(0.999)(0.05)=0.0509

which is nearly the same as previous thus P(S|T)= 0.018644

(c) Now P(T'|S')=0.998

first calculate P(T)=P(S)P(T|S)+P(T|S`)P(S`) = (0.001)(0.95)+(0.002)(0.999)=0.0029

now P(S|T)=P(T|S)P(S)/P(T)=(0.001)(0.95)/0.0029=0.32758

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