Question

28. A diagnostic test has a probability 0.95 of giving a positive result when applied to a person suffering from a certain disease, and a probability 0.10 of giving a (false) positive when applied to a non-sufferer. It is estimated that 0.5 % of the population are sufferers. Suppose that the test is now administered to a person about whom we have no relevant information relating to the disease (apart from the fact that he/she comes from this population). Calculate the following probabilities:

(a) That the test result will be positive.

(b) That, given a positive result, the person is a sufferer.

(c) That, given a negative result, the person is a non-sufferer.

(d) That the person will be misclassified.

(e) That the person is not suffer, if the test is negative.

Answer 1

The data can be presented as below:

True Disease
No disease		Disease
P(D')=0.995		P(D)=0.005
99.5 out of 100		0.5 out of 100
Test positive	Test negative	Test positive	Test negative
10.00%	90.00%	95.00%	5.00%
9.95	89.55	0.475	0.025

Let D be the event having the disease = D= P(D) = 0.5% = 0.005

Probability of not having disease = D' = P(D') = 100%-0.5% = 99.50% = 0.995

Let T be the event of a person testing positive for the disease

Given that the person is suffering, probability that test is positive = P(T/D) = 0.95

Given that the person is not suffering, probability that the test is positive = P(T/D') = 0.10

(a) Probability that the test result will be positive:

P (T) = P(T/D) * P(D) + P(T/D') * P(D') = (0.95 * 0.005) + (0.10 * 0.95) = 0.10425

P(T) = 0.10425

(b) given a positive result, the person is a sufferer.

P(D/T) = P(T/D) * P(D) / [ P(T/D) * P(D) + P(T/D') * P(D')]

P(D/T) = 0.95* 0.005 / [ (0.95* 0.005) + (0.1* 0.995) = 0.0455

P(D/ T) = 0.0455

(c) given a negative result, the person is a non-sufferer.

P(D' / T') = P(T' / D') * P(D') / P(T') = 0.9 * 0.995 / (1-0.10425) = 0.997

P(D' / T') = 0.997         

(d) the person will be misclassified. (tested positive when he doesn't have disease or tested negative when he has the disease)

P(Misclassified) = P (T D') + P(T' D) = P(T/ D') * P(D') + P(T'/D) * P(D) = 0.09975

(e) the person is not suffer, if the test is negative. = P(D' / T')

same as option c.