Question

In a certain community, 4 percent of all adults over the age of 50 have tuberculosis (T.B.). A health service in this community correctly diagnoses 98 percent of all persons with T.B. as having the disease, and incorrectly diagnoses 3 percent of all persons without T.B. as having the disease. Find the probabilities that,

a) the community health service will diagnose an adult over 50 as having T.B.,

b) a person over 50 diagnosed by the health service as having T.B. actually has the disease.

Answer 1

Let, A be the event that the person is diagnosed with TB

Let, E1 be the event that person actually has TB

Let, E2 be the event that person does not have TB

Probability person having TB , P(E1) = 4% = 0.04

Probability person does not have TB , P(E2) = 1 - P(E1) = 1 - 0.04 = 0.96

Probability person having TB and diagnosed correctly, P[ A | E1 ] = 98% = 0.98

Probability person does not have TB and diagnosed incorrectly, P[ A | E1 ] = 98% = 3% = 0.03

a) Probability that the community health service will diagnose an adult over 50 as having T.B, P(A)

P(A) = P(E1)*P[ A | E1 ] +  P(E2)*P[ A | E2 ] = 0.04*0.98 + 0.96*0.03 = 0.0392 + 0.0288 = 0.068 = 6.8%

P(A) = 0.068 = 6.8%

b) Probability a person over 50 diagnosed by the health service as having T.B. actually has the disease = P[ A | E1 ]

P[ A | E1 ] = P(E1)*P[ A | E1 ] / P(A) ( using Baye's theorem )

P[ A | E1 ] = 0.04*0.98 / 0.068 = 0.0392 / 0.068 = 0.5765

P[ A | E1 ] = 0.5765 = 57.65%