Question

A test for tuberculosis was given to 1000 subjects, 8% of whom were known to have tuberculosis. For the subjects who had tuberculosis, the test indicated tuberculosis in 90% of the subjects, was inconclusive for 7%, and negative for 3%. For the subjects who did not have tuberculosis, the test indicated tuberculosis in 5%, was inconclusive for 10%, and was negative for the remaining 85%. What is the probability of a randomly selected person having tuberculosis given that the test indicates tuberculosis

Answer 1

Answer:

the probability is 9/71 ( 12.67%)

Step-by-step explanation:

defining the event Te= the test indicates tuberculosis , then the probability is:

P(Te)= probability of choosing a subject that has tuberculosis * probability that the test indicates tuberculosis given that a subject with tuberculosis was chosen + probability of choosing a subject that has not tuberculosis * probability that the test indicates tuberculosis given that a subject that has not tuberculosis was chosen = 8/1000 * 90/100 + 992/1000 * 5/100 = 0.0568

After this,  we can use the theorem of Bayes for conditional probability. Then defining the event Tu= choosing a subject that has tuberculosis, we have

P(Tu/Te)= P(Tu∩Te)/P(Te) = 8/1000 * 90/100 / 0.0568 = 9/71 ( 12.67%)

where

P(Tu∩Te) = probability that a subject with tuberculosis is chosen and the test indicates tuberculosis

P(Tu/Te)= probability that a subject with tuberculosis was chosen given that the test indicates tuberculosis