Question

Suppose four out of every 10,000 individuals in a country has a rare disease. a test for the disease exist. individuals known to have the disease test positive 96% of the time individuals known to be free of the disease test negative 96% of the time. Suppose an individual from the country is selected at random( with all being equally likely to be selected) and given the test . Given that the individuals test positive what is the probability that he in fact has a disease .Give the probability as a decimal number to six decimal places

Answer 1

Answer)

Lets say we have a sample of 10,000

Now 4 out of these 10,000 has the disease

And it is mentioned that if the individual has a disease then, he will test positive 96% of the time

0.96*4 = 3.72

And it is also mentioned that if the person does not have disease then he will be tested negative 96% of the time that is 4% of the time results will be positive

Persons with no disease = 10000-4 = 9996

Positive results = 0.04*9996 = 399.84

Probability is given by favorable/total

Total = positive tests = 3.72 + 399.84

Favorable = person has the disease = 3.72

Required probability = 3.72/(3.72+399.84) = 0.00921796015 = 0.009218