Question

A new medical test has been designed to detect the presence of the mysterious brainlesserian disease. among those who have the disease, the probability that the disease will be detected by the new test is 0.9. however, the probability that the test will erroneously indicate the presence of the disease in those who do not actually have it is 0.06. it is estimated that 18 % of the population who take this test have the disease. if the test administered to an individual is positive, what is the probability that the person actually has the disease?

Answer 1

Introduce the following notations of events:
A "The person has the disease"
B "The test is positive", C "the test is erroneous"
Our formula will be then:
$P(A)=P(A!B)P(B)+P(A!C)P(C)\\\text{Using the given value we get the formula:}\\0.18=P(A!B)0.9+0.06\times0.1\\\text{Solving the above equation for P(A!B) we get:}\\P(AB):0.19$
So the probability we are looking for is 0.19