A new medical test has been designed to detect the presence

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 0.78. However, the probability that the test will erroneously indicate the presence of the disease in those who do not actually have it is 0.12. It is estimated that 15 % 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?

Solution

suppose

T = event that new test shows detection of disease

A = perosn has actually disease

B = person do not has diserase

we are given that

p(A) =0.15

p(T/A) = 0.78

p(T/B)= 0.12

thus p(B) =0.85 [= 1-P(A) ]

Now we have to find

p(A/ T) = [ p(A) p(T/A) ] / [ p(A) p(T/A) + p(B) p(T/B) ] (BY BAYS PROBABILITY THEOREM )

= 0.15*0.78/ (0.15*0.78 + 0.85*0.12)

= 0.534246575