Poisson Distribution A laboratory blood test is 95 percent effective in detectin

Question

Poisson Distribution

A laboratory blood test is 95 percent effective in detecting a certain disease when it is, in fact, present. However, the test also yields a "false positive" result for 1 percent of the healthy persons tested. (That is, if a healthy person is tested, then, with probability 0.01, the test result will imply he/she has the disease.) If 0.5 percent of the population actually has the disease, what is the probability a person has the disease given that his/her test result is positive?

Explanation / Answer

let

A is tested person has disease
B test result is +ve

now

we need P(A/B) =

given are


P(B/A) = probablity that test result +ve when person has disease = 0.95
P(B/A') = probablity that test result false +ve when person has not disease = 0.01

P(A) = 0.005       P(A') = 1 - 0.005 = 0.995

now apply bayes rule we get

P(A/B) = P(B/A) P(A) / ( P(B/A) P(A) + P(B/A') P(A')

=   (0.95*0.005) / ( 0.95*0.005 + 0.01*0.995)

= 0.323

= 32.3 %

32% of those people whose test results are positive actually have the disease

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