=-Probability problem. Will choose BEST answer!!-=?

Question

If a certain disease is present, then a blood test will reveal it 95% of the time. but the test will also indicate the presence of the disease 2% of the time when in fact the person tested is free of that disease; that is, the test gives false positive 2% of the time. If 0.3% of the general population actually has the disease, what is the probability That a person chosen at random form the population has the disease given that he or she tested positive?

Pathomaniac · Accepted Answer

this is a conditional probability problem

let event A = person has the disease
let event B = test was positive


P(A) = .003
P(B) = false positive + true positive = (.997)(.02) + (.003)(.95) = .01994 + .00285 = 0.02279


conditional probability is given by:

P(A|B) = P(A n B) / P(B) = (.003)(.95) / (0.02279) = .125055

If you test positive, then there is only a ~12.5% chance that you have the disease!

This answer is initially surprising, but makes sense on reflection. There are two ways you could test positive. First, it could be that you are sick and the test is correct. Second, it could be that you are healthy and the test is incorrect. The problem is that almost everyone is healthy; therefore, most of the positive results arise from incorrect tests of healthy people!

cidyah · Answer

D = a person has the disease
D' = a person doesn't have the disease
R = correct positive diagnosis
P(D)=0.003
P(D')=0.997
P(R/D)=0.95
P(R/D') = 0.02

Applying Bayes' Theorem,

P(D/R) = P(D)P(R/D) / [P(D)P(R/D)+P(D')P(R/D')]
=(0.003)(0.95) / [ (0.003)(0.95) + (0.997) (0.02)]
=0.12505

Anonymous · Answer

.125055

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=-Probability problem. Will choose BEST answer!!-=?

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