# Given a positive test result, find the probability that test taker has the disease.

#### Pythagonacci

Hi All,

I have a probability problem that I've chased around until I'm sick of looking at it. Here goes.

The incidence of a disease is .012. A test, with sensitivity = .86, and specificity =.88. Giben a positive test result, what is the probability that the test taker has the disease.

The solution, that I can't get to, is 8%.

While I'm not sure I've dine anything right, I am sure that I'm having some trouble applying the specificity and sensitivity.

Thanks for lookin in!

Last edited by a moderator:

#### downthesun01

You have to use Bayes' Theorem:

pr(D)= the probability that the person has the disease = 0.012
pr(H)= the probability that the person does not have the disease = 0.988
pr(+|D)= the probability that the person will test positive for the disease given that they have it(sensitivity) = 0.86
pr(+|H)= the probability that the test will return positive given that the person does not have the disease (1-specificity)=0.12

Using Bayes' Theorem:

$$\displaystyle pr(D|+)=\frac{pr(+|D)*pr(D)}{(pr(+|D)*pr(D))+(pr(+|H)*pr(H))}$$

$$\displaystyle pr(D|+)=\frac{0.86*0.012}{(0.86*0.012)+(0.12*0.998)}=$$$$\displaystyle \frac{0.01032}{0.01032+0.11976}=$$$$\displaystyle \frac{0.01032}{0.13008}=\frac{1032}{13008}=$$$$\displaystyle \frac{43}{542}\approx 0.079 \approx 8\%$$

Thanks!