# Conditional Probability on Disease

A man living in a country where only 1 out of 1000 people has the virus A. There is a test available that gives a positive result 5% of the time when the patient does not have virus A and a negative result 1% of the time when the patient does have Virus A. Otherwise, it gives correct results. Recall that past computation showed that the man ’s chance of having the virus, conditional on a positive test, is less than 1.9%. Assume all tests are independent.

Let the conditional probability computed 1.9% serve as the new prior. Compute the new probability that he has the virus A (new posterior) based on his receiving a second positive test.

So my thought process was the following :

P(A) = 0.0001 P(Ac) = 1- 0.0001 P(+/Ac) = 0.05 P(-/Ac) = 0.01

I am asked to calculated to P(A/+) = P(+/A)*P(A) / P(+)

P(+) = P(+/A)*P(A) + P(+/Ac)*P(Ac)

My question is I am still confused on how I can get P(+/A).

Could you please help me?

• You started talking about men but later mention "patients". The size of test errors (1% and 5%) refers to men (patients=men)? Or not (patients = men+women) ? Commented Apr 17 at 8:48

## 1 Answer

$$P(+ | A) = 1 - P(- | A) = 1 - 0.01 = 0.99$$