probability puzzle

Question

Assume you go to the doctor and you are tested for the dreaded XYA disease. He informed you that you tested positive. Since the the test has a false positive rate of 5% he tells you there is only a 5% chance you don't have the disease. Overall the incidence rate of the disease in the general population is 1% or 1 in 100 people actually contract this disease. Did the doctor get his statistics correct? (Hint:  he did not) if not what is the real chance that you have the dreaded XYZ disease based on your positive test results. 

Answer 1

Jean,

       I think there is a piece of information missing, the efficiency of the test, i.e.p(positive|disease) I'll assume it's a good test and that the efficiency is 90%. So if you have the disease there is a 90% chance you will test positive.

       The chance that you have the dreaded XYZ disease given that you tested positive, p(disease|positive)=a/(a+b) where a=number of people who test positive and have the disease and b=number of people who test positive and do not have the disease.

 

Suppose we test 1000 people we know that 10 people will have the disease (1% incident rate) and 990 will not have the disease of the 10 who have the disease 9 (90% of 10) will test positive this is a, and of the number who do not have the disease  (5% of 990) or 50 will test positive this is b

 

So the probability that you have the disease given that you tested positive p(disease|positive)=9/(9+50)=.1525

 

The result of the test is: before you took the test you had a 1% chance of having XYZ, after the test you have a 15% chance of having XYZ. Looked at another way you have an 85% chance of not having XYZ.

 

I hope this helps. Many times people get very nervous when testing positive and do not do the calculations that take the efficiency of the test and the false positive rates into account so they can make a reasoned decision.

 

Jim