Question

from a national survey it is known that approximately 12% individuals over
50years old suffer from diabetes. The person who has confirmed diabetes,
30% of the cases test shows correct result. When the individuals were having
no diabetes, 4% were reported to have disease in the test. What is the
probability that an individual has this disease given that test indicates its
presence?

Answer 1

The survey showed that 12% actually had diabetes, so 88% did not.

Now we create some variables in the form actual(test result):

Let p(+) be the percentage of those with diabetes and tested positive.

Let p(-) be the percentage of those with the disease but who tested negative.

So it appears that p(-)=4%, because 4% were incorrectly tested as negative, whereas, in fact, they actually had the disease.

Let n(+) be the percentage of those with no diabetes but who tested positive. 

Let n(-) be the percentage of those with no diabetes and who tested negative. 

The test was correct for p(+)+n(-)=30% and incorrect for p(-)+n(+)=70%.

n(+)=70-4=66%. So 66% without diabetes were incorrectly tested as positive. 

Also, p(+)+p(-)=12%, n(+)+n(-)=88%. From this can be concluded that:

p(+)=12-4=8%, n(-)=88-66=22%.

To summarise:

p(+)=8%, p(-)=4%, n(+)=66%, n(-)=22%, which total 100%.

The test diagnoses positive for n(+)+p(+)=74%, but only p(+)=8% actually have the disease, so the probability of correct diagnosis is 8/74=0.109 or 10.9%. That is, out of the 74% who test positive only 8% of these results (10.9%) would be correct.