The prevalecne rate of a rare disease in a population is 2.5%. A test is propose

Question

The prevalecne rate of a rare disease in a population is 2.5%. A test is proposed by a lab for this diesase that is supposed to be positiive to indicate that presence of the disease and negative otherwise. In the lab the test is given to a person who has this disease, the test is positive 99% of the time. When the test is given to a person who does not have this disease, the test is negative 98% of the time. If the test is give to a random patient and shows a positive result, what is the chance that the paitent acutally has this disease.

Explanation / Answer

Ans:

Given that

P(disease)=0.025

P(no disease)=1-0.025=0.975

P(positive/disease)=0.99

P(negative/no disease)=0.98

P(positive/no disease)=1-0.98=0.02

P(positive)=P(positive/disease)*P(disease)+P(positive/no disease)*P(no disease)

=0.99*0.025+0.02*0.975=0.04425

We have to find:

P(disease/positive)=P(positive/disease)*P(disease)/P(positive)

=0.99*0.025/0.04425=0.5593

55.93%

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