Deadly XYZ virus

Imagine that XYZ viral syndrome is a serious condition that affects one person in 1,000. Imagine also that the test to diagnose the disease always indicates correctly that a person who has the XYZ virus actually has it. Finally, suppose that this test occasionally misidentifies a healthy individual as having XYZ. The test has a false-positive rate of 5 %, meaning that the test wrongly indicates that the XYZ virus is present in 5 % of the cases where the person does not have the virus.

Next we choose a person at random and administer the test, and the person tests positive for XYZ syndrome. Assuming we know nothing else about that individual's medical history, what is the probability that the individual really has XYZ?

If the probability can be written as a b \dfrac{a}{b} , where a a and b b are coprime integers. give your answer as a + b a+b .


The answer is 1039.

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1 solution

Mark Hennings
Nov 21, 2018

The probability of a positive diagnosis is 1 × 1 1000 + 1 20 × 999 1000 = 1019 20000 1 \times \tfrac{1}{1000} + \tfrac{1}{20} \times \tfrac{999}{1000} \; = \; \tfrac{1019}{20000} The probability of a patient having the disease, and being diagnosed positively, is 1 × 1 1000 = 1 1000 1 \times \tfrac{1}{1000} \; = \; \tfrac{1}{1000} Thus the probability that a person who has had a positive diagnosis actually has the disease is 1 1000 ÷ 1019 20000 = 20 1019 \tfrac{1}{1000} \div \tfrac{1019}{20000} \; = \; \tfrac{20}{1019} making the answer 20 + 1019 = 1039 20 + 1019 = \boxed{1039} .

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