Incongruent test

Probability Level pending

Suppose that 40% of the population of the community of Pathogen (population 60 00) have a virus. A test accurately detects the virus in 95% of the cases when the person has the virus , but FALSELY detects the virus in 10% of the cases when the person does not have the virus. EACH person is tested TWO times. If the two tests do not agree, the results are said to be INCONGRUENT . How many of the results would you expect to be INCONGRUENT?

note: Each of the 60 00 persons is given the test two times.


The answer is 876.

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1 solution

Josh Speckman
Jul 19, 2015

We note that 2400 2400 people have the virus, and 3600 3600 do not. For each person who has the virus, the probability that the results are incongruent is 19 20 1 20 + 1 20 19 20 = 19 200 \dfrac{19}{20} \dfrac{1}{20} + \dfrac{1}{20} \dfrac{19}{20}= \dfrac{19}{200} . This is true because we need both a positive and a negative result in order for our results to be incongruent, and these can happen in any order. Thus, we expect that about 2400 19 200 = 228 2400 \cdot \dfrac{19}{200} = 228 of these will have the virus. Similarly, for those without the virus, the probability of incongruent results is 9 10 1 10 + 1 10 9 10 = 9 50 \dfrac{9}{10} \dfrac{1}{10} + \dfrac{1}{10} \dfrac{9}{10}= \dfrac{9}{50} , and since there are 3600 3600 of these, we expect incongruent results on 3600 9 50 = 648 3600 \cdot \dfrac{9}{50} = 648 of these. Thus, we expect a total of 648 + 228 = 876 648+228=\boxed{876} people to have incongruent results.

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