There is a rare disease that only happens to 1 out of 100,000 people. A test sho

Question

There is a rare disease that only happens to 1 out of 100,000 people. A test shows positive 99% of times when applied to an ill patient and, 1% of times when applied to a healthy patient. Please answer the following questions.

1. What is the probability for you to have the disease given that your test result is positive?

2. What is the probability for you to have the disease when you did two tests and both of them show positive? Assume that two tests are conducted independent.

3. Assume that the patient keeps on trying the test, what is the minimum number of tests that the patient has to try to be 99% percent sure that he is actually ill? Assume that all tests are conducted independently.

Explanation / Answer

probabilty of disease P(D) =1/100000 =0.00001

and of not having disease =P(ND) =1-0.00001 =0.99999

proabbailty of +ve test result, given person has disease =P(+ve|D) =0.99

proabbailty of +ve test result, given person does not have disease =P(+ve|ND) =0.01

1) hence probabilty of +ve test result =P(+ve) =P(D)*P(+ve|D)+P(ND)*P(+ve|ND)

=(0.00001*0.99+0.9999*0.01)= 0.010009

therefore probability for you to have the disease given that your test result is positive =P(D|+ve) =P(D)*P(+ve|D)/P(+ve) =(0.00001*0.99)/0.010009 =0.000989

2) probability for you to have the disease when you did two tests and both of them show positive =1-P(not having the disease) =1-(1-0.000989)*(1-0.000989) =0.001977

3) let n be the number of test hence

probabilty of having disease with n test +ve =1-(1-0.000989)n

0.99<1-(1-0.000989)n

(0.999011)n<0.01

nln(0.999011)<ln(0.01)

n>4653.571

n=4654

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