SOLUTION: A certain disease has an incidence rate of 0.6%. If the false negative rate is 4% and the false positive rate is 1%, compute the probability that a person who tests positive actual

Let 

A = person tests positive (affirmative test) 

N = person tests negative

D = person has the disease

H = person does not have the disease (healthy)



We wish to find  P(D|A).  The Bayes setup for this problem is:



P(D|A) = P(D)*P(A|D) / (P(D)*P(A|D) + P(H)*P(A|H))



What was given? 

P(D) is given as 0.6% = 0.006

P(N|D) = 4% = 0.04  (false negative rate)

P(A|H) = 1% = 0.01  (false positive rate)



We compute:

P(A|D) = 1-P(N|D) = 1-0.04 = 0.96

P(H) = 1-P(D) = 1-0.006 = 0.994



P(D|A) = (0.006*0.96) / ((0.006)*(0.96) + (0.994)*(0.01)) = 0.00576 / 0.0157 = 0.3669 or about 36.7%





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Check

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An example to sanity-check the above answer...



Given a population of 1000000 people, we expect 0.006*1000000 = 6000 to have the disease and 1000000-6000 = 994000 to NOT have it.  



I. The expected number who test positive and have the disease:

  6000*(1-0.04) = 5760



II. The total number of positive tests expected is the number who have the disease and test positive, PLUS the number who do not have the disease who also test positive:



  5760 + 994000(0.01) = 5760 + 9940 = 15700  





The fraction (I)/(II) should match the probability computed above:

   5760/15700 = 0.3669  (ok)