Question 1170690
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A certain disease has an incidence rate of 0.2%. If the false negative rate is 5% and the false positive rate is 5%, 
compute the probability that a person who tests positive actually has the disease.
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This problem is on conditional probability, and I will put things in order.


Originally, we have a whole population of X people.

0.2% of them, or 0.002*X, have the disease.

The rest, or 0.998*X, do not have the disease (are healthy).


The number of those who actually has the disease AND has test positive is  n = (1-0.05)*(0.002*X) = 0.95*0.002*X.


The number of those who are test positive is the sum  N = 0.95*(0.002*x) + 0.05*(0.998*X).


In this sum, first  addend comes from the set of people having the disease; 
             second addend comes from the set of people who do not have the disease.


The problem wants you find the conditional probability, which is the ratio n/N.  It is 

    P = {{{n/N}}} = {{{(0.95*0.002*X)/(0.95*0.002*X + 0.05*0.998*X)}}}.


Now reduce X in the numerator and in the denominator.  You will get

    P = {{{n/N}}} = {{{(0.95*0.002)/(0.95*0.002 + 0.05*0.998)}}} = {{{0.0019/(0.0019 + 0.0499)}}} = {{{0.0019/0.0518}}} = 0.03668  (rounded), or 3.668%.

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Solved.