About 05 of the American population carries HIV For those wh

About 0.5% of the American population carries HIV. For those who carry HIV, the test results are positive (+) in 98.7% of the cases. For those who do not carry HIV, the test results are negative (-) in 94.6% of the cases. Given that a person have negative (-) test result what is the probability that he/she actually does not carry HIV. Please show how answers were found

Solution

Let American population be x

% of American population that carries HIV = 0.5x

% of population who carry HIV and for whom test results are positive = 0.987 * 0.5x

% of population who carry HIV and for whom test results are negative = (1 - 0.987) * (0.5x) = 0.013 * 0.5x

% of population who do not carry HIV and for whom test results are negative = 0.946 * (x - 0.5x) = 0.946 * 0.5x

Given that a person have negative (-) test result what is the probability that he/she actually does not carry HIV is given by =

% of population who do not carry HIV and for whom test results are negative / (Total population for whom test is negative)

= (0.946 * 0.5x) / [(0.946 * 0.5x) + (0.013 * 0.5x)]

= 98.6%