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A certain disease has an incidence rate of 0.5%.
The false-negative rate on a test for the disease is 7%;
the false positive rate is 5%.
Compute the probability that a person who tests positive actually has the disease.
(You may find it useful to construct a probability contingency table.)
The probability a person who tests positive actually has the disease is
Give your answer accurate to at least 3 decimal places
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This problem is about calculating the CONDITIONAL PROBABILITY.
They want you calculate this ratio .
Let X be the total population.
Then the number of those who has the disease is 0.005*X; the number of those who has no the disease is 0.995*X.
The number of those who has the disease and tests positive is 0.005*(1-0.07)X = 0.005*0.93*X.
To calculate this amount, I excluded from 0.005X, the number of sick persons, the value 0.005X*0.07,
which represents the part of sick who are tested as healthy (i.e. are false-negative).
The number of those who has no the disease and (or "but") tests positive is 0.995*0.05*X (false test positive).
So the number of those who tests positive is the sum 0.005*0.93*X + 0.995*0.05*X.
They want you calculate the conditional probability
P = = = cancel X in the numerator and in the denominator =
= = = use your calculator = 0.085478 (rounded). ANSWER
ANSWER. This probability is 0.085478.
Solved.