A certain disease has an incidence rate of 0.3%. If the false negative rate is 6% and the false positive rate is 2%, compute the probability that a person who tests positive actually has the disease.

Answer 1

Solve using a probability tree:

(sick)/0.003*0.9=0.00282 (positive)
/0.003\0.003*0.06=0.00018 (false negative)
1
\0.997/0.997*0.02=0.01994 (false positive)
(not \0.997*0.98=.97706 (negative)
sick)

check: .00282+.00018+0.01994+.97706=1.0000 √

"the probability that a person who tests positive actually has the disease."
Looking at the tree diagram,
the probability of a person testing positive is 0.00282+0.01994=0.02276.
Probability of a person testing positive AND having the disease is 0.00282.

Therefore the conditional probability of a person having the disease GIVEN that he tests positive is
P(D|+)=P(D∩+)/P(+)
=0.00282/0.02276
where
D=event that the person has the disease
+=event that the person tests positive.