SOLUTION: Suppose there is a test for a disease that correctly gives positive results for 95% of those having the disease, and correctly gives negative results for 90% of those who do not ha

Question 178988: Suppose there is a test for a disease that correctly gives positive results for 95% of those having the disease, and correctly gives negative results for 90% of those who do not have the disease. Suppose also that the incidence of the disease is 1%. If a person tests positive for this disease, what is the chance that the person has the disease? Find the exact percentage.
Answer by kev82(151) (Show Source): You can put this solution on YOUR website!
TP = test positive
TN = test negative
D = diseased
O = not diseased
P(TP | D) = 0.95
P(TN | O) = 0.9
P(D) = 0.01
And obviously (probabilities sum to one)
P(TN | D) = 0.05
P(TP | O) = 0.1
P(O) = 0.99
Want to know:
P(D | TP)
By definition of conditional probability
P(D | TP) = P(D ^ TP)/P(TP)
P(TP | D) = P(TP ^ D)/P(D)
P(D | TP) = P(TP | D)*P(D)/P(TP)
From set theory
P(TP) = P(TP ^ O) + P(TP ^ D)
Conditional probability again
P(TP) = P(TP | O)*P(O) + P(TP | D)*P(D)
Substitute
P(D | TP) = P(TP | D)*P(D)/( P(TP | O)*P(O) + P(TP | D)*P(D) )
0.95*0.01 / ( 0.1*0.99 + 0.95*0.01 ) = 19/217
I'm pretty sure that's right, but last time I did this was 1999!
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