A certain clinical test is 98% effective in detecting a disease when
the disease is in fact present. However, the test also shows a false
positive in 1% of cases where the patient tested is in fact healthy.
If 0.4% of the population actually suffers from the disease, what is
the probability that a person who tests positive actually has the
disease?
I would like to know how to do it in the way shown if my answer is incorrect as that is what we are asked to do in the exam. In other words we are not allowed to blow the population up by *1000.
P(B) = population who has the disease
P(B1) = .004
P(B2) = Population that does not have the disease but test positive
P(B2) = 1 − P(B1) = .996
P(A) Test is positive
P(A|B1) = .98
P(A|B2) = .01
Then
(.996∗.004)/(.996∗.004)+(.996∗.01)
= 27 (.28)
This figure seems off to me, it seems to good to be true if the test is positive you actually have the disease 28% of the time.
Are my calculations correct or did I miss something?
Best Answer
$H$=person is healthy
$T$=person is tested positive
You want to calculate $P(H^c|T)$ We have $$P(T|H^c)=0.98$$ $$P(T|H)=0.01$$ and $$P(H)=0.996$$
Using conditional probability definition, $$P(H^c|T)=\frac{P(T\cap H^c)}{P(T)}=\frac{P(T|H^c)P(H^c)}{P(T)}$$
Total probability gives : $$P(T)=P(T \cap H)+P(T\cap H^c)=P(T | H)P(H)+P(T| H^c)P(H^c)$$
Replacing, we have $$P(H^c|T)=\frac{P(T|H^c)P(H^c)}{P(T | H)P(H)+P(T| H^c)P(H^c)}=\frac{0.98*0.004}{0.01*0.996+0.98*0.004} \approx 28.24\%$$
Now you can understand why $1\%$ false positive is unacceptable