Conditional Probability Problem

[pocketjacks]

Suppose that if a person with tb is given a chest X-ray the probability that his condition will be detected is .95, and that if a person without tb is given a chest X-ray the probability that he will be diagnosed incorrctly as having tb is .002. Suppose, furthermore, that .1 per cent of the adult residents of a certain city have tb. If one of these persons selected at random is diagnosed as having tb on the basis of a chest X-ray, what is the probability that he actually has tb?

Why is the answer not just the complement of .002 which is .998? Or is it?

Do you really have to consider the last detail of the problem?

Thank you very much in advance.

[CT11]

I get ~0.3223.
Its a standard application of Bayes rule and you do need all of the information you give.

[CT11]

In case you want to know where that came from.
Let A and B be events in the same probability space.
Let B be the event that a person has TB. Given no information about this person P(B) = 0.001 (given by you).
Let A be the event that the same person tests positive.

You have also given that P(A|B)=0.95 and P(A|B')=0.002 where B' is the complement of B.

By the law of total probability
P(A) = P(A|B)P(B)+P(A|B')P(B') (1)

where P(B')=1-P(B) (2)
by the axioms of probablillity.

The last sentence poses the question what is P(B|A)?
by Bayes rule:
P(B|A) = P(A|B)P(B)/P(A) (3)
by substituting (1) and (2) into (3) we get an equation with all know probabilities.

I punched it into my calculator and got the number above

[shadowclown]

You are correct CT11.

pocketjacks if you do not understand CT11's math used to solve this problem you can think of it litteraly.

Say for instance you have 10000 people in your town. Your problem tells us .1% has the disease. So (10000)(.001) = 10 people.

So 9990 people do not have the disease and 10 people do.

Your problem then states when a person that has the disease takes the test they will be found positive 95% of the time. So (10)(.95) = 9.5 people will be found with the disease.

It then goes on to say that of the people without the disease the test will produce a false positive .2% of the time. So (9990)(.002) = 19.98 people.

So when the town of 10000 people are tested (9.5+19.98) = 29.48 people are found with the disease.

The question is what is the probability that the person testing positive actually has the disease and its simply (9.5 / 29.48) = 32.23%