A test for a certain diseases has a probability 4/5 of detecting the disease when it is present and the probability of 1/10 of falsely detecting it when it is not present. The proportion of the people afflicted by the disease is 15%. If a person from the population is randomly selected and gives a positive result to the test, what is the probability that this person has the disease?

Question

anonymous · Answer

So the test has a 4/5 = 8/10 probability of being a true positive and 1/10 of it being false positive. So the total probability of the test being any type of positive is 9/10. The person got a positive test. The probability of getting a TRUE positive from a true test is the probability of getting a true false divided by the probability of getting any type of positive.

anonymous · Answer

The answer is 24/41...how did they get that?

anonymous · Answer

Hmmm....

anonymous · Answer

I'm stumped.

phi · Answer

The way you get a positive result is:
0.15* 0.8  (15 % chance of having the disease times the prob the test picks it up)
plus
0.85* 0.1  (you don't , but you get a false positive)
that adds up to 0.205
the first expression (when you have the disease) is 0.12
so .12 out of 0.205 times you really have the disease
the ratio of the two is 0.12/0.205  or   24/41

anonymous · Answer

Oh, so there is some sort of law of total probability?

anonymous · Answer

how do i know that i dont have to apply bayes theorem?

phi · Answer

You can make a chart with all 4 possibilities.  Maybe that helps to see what is going on

anonymous · Answer

like a probability tree?

phi · Answer

P(a|b)  = P(a n b)/ P(b)  (n means intersection)

prob of a given b  corresponds to P(sick) given positive test.

anonymous · Answer

so its conditional probability question...is it advisable to just a diagram for conditional probability questions?

phi · Answer

That is a good strategy.  Practice, practice, practice helps a lot.

anonymous · Answer

anyways thanks!