What Is the Probability That a Positive Test Indicates Actual Disease?

AI Thread Summary
The discussion centers on calculating the probability that a person who tests positive actually has the disease, given a prevalence of 1 in 1000 and a 5% false positive rate. Participants suggest using Bayes' theorem to solve the problem. The poster attempts to define probabilities, noting that out of 1,000 tests, they expect 50 false positives and 1 true positive. This leads to the conclusion that the likelihood of a positive test indicating actual disease is low. The conversation emphasizes the importance of understanding conditional probabilities in interpreting test results.
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Homework Statement



Approx. 1/1000 has a disease, and the method of testing has a 5 % false positive rate. If a random person tests positive, what is the probability that he has the disease?

Homework Equations



I'm pretty sure Baye's theorem is the thing to use.

The Attempt at a Solution



I'm having trouble defining my probabilities. I've defined that P(sick) = 0,001 and P(not sick|+) = 0,05. But then I'm stuck...
 
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You could try the problem this way:

Let's say we test 1,000 people:
How many false positives do we expect to get?
How many real positives do we expect to get?
 
Well, wouldn't that be 50 and 1?
 
So, how likely is a positive test to indicate disease?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

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