A disease hits with 1/1000 frequency. There is a test that produces 5% false positives. People are tested randomly without regard to showing symptoms or not. What is the probability that someone who tests positive is sick?

In the book he solves it using the frequentist approach, the answer is 1/51 = 0.0196. I have been trying using conditional probability (and failing :( ). This is where I got so far: Let S=sick, NS=not sick, TY=test says 'yes', TN=test says 'no'.

P(S) = 0.001
P(NS) = 0.999
P(TY|NS) = 0.05
P(TN|NS) = 0.95

From this we also get:
P(TY∩NS) = 0.04995
P(TN∩NS) = 0.94905
- though I don't know what to do with this, if anything.

There is no information on what the test says when people are sick but there is no talk of 'false negatives', so maybe (?) we can assume that:
P(TY|S) = 1
P(TN|S) = 0, and so
P(TY∩S) = 1
P(TN∩S) = 0

We are looking for P(S|TY)
Which all looks fine: sick/non-sick, test-says-yes/test-says-no are mutually exclusive and collectively exhaustive - the problem is that I cannot get to that 0.0196 :(. If I try Bayes Theorem, I end up with (apologies for the typing, I don't know how to use LaTex):

P(S|TY) = P(TY|S)P(S)/[P(TY|S)P(S) + P(TY|NS)P(NS)]
but this gives 0.050, so it's obviously not the right answer. Where am I going wrong? Or can this not be solved this way at all?


1 Answer 1


Your application of Bayes formula is correct. So, if assumed zero false negative, i.e. $P(TY|S)=1$: $$P(S|TY)=\frac{1\times0.001}{1\times 0.001+0.999\times0.05}=\frac{1}{50.95}\approx 0.0196$$

which is very close to $1/51$, so I believe the author just made an approximation by saying $1/51$.

  • $\begingroup$ Thank you, @gunes! It's totally maddening that it's the Maths of adding/multiplying/dividing a few numbers with a calculator that has somehow defeated me all along. (I was up with this till about midnight last night!) [facepalm] $\endgroup$
    – Reader 123
    Jun 5, 2020 at 14:50

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