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Bound_up comments on Open Thread, Jun. 8 - Jun. 14, 2015 - Less Wrong Discussion
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Comments (153)
Did I use Bayes' formula correctly here?
Prior: 1/20
12/20 chance that test A returns correctly +
16/20 chance that test B returns correctly +
12.5/20 chance that test C returns correctly +
Odds of correct diagnosis?
I got 1/2
Let's assume that every test has the same probability of returning the correct result, regardless of what it is (e.g., if + is correct, then Pr[A returns +] = 12/20, and if - is correct, then Pr[A returns +] = 8/20).
The key statistic for each test is the ratio Pr[X is positive|disease] : Pr[X is positive|healthy]. This ratio is 3:2 for test A, 4:1 for test B, and 5:3 for test C. If we assume independence, we can multiply these together, getting a ratio of 10:1.
If your prior is Pr[disease]=1/20, then Pr[disease] : Pr[healthy] = 1:19, so your posterior odds are 10:19. This means that Pr[disease|+++] = 10/29, just over 1/3.
You may have obtained 1/2 by a double confusion between odds and probabilities. If your prior had been Pr[disease]=1/21, then we'd have prior odds of 1:20 and posterior odds of 1:2 (which is a probability of 1/3, not of 1/2).
Kindly, indeed.
Thank you. I believe I've got it down now.
Prior:1/101
Test: Correct positive 95%
False positive 20%
1 of the 101 has the disease, with 95% probability of receiving a positive reading, denoting 1 x .95 = .95
And 100 don't have the disease, each with a 20% probability of a positive reading, denoting 100 x .2=20
.95 + 20 = 20.95
.95 / 20.95 = .045, denoting a 4.5% chance that someone receiving a positive reading has the disease.
Thank you again :)
I am assuming you want the posterior probability of disease given three positive tests. You're going to need more information - unless you provide either the specificities or the likelihood ratios, the question cannot be answered.
I may have used weak phrasing.
Each test returns positive. The frequency out of 20 go's that each does so correctly is indicated respectively by the "12," "16," and "12.5."
So really, it's the odds of actually having the disease, given the three positive test results, I guess. Would it be 1/2 under those circumstances?
Thank you for your assistance.
You also need to consider the odds of a false positive for each test.