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).
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 :)