callipygian
Well-Known Member
I didn't watch the video, but I'll throw in my analysis.
I'm going to assume "99% accurate" means 99% of the counters/diseased (they're one and the same, anyway :laugh: ) are identified, and 99% of the non-counters are cleared.
So if X fraction of the population are counters, then the fraction of the population who tests positive are 0.99*X+0.01*(1-X) = 0.01+0.98*X. The percentage of people who are counters who test positive would be f = 0.99*X/(0.01+0.98*X).
For an X of 0.01, f would equal 0.5, or a 50% chance that the postive is a counter.
For an X of 0.1, f would equal 0.99/1.08, or a 92% chance.
Interestingly enough for the opposite f' (defined as the probability of actually being a non-counter if the test is negative), it's virtually certain.
For an X of 0.01, f' would equal 0.99*0.99/(0.99-0.98*.01) = 0.9999!
For an X of 0.1, f' = 0.99*0.9/(0.99-0.98*.1) = 0.9989!
So it's relatively simple to clear people, but relatively difficult to prove them guilty.
I'm going to assume "99% accurate" means 99% of the counters/diseased (they're one and the same, anyway :laugh: ) are identified, and 99% of the non-counters are cleared.
So if X fraction of the population are counters, then the fraction of the population who tests positive are 0.99*X+0.01*(1-X) = 0.01+0.98*X. The percentage of people who are counters who test positive would be f = 0.99*X/(0.01+0.98*X).
For an X of 0.01, f would equal 0.5, or a 50% chance that the postive is a counter.
For an X of 0.1, f would equal 0.99/1.08, or a 92% chance.
Interestingly enough for the opposite f' (defined as the probability of actually being a non-counter if the test is negative), it's virtually certain.
For an X of 0.01, f' would equal 0.99*0.99/(0.99-0.98*.01) = 0.9999!
For an X of 0.1, f' = 0.99*0.9/(0.99-0.98*.1) = 0.9989!
So it's relatively simple to clear people, but relatively difficult to prove them guilty.