= 27d567bc2d48d9f40e9ccc20ea370b15)
Manfred comments on Open thread, July 21-27, 2014 - Less Wrong Discussion
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (160)
Question about Bayesian updates.
Say Jane goes to get a cancer screening. 5% prior of having cancer, the machine has a success rate of 80% and a false positive rate of 9%. Jane gets a positive on the test and so she now has a ~30% chance of having cancer.
Jane goes to get a second opinion across the country. A second cancer screening (same success/false positive rates) says she doesn't have cancer. What is her probability for having cancer now?
Doing this with probabilities is a bit more complicated than what Coscott did, but to illustrate it anyhow...
where A is cancer and and C are the two test results, P(A|BC)=P(A) P(BC|A) / P(BC). P(A) is our prior of 5%. Because B and C are independent, P(BC|A) is just 0.8 * 0.2.
P(BC) is where using probabilities is more complicated than using odds, because it's not the probability of false positives, it's the total prior probability of seeing B and then C. Using the product rule, P(BC) = P(B)*P(C|B). Then splitting the possibilities up into cancer and not-cancer, this becomes (P(AB)+P(¬A B))*(P(AC|B)+P(¬A C|B)). Because B and C are independent, the second part becomes (product rule) P(A|B)*P(C|A)+P(¬A|B)*P(C|¬A) - note that even when we added both pieces of evidence at once, we still have to calculate the intermediate probability P(A|B)! Stupid non-time-saving grumble grumble.
Anyhow, if you plug in the numbers, it's ~0.094.