P(S|~M) vs. P(S)

Someone (If I remember faces right, Brandon Mechtley)  asked whether it is really easier to assess P(S|~M) compared to P(S). I tried to venture a weak answer in the class.

In general, there don't seem to be good and convincing arguments that it will always be easier to assess P(S|~M) compared to P(S). (So I oversold my
case :-(

The most reasonable explanation as to why we look at P(S|~M) type probabilities rather than P(S)  that I can offer on further reflection is that we are interested in computing posterior probability distribution of a variable (rather than just its maximal probable value).  If the patient's disease can be one of 7 types. There may be a prior distribution over these diseases, and after seeing some evidence, the doctor wants to get the posterior *distribution* on the diseases (not just the most likely disease but the distribution). If we are doing, we will anyway be needing
probabilities of type P(S|disease= di)  (note that P(S|~M) can be seen as P(S|M=false)).


I added a slide to the class notes making this point.

cheers
Rao