Given the probability to tag a b-jet by:
e = # b-jets tagged / # total b jetsthe probablity per EVENT to tag 1, 2 or more than 1 b-jets, will be given by
P(1 tag) = e.(1-e) + e.(1-e) P(2 tag s) = e^2 P(>=1 tags) = P(1tag) + P(2 tags)if the process is independent (i.e. tagging the second b-jet is independent from having tagged the first one)
Based on e = 0.49, the tagging probablitity per event for NON-CORRELATED tagging should be:
Which is in agreement to what we found.