Prove that the probability of occurrence of all of a given set of events A(1) through A(n) is equal to the sum of the individual probabilities minus the sum of the probabilities of all pairs of events, A(i) OR A(j) plus the sum of all triples of events, A(i) OR A(j) OR A(k), ..., plus (1)^(n1) times the ntuple A(i) OR ... OR A(n).
Prove for the specific cases of n = 3 and n = 10, and the general case.
(In reply to
re(2): Done !!!! by Gamer)
Yes, you could say that +, in representing OR, adds only the parts that aren't already there, as A OR B includes A AND B but has the part of A not in B and the part of B not in A.
Looking at your algorithm it also looks as if you are going for A AND B AND C in terms of ORed pairs and triples, while what is sought is A OR B OR C in terms of ANDed pairs and triples, and to prove that the algorithm is true. In fact, the ORed compound given in terms of ANDed compounds is found in probability texts, but the other way around, I haven't seen.
Yes I did see the same post 3 times. I can only hypothesize it's the result of trying to split a long post into three parts, but getting the same part each time.

Posted by Charlie
on 20030315 19:06:04 