Re: p(A1 U A2 U ...UAk) = p(1) + p(2) + ... + p(k)

Bob O´Hara wrote

*** Anyway, the problem is that the Binomial theorem assume that the events are independent (that's why you can take (1-p + p) to the kth power). But you assume that the events are mutually exclusive, which is impossible if independence is also assumed, except in the trivial case.***


The expression on right
____p(1) + p(2) + ... + p(k)

is referring to mutually exclusive events that are independent (yes Bob). In fact
____or 1 success occurs
____or 2 successes
..
..
..
____or k successes



Certinly he occurrence of exactly j events excludes that occurs
1,...,(j-1),(j+1),..., k but IMO this have nothing to do with INDEPENDENCE because the context is ONE ONLY random experiment and its Sample Space which has ONLY an *enumerative* role - what are all possible results that came out from the experiment.
The Bernoilli trials are INDEPENDENT because the result of one of them IS NOT INFLUENCED by what happened before. Simple, isn´ít ?

(furthermore the assumption of Independence in Bernouillan trials can be found in every elementary text on Probability).

____licas (Luis A. Afonso)
.