Between classical and meridian probability theories

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Michael Atovigba

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Nov 20, 2015, 7:14:57 AM11/20/15
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There is a problem with the classical probability theory. If we toss a
coin we have 0.5 of obtaining a head and 0.5 of not having a head,
using the classical theory. so no matter the number of times a coin is
thrown both the probability sum for having a head and having a tail,
and that of not having a tail and not having a head sum to 1 as
specified by the forefathers of probability theory.
However, if we have a dice of six faces we have the probability of
having a one, or two or ,,,,or six will each be 1/6 and the total sum
of the probabilities will be one as specified; but having the
probability of not a one, not a two, ... not a six which is each 5/6
will end up with 5 as sum, which contradicts the specification of the
probability theory as stated by its forefathers.
This is the crash of the classical theory of probability.
Luckily the meridian probability theory, has come handy to save the
situation. a BSU don Michael Atovigba tried to discover the geometric
proof of the probability theory and ended up with the proof stated in
form of the meridian probability theory. see

https://groups.yahoo.com/neo/groups/geometric-proof-of-the-probability-theory/info
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