Re: p(A1UA2UA3)<= p(A1)*p(A2)*p(A3)


"Luis A. Afonso" wrote:
Correction

Instead of p(1)*p(2)*p(3) read (I am so sorry)

_____p(1) + p(2) + p(3)

_____a real and self-evident misprinting...

______licas (Luis A. Afonso)

Your misprint was your INEQUALITY!

Your correction is WRONG because p(A1 U A2 U A3) is p(1) + p(2) + p(3)
only if A1, A2, A3 are mutually exclusive.

But P(A1 U A1 U A3) is always .GE. p(A1 <and> A2 <and> A3)

and the proof does NOT involve any coin tossing example.
The inequality is ALWAYS true, for any events A1, A2, and A3.

Now proof it, without any actual example of events or any GBASIC
program.

Furthermore, if A1, A2, and A3 are mutually independent, then your
statement in the SUBJECT is correct except for the wrong inequality.

-- Bob.

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