[PairTheBoard]

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Regardless, your original conditions still imply that

P( (a U b U d U e) intersect (a U c) ) = .2


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Frankly, I'm getting rather bored with your simply stating that this is the case. You may be right, but "PairTheBoard said so" does not constitute a proof in my mind.

There is obviously confusion over this point, so if you can prove what you claim, why don't you just do it?


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Given any two events E,F of a Sample Space, it's well known that,

* P(E union F) = P(E)+P(F)-P(E intersect F)

Letting E=(a U b U d U e)
and F=(a U c)
the proof follows from the fact that
(a U b U d U e) U (a U c) = (a U b U c U d U e)
and your assumption that
P(a U b U c U d U e) = 1

ie. P(E union F) = P(E) + P(F) - P(E intersect F)
so, 1 = .8 + .4 - P(E intersect F)
so, P(E intersect F) = .8 + .4 - 1 = .2

To see the * equation holds for any two Events E,F, using the set notation,
A-B = {x: x is in A and x is not in B)
we have the following partitions into mutually exlusive events,

E = (E-F) union (E intersect F)
F = ((F-E) union (E intersect F)
(E U F) = (E-F) union (F-E) union (E intersect F)
so that,

P(E-F) = P(E) - P(E intersect F)
P(F-E) = P(F) - P(E intersect F)
and
P(E union F) = P(E-F) + P(F-E) + P(E intersect F) =
= P(E)-P(E intersect F) + P(F) - P(E intersect F) + P(E intersect F) =
= P(E) + P(F) - P(E intersect F)

PairTheBoard