Re: thank you
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ahhh.... Outstanding counter-example, thanks. I'm convinced there is no unique solution.
I'd like to hear more about your claim that a must not be greater than .2... can you explain this further? I don't see the conflict in the equations you mentioned.
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Sorry, I misstated it in my answer. You need the second equation as well. The second and fourth together tell us that the probability of c U (not (a U b U d U e)) is 0.2 (since taking c out reduces the probability from 1 to 0.8). Therefore P(c U (not a)) >= 0.2. P(a U c) = P(a) + P(not a U c) = 0.4, so P(a) <= 0.2.
If you only want to prove there's no unique solution, it's easy to see that:
P(a) = P(b) = P(c) = P(d) = P(e) = 0.2 is a solution with everything mutually exclusive and,
P(a) = P(d) = 0, P(b) = P(c) = P(e) = 0.4 is another solution with b and c mutually exclusive and P(not b U c U not e) = 0.2.
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