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Comments (72)
While A and B being mutually exclusive seems reasonable, I don't think it holds for C and D. And I'm pretty sure that it doesn't hold at all for E and F.
If I remember visualising 2+2=3 yesterday and 2+2=4 the day before, then E and F are both simultaneously true.
These three statements, taken together, are impossible. Consider:
Over the 0.75 probability space where C is true (second statement), A is only true in half that space (third statement). Thus, A is false in the other half of that space; therefore, there is a probability space of at least 0.375 in which A is false. Yet A is only false over a probability space of size 0.25 (first statement).
In your calculations further down, you use the value P(C) = (.75.50+.250) = 0.375; using that value for P(C) instead of 0.75 removes the contradiction.
Similarly, the following set of statements lead to a contradiction, considered together:
The first and third comments are correct. I made some errors in first typing it up that shouldn't take away from the argument that are now fixed. The third comment is an actual mistake that has also been fixed.
This is wrong. P(C|A) is read as C given A, which is the chance of C, given that A is true. You're mixing it up with P(A|C). However, if you switch A and C in your paragraph, it becomes a valid critique, which I've fixed, substituting the correct values in. Thanks. (Did I mess anything else up?)
I'm starting to appreciate mathematicians now :)
You need to escape your * symbols so they output correctly.
You're right, I had that backwards.
Hmmm....
You have two different values for P(F). Similarly, the value P(E)=0.70 does not match up with P(C), P(D) and the following:
None of which is going to affect your point, which seems to come down to the claim that there exist possible events A, B, C, D, E and F such that P(A|C) = 1.