- #71
Agent Smith
- 336
- 36
@Dale and @FactChecker I think the discussion we've had is adequate for my level. Thank you.
One last question. For Bayes' theorem given as ##P(H|D) = \frac{P(H) \times P(D|H)}{P(D)}##, the Wikipage says that ##P(D) \ne 0## (division by ##0##). But I noticed that since ##P(\neg A) = 1 - P(A)##, if ##P(D) = 0##, we know that ##P(\neg D) = 1##. Can't we use this knowledge to solve the division by ##0## problem?
So we could do ##P(H|\neg D) = \frac{P(H) \times P(\neg D|H)}{P(\neg D)} = \frac{P(H) \times P(\neg D|H)}{1} = P(H) \times P(\neg D|H)##
One last question. For Bayes' theorem given as ##P(H|D) = \frac{P(H) \times P(D|H)}{P(D)}##, the Wikipage says that ##P(D) \ne 0## (division by ##0##). But I noticed that since ##P(\neg A) = 1 - P(A)##, if ##P(D) = 0##, we know that ##P(\neg D) = 1##. Can't we use this knowledge to solve the division by ##0## problem?
So we could do ##P(H|\neg D) = \frac{P(H) \times P(\neg D|H)}{P(\neg D)} = \frac{P(H) \times P(\neg D|H)}{1} = P(H) \times P(\neg D|H)##
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