SarahNibs comments on [Link] The Bayesian argument against induction.

SarahNibs 19 Jul 2011 0:31 UTC
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P(Av~B|B) does not equal P(A). P((Av~B) & B) equals P(A).

Edit: it doesn’t, of course. P(Av~B|B) = P(A|B) and the other thing I said is just silly.
What links here?
- Zack_M_Davis's comment on [Link] The Bayesian argument against induction. by Peterdjones (19 Jul 2011 18:54 UTC; 3 points)
- Zack_M_Davis 19 Jul 2011 0:56 UTC
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  P(Av~B|B) does not equal P(A) [...] P(Av~B|B) = P(A|B)
  
  Whoops! You’re right. I’ll edit the comment to reflect this correction.
- Manfred 19 Jul 2011 7:23 UTC
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  Just for general convenience: P((A+~B)B) = P(AB).