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Given $P(A|B) = x$, what can you say about $P(\overline{A}|B)$?

Can we say $1 - x$?

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    $\begingroup$ Possible duplicate of Negation of Bayes' theorem. $\endgroup$ – johny Feb 13 '17 at 23:24
  • $\begingroup$ The proposed duplicate is a fairly circuitous presentation of ideas at best, starting with an incorrect statement of conditional probability (or of Bayes' Theorem, hard to say). I don't think it makes a compelling duplicate. $\endgroup$ – hardmath Feb 14 '17 at 1:25
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Yes we can. The same question is answered in this answer.

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  • $\begingroup$ If that answers this question as well, you should flag this one as a duplicate. $\endgroup$ – Glorfindel Feb 2 '17 at 13:12

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