Given $P(A|B) = x$, what can you say about $P(\overline{A}|B)$?
Can we say $1 - x$?
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1$\begingroup$ Possible duplicate of Negation of Bayes' theorem. $\endgroup$– johnyFeb 13, 2017 at 23:24
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$\begingroup$ 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. $\endgroup$– hardmathFeb 14, 2017 at 1:25
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$\begingroup$ If that answers this question as well, you should flag this one as a duplicate. $\endgroup$ Feb 2, 2017 at 13:12