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In the slides of my professors there is written that

$$ P(A|B,X) = \dfrac{P(B|A,X) P(A,X)}{P(B,X)} $$

Is it correct? And how it is so? The Bayes theorem states that

$$ P(A|B) = P(B|A) \dfrac{P(A)}{P(B)} $$

So by the Bayes theorem I think that it should be that:

$$ P(A|B,X) = P(B,X|A) \dfrac{P(A)}{P(B,X)} $$

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    $\begingroup$ Because "$,X$" appears everywhere in the first formula, it is superfluous, so drop it: what remains? $\endgroup$ – whuber Dec 17 '19 at 17:01
  • $\begingroup$ @whuber Thanks, I didn't know about that rule that if it appears everywhere we can drop it. Why we can do that? $\endgroup$ – raffaem Dec 17 '19 at 17:03
  • $\begingroup$ Because it's superfluous: explicitly writing "$X$" as a condition of every probability (in this general context) tells us exactly nothing. $\endgroup$ – whuber Dec 17 '19 at 17:06
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\begin{equation} P(A, B, X) = P(A|B,X)P(B,X) \end{equation} This is obvious, if you think of $A \& X$ as a single event, say $C$. Similarly \begin{equation} P(A, B, X) = P(B|A,X)P(A,X) \end{equation} And your relation immediately follows.

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