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:

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

• Because "$,X$" appears everywhere in the first formula, it is superfluous, so drop it: what remains? – whuber Dec 17 '19 at 17:01
• @whuber Thanks, I didn't know about that rule that if it appears everywhere we can drop it. Why we can do that? – raffaem Dec 17 '19 at 17:03
• Because it's superfluous: explicitly writing "$X$" as a condition of every probability (in this general context) tells us exactly nothing. – whuber Dec 17 '19 at 17:06

$$\begin{equation} P(A, B, X) = P(A|B,X)P(B,X) \end{equation}$$ 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.