Conditional probability - mistake in interpretation? Assume 3 events $A, B, C$ with success rates $p_1, p_2, p_3$. Let $X$ define an event, where exactly one of the three events had a success.
For me it's clear that $P(X) = P(X|A)P(A) + P(X|B)P(B) + P(X|C)P(C)$. 
Further, given that we know that $X$ holds (exactly one success was seen), the probability that $A$ was the one successful is $$ P(A|X) = p_1 \cdot (1-p_2) \cdot (1-p_3)$$ 
On the other hand, $$ P(X|A) = \frac{P(A|X) P(X)}{P(A)}$$
Logically it seems that $P(X|A)$,  the probability that exactly one success was seen, given that A was successful, is equal to $p_1 \cdot (1-p_2) \cdot (1-p_3)$, because it depends on the other two events failing. But this is equal to $P(A|X)$. 
Clearly, $P(X) \neq P(A)$, hence I'm wrong somewhere. 
Probably in the interpretation of the probability $P(X|A)$.  What am I missing?

Careful! By your same formula we have that 
$$P(A|X)=\frac{P(A\cap X)}{P(X)} = \frac{P(X|A)P(A)}{P(X|A)P(A) + P(X|B)P(B) + P(X|C)P(C)}$$
which is not equal to $p_1(1-p_2)(1-p_3)$.
The denominator in the above formula is exactly what you would interpret as "given that we know that $X$ holds". Indeed the denominator contains all possible outcomes. This is both the beauty and subtleness of Bayes' theorem.
