Does this equation hold? $ P(A \cap B \cap C | D \cap E) = P(A \cap B \cap (C | D \cap E)) $

Question

Does this equation hold? Can we treat conditioning and intersection operation associative?

Answer 1

Conditioning is not an operation on events. Let $\Omega$ denote the sample space and $\mathcal F\subseteq\mathcal P(\Omega)$ denote the set of events (a $\sigma$-algebra on $\Omega$). While intersection takes two events $A,B\in\mathcal F$ to a new event $A\cap B\in\mathcal F$, conditioning is an operation on probability measures:

Given any event $B\in\mathcal F$ with $P(B)\neq 0$, it takes the probability measure $P\colon \mathcal F\to[0,1]$ to a new probability measure $P({-}\mid B)\colon \mathcal F\to [0,1]$ given by $$ P(A\mid B) = \frac{P(A\cap B)}{P(B)}. $$

Here "$A\mid B$" is not an event or anything considered on its own. The notation $P(A\mid B)$ should be read as the probability of $A$ under the conditioned measure $P({-}\mid B)$.

Hence, an expression of the kind $P(X\cap(Y\mid Z))$ is meaningless, since it is not of the kind $P(\ldots\mid B)$ and the notation "$Y\mid Z$" on its own has no meaning.