While working through a problem, I encountered $P(A'\cap B' \cap C'|A')$. The confusion I'm having is when breaking it apart using the conditional probability formula. The way I'm thinking about it is
$$P(A'\cap B' \cap C'|A') = \frac{P((A' \cap B' \cap C') \cap A')}{P(A')} = \frac{P(A')}{P(A')} = 1$$
Now the issue as you might see is that it doesn't work like that but I can't seem to find an explanation as to why it doesn't behave like this. Would anyone be able to explain why this is incorrect?
EDIT : I figured out my issue. I was looking at the intersection between the two sets and was thinking to myself " What's common between those two sets? A' ", so I was thinking the intersection was just A'. But that's not right, because $(A' \cap B' \cap C')$ isn't all of $A'$, it's only the portion of $A'$ that also intersect with $B'$ AND $C'$. Thanks so much for the answers and comment. I knew I was missing something, I just wasn't quite sure what.
Okay you were on the right path but took a wrong turn.
Why, $A'\cap(A'\cap X)$ is not necessarily $X$ unless $X\subseteq A'$.
However, $A'\cap(A'\cap X)$ is always $A'\cap X$, so...
$$\begin{align}\mathsf P(A'\cap X\mid A')&=\dfrac{\mathsf P(A'\cap (A'\cap X))}{\mathsf P(A')}\\[1ex]&=\dfrac{\mathsf P((A'\cap A')\cap X)}{\mathsf P(A')}\\[1ex]&=\dfrac{\mathsf P(A'\cap X)}{\mathsf P(A')}\\[1ex]&=\mathsf P(X\mid A')\end{align}$$
When $X=B'\cap C'$ we have $$\mathsf P(A'\cap B'\cap C'\mid A')=\mathsf P(B'\cap C'\mid A)$$