Probability of circuit operation

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I have a question, which is relate to the following topic.

Assuming the following circuit, the probability that the entire system works is given by: $$P[X]\cup P[Y] = P[X] + P[Y] - P[X\cap Y],$$

given P[X] is defined by the branch AB and P[Y] by CDE. Now, what is the probability that A is not working given the system works?

I would write it as

$$P[A'|Z] = \frac{P[A'\cap Z]}{P[Z]},$$ where $P[Z]=P[X]\cup P[Y].$ Intuitively, one could consider $P[A']\cap P[Z] = P[A']P[Z]$. However, this only holds when $A'$ and $Z$ are independent. Since the system works independently from branch $AB$ when $C$, $D$ and $E$ are working, one could use $P[A'\cap (C\cap D \cap E)] = P[A']P[C]P[D]P[E]$. Finally, I reach my question. Why should one still use $P[Z]$ in the denominator, i.e., $$P[A'|Z] = \frac{P[A'\cap (C\cap D \cap E)]}{P[Z]}\quad (1)$$ rather than $$P[A'|C\cap D \cap E] = \frac{P[A'\cap (C\cap D \cap E)]}{P[C\cap D \cap E]}\quad (2)?$$

Equation (1) must be used in order to reach the same answer as given in the book. However, it bugs me that the conditional probability equation requires an "adaptation".

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The question is to calculate the probability that A is not working knowing the system works so $$P[A'|Z] = \frac{P[A'\cap (C\cap D \cap E)]}{P[Z]}\quad (1)$$

as you mentioned it.

The second equation you wrote is literally what is the probability that A is not functioning knowing the bottom branch is working. You would agree that the bottom branch working or not has nothing to do with A working or not, hence those two events are independent and you get simply:

$$P[A'|\cap (C\cap D \cap E)] = P[A'] * P[C\cap D \cap E]\quad (2)$$