We are given the events $A$, $B$, $C$ and I, where $A$, $B$, $C$ are independent. We know the following probabilities: $P(I)$, $P(A|I)$, $P(B|I)$, $P(C|I)$, $P(A|\overline I)$, $P(B|\overline I)$, $P(C|\overline I)$. I have to find $P(\overline A \cap \overline B \cap \overline C)$ but i do not understand which and why is the correct way to calculate it: $$\begin{align}P(\overline A \cap \overline B \cap \overline C) &= P(\overline A)\cdot P(\overline B)\cdot P(\overline C)\\[2ex]& = {{[P(\overline A|I)P(I) + P(\overline A|\overline I)P(\overline I)]} \\ \cdot {[P(\overline B|I)P(I) + P(\overline B|\overline I)P(\overline I)]} \\ \cdot{[P(\overline C|I)P(I) + P(\overline C|\overline I)P(\overline I)]}}\end{align}$$ or $$\begin{align}P(\overline A \cap \overline B \cap \overline C) &= P(\overline A \cap \overline B \cap \overline C | I)\cdot P(I) + P(\overline A \cap \overline B \cap \overline C| \overline I)\cdot P(\overline I) \\[2ex]&= P(\overline A|I)\cdot P(\overline B|I)\cdot P(\overline C|I) \cdot P(I) + P(\overline A|\overline I)\cdot P(\overline B|\overline I)\cdot P(\overline C|\overline I) \cdot P(\overline I)\end{align}$$
I tried to simulate this problem and I found out that the second equation is the correct one, but i don't understand why the first one is wrong