The rare event approximation for event $A$ and $B$ means the upper-bound approximation $P(A\cup B)=P(A)+P(B)-P(A\cap B)\leq P(A)+P(B)$. Now by inclusion-exclusion-principle $$P(A\cup B\cup \neg C)= P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap \neg C) -P(B\cap\neg C) +P(A\cap B\cap \neg C) \leq P(A)+P(B)+P(C)+P(A\cap B\cap \neg C)$$
and now by Wikipedia, this is the general form
so does the rare-event-approximation means removal of minus terms in the general form of inclusion-exclusion principle aka the below?
$$\mathbb P\left(\cup_{i=1}^{n}A_i\right)\leq \sum_{I\subset\{1,3,...,2h-1\}; |I|=k}\mathbb P(A_I)$$
where $2h-1$ is the last odd term in $\{1,2,3,...,n\}$.
Example
For example in the case of three events, is the below true rare-event approximation?
$$P(A\cup B\cup \neg C) \leq P(A)+P(B)+P(\neg C)+P(A\cap B\cap \neg C)$$
P.s. I am studying probability-risk-assessment course, Mat-2.3117.
Removing all the negatives certainly gives an upper bound. But if one looks at the logic of the inclusion-exclusion argument, whenever we have just added, we have added too much (except possibly, at the very end). So at any stage just before we start subtracting again, our truncated expression gives an upper bound.
Thus one obtains upper bounds by truncating after the first sum, or the third, or the fifth, and so on.