Given events $A_1,...,A_n$. We say that $i\leq n$ "is good" iff $A_i$ occurs and $i$ is bad iff $A_i^c$ occurs. We are given $\delta_1,...\delta_n\in[0,1]$ s.t. $\mathbb P[i \text { is good}]\geq1-\delta_i$. The question: Bound $\mathbb P[\exists i\leq n:i\text{ is good}]$.
Attempt:
$$\mathbb P[\exists i\leq n:i\text{ is good}]=1-\mathbb P[\forall i\leq n:i\text{ is bad}] $$ So it suffices to bound $$\mathbb P[\forall i\leq n:i\text{ is bad}] $$ where we know $$\forall i\leq n: \mathbb P[i\text{ is bad}]\leq \delta_i $$ I don't know how to proceed because it is not given that $(A_i)$ are independent.
In other words, you want to bound $ P[ \bigcup_i A_i]$. A standard way of doing this is to use the Bonferroni inequalities, which give both upper and lower bounds to the probability of $\bigcup_i A_i$. This can be viewed as a version of the principle of inclusion and exclusion.
Among them, the simplest lower and upper bounds are $$ \sum_{i}P[A_i] - \sum_{i<j} P[A_i\cap A_j]\le P[ \bigcup_i A_i] \le \sum_i P[A_i].$$
How useful this is in practice depends a lot on the particulars of the situation. The upper bound displayed here (often called the union bound or Boole's inequality) is often very effective if the individual $A_i$ have small probabilities.