# Why does $P\left(\bigcap\limits_{i \geq 1} A_i\right) \geq 1- \sum\limits_{i \geq 1} \alpha_i$ hold?

Given a probability space $(\Omega,F,P)$.

Let $A_1, A_2, ..$ be events, $A_i\in F$.

Let $P(A_i) \geq 1-\alpha_i$ where $i \geq 1$ and $0 \leq \alpha_i \leq 1$.

Why does the following hold:

$$P\left(\bigcap\limits_{i \geq 1} A_i\right) \geq 1- \sum\limits_{i \geq 1} \alpha_i$$

Thanks a lot in advance :)

EDIT:

I see, so:

1. If $P(A_i) \geq 1-\alpha_i$, then $P\left(\overline{A_i}\right) \leq \alpha_i$.
2. then $P\left(\bigcup\limits_{i \geq 1} \overline{A_i}\right) \leq \sum\limits_{i \geq 1} P(\overline{A_i}) \leq \sum\limits_{i \geq 1} \alpha_i$
3. and because $P\left(\overline{\bigcup\limits_{i \geq 1} \overline{A_i}}\right ) = P\left(\bigcap\limits_{i\geq 1} A_i\right)$ it holds $P\left(\bigcap\limits_{i \geq 1} A_i\right) \geq 1- \sum\limits_{i \geq 1} \alpha_i$?
• Hint: use $P(\overline{A_i})\le \alpha_i$ – Gribouillis Sep 10 '17 at 10:54
• and $P\left(\bigcup_j B_j\right) \le \sum_j P(B_j)$ – Henry Sep 10 '17 at 10:57
• @Gribouillis Edited, is that correct? :) – user419869 Sep 10 '17 at 11:07
• @Derping s almost correct, $P$ must not be overlined, only what's inside $P()$. – Gribouillis Sep 10 '17 at 11:18
• @rtybase now it should be fine^^ sorry – user419869 Sep 10 '17 at 11:44

Perhaps more compactly, $$P\left(\bigcap_i A_i\right) = 1 -P\left[\left(\bigcap_i A_i\right)^c\;\right] \ge 1 - P\left(\bigcup_i A_i^c\right) \ge 1 - \sum_iP(A_i^c) \ge 1 - \sum_i \alpha_i.$$