
I'm trying to write a proof for this, but I don't know how to get started.
Would proof by induction be the easiest way?
If you could break it down into general steps I could wrap my head around, I would really appreciate it. Thanks.

I'm trying to write a proof for this, but I don't know how to get started.
Would proof by induction be the easiest way?
If you could break it down into general steps I could wrap my head around, I would really appreciate it. Thanks.
Using finite subadditivity of probability, we have
$1 - \mathbb{P}\left( {\bigcap\limits_{i = 1}^n {{E_i}} } \right) = \mathbb{P}\left( {{{\left( {\bigcap\limits_{i = 1}^n {{E_i}} } \right)}^C}} \right) = \mathbb{P}\left( {\bigcup\limits_{i = 1}^n {E_i^C} } \right) \leqslant \sum\limits_{i = 1}^n {\mathbb{P}\left( {E_i^C} \right)} = \sum\limits_{i = 1}^n {\left( {1 - \mathbb{P}\left( {{E_i}} \right)} \right)} = n - \sum\limits_{i = 1}^n {\mathbb{P}\left( {{E_i}} \right)} $
which implies $\mathbb{P}\left( {\bigcap\limits_{i = 1}^n {{E_i}} } \right) \geqslant 1 - n + \sum\limits_{i = 1}^n {\mathbb{P}\left( {{E_i}} \right)} = \sum\limits_{i = 1}^n {\mathbb{P}\left( {{E_i}} \right)} - \left( {n - 1} \right)$