Question

Somebody please assist me here

Answer 1

The base case of $n=1$ is trivially true, since

$\displaystyle P\left(\bigcup_{i=1}^1 E_i\right) = P(E_1) = \sum_{i=1}^1 P(E_i)$

but I think the case of $n=2$ may be a bit more convincing in this role. We have by the inclusion/exclusion principle

$\displaystyle P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1 \cup E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le P(E_1) + P(E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le \sum_{i=1}^2 P(E_i)$

with equality if $E_1\cap E_2=\emptyset$.

Now assume the case of $n=k$ is true, that

$\displaystyle P\left(\bigcup_{i=1}^k E_i\right) \le \sum_{i=1}^k P(E_i)$

We want to use this to prove the claim for $n=k+1$, that

$\displaystyle P\left(\bigcup_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)$

The I/EP tells us

$\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cup E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right)$

and by the same argument as in the $n=2$ case, this leads to

$\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1})$

By the induction hypothesis, we have an upper bound for the probability of the union of the $E_1$ through $E_k$. The result follows.

$\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^k P(E_i) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)$