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)`