Probability of the union of independent events proof (induction argument)

**Jame** · March 9th 2011, 04:44 PM

I am trying to prove the following:

If $E_1, E_2, ... , E_n$ are mutually independent events, then
$ P(E_1 \cup E_2 \cup ... \cup E_n) = 1 - \Pi^{n}_{i=1} [1 - P(E_i)| $

This is true for $n=2$

$ P(E_1 \cup E_2) = 1 - P(E_1^{C} \cap E_2^{C})$ by DeMorgan's Law

and since $E_1$ and $E_2$ are independent, $E_1^{C}$ and $E_2^{C}$ are independent

Therefore $P(E_1 \cup E_2) = 1 - P(E_1^{C} \cap E_2^{C}) = 1-P(E_1^{C})P(E_2^{C})$

Assume true for n
$P(E_1 \cup E_2 \cup ... \cup E_n) = 1 - P(E_1^{C} \cap E_2^{C} \cap ... \cap E_n^C) = 1-P(E_1^{C})*P(E_2^{C})*...*P(E_n^{C})$

Then $P(E_1 \cup E_2 \cup ... \cup E_n \cup E_{n+1}) = P((E_1 \cup E_2 \cup ... \cup E_n) \cup E_{n+1}) $

Thus by the theorem for n=2
$ P((E_1 \cup E_2 \cup ... \cup E_n) \cup E_{n+1}) = 1-P((E_1^{C} \cap E_2^{C} \cap ... \cap E_n^{C}) \cap E_{n+1}) = 1- P(E_1^{C} \cap E_2^{C} \cap ... \cap E_n^{C})P(E_{n+1}^{C})$

But by induction hypothesis
$1- P(E_1^{C} \cap E_2^{C} \cap ... \cap E_n^{C}) = 1-P(E_1^{C})*P(E_2^{C})*...*P(E_n^{C})$

Thus
$P(E_1 \cup E_2 \cup ... \cup E_n \cup E_{n+1}) = P((E_1 \cup E_2 \cup ... \cup E_n) \cup E_{n+1}) = 1-P(E_1^{C})*P(E_2^{C})*...*P(E_n^{C})*P(E_{n+1}^C})$

Which is what we want.

I was wondering if I proved the base case correctly and applied the induction hypothesis correctly. Thank you for your help.

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