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

• Mar 9th 2011, 04:44 PM
Jame
Probability of the union of independent events proof (induction argument)
I am trying to prove the following:

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

This is true for \$\displaystyle n=2\$

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

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

Therefore \$\displaystyle 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
\$\displaystyle 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 \$\displaystyle 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
\$\displaystyle
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
\$\displaystyle 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
\$\displaystyle 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.