Properties of Independent event in Probability

Question

If $A_1,.....A_n$ are independent event how can I prove that

$\hspace{2cm} P(A_1\cup A_2\cup......\cup A_n)=1-\prod_{i=1}^n (1-P(A_i))$

Answer 1

Hint Use the identity $$ A_1\cup A_2\cup......\cup A_n=(A_1^c\cap A_2^c\cap......\cap A_n^c)^c $$ to write $$ P(A_1\cup A_2\cup......\cup A_n)=1-P(A_1^c\cap A_2^c\cap......\cap A_n^c) $$ and proceed from there.

Answer 2

Proof: Inducting on n.

It suffices to check for $n = 2$, which holds by comparing the formula with the inclusion-exclusion formula [$P(A \cup B) = P(A) + P(B) - P(A \cap B) = P(A) + P(B) - P(A)P(B)$ --- since A and B are independent events].

Suppose the formula holds for n. Let $B = A_{1} \cup \cdots \cup A_{n}$. Then

$P(B \cup A_{n+1}) = 1 - (1- P(B))(1 - P(A_{n +1}))$

$ = 1 - \left( 1 - \left(1 - \prod_{i=1}^{n} \left(1 - P(A_{i})\right) \right) \right) \left(1 - P(A_{n+1})\right) = 1 - \prod_{i=1}^{n+1}(1 - P(A_{i}))$.

The result holds by induction.