Addition or probability

Question

As shown, a is the sum of all probabilities, b is the exact probability of occurrence. Can I show $a$ is always greater or equal to $b$?

$a=\sum_{i=0}^n A_i $

$b=(1-\prod_{i=0}^n (1-A_i))$

$0\leq A_i \leq 1$

prove:

$a\geq b$

Answer 1

Use induction. The inequality is obvious if $n=0$. Suppose we know that $1-\prod_{i=0}^{n} (1-A_i)\leq \sum_{i=0}^{n} A_i$. Then $1-\prod_{i=0}^{n+1} (1-A_i)=1-\prod_{i=0}^{n} (1-A_i) (1-A_{n+1}) \leq 1- [1-\sum_{i=0}^{n} A_i] (1-A_{n+1}) $ by induction hypothesis. Just simplify; you should be able to complete the proof now.