Question

Show that $0\le a_i\le 1,i=1,2,\dots ,$then

$\sum _{i=1}^{\infty }\left[a_i\prod _{j=1}^{i-1}\left(1-a_j\right)\right]+\prod _{i=1}^{\infty }\left(1-a_i\right)=1$

Short answer

Observe infinitely many experiments. probability of success in the $i$-th experiment is $a_i$, and probability of failure is $\left(1-a_i\right)$with probability $1$, mutually exclusive events:

that the first success is on the $i$-th experiment - for $i=1,2,3,...,$or extreme case that no success occurred.

Step-by-step solution

Find Mutually exclusive.

From the intervals$[0,1]$, there are an unlimited number of possibilities.

$a_1,a_2,a_3,\dots$

prove:

$\sum _{k=1}^{\infty }\left[a_k\prod _{i=1}^{k-1}\left(1-a_i\right)\right]+\prod _{i=1}^{\infty }\left(1-a_i\right)=1$

A series of separate experiments is repeated an endless number of times.

Each time the chances of success is denoted by $a_i\in [0,1]$, the probability of success is denoted by $i=1,2,3,$...

Because all of these events are independently exclusive, the first success occurs in the $k$-th experiment with probability $1$.

For $k=1,2,3,...$or all experiments, the first success occurs in the $k$-th experiment with probability $1$

$1=\sum _{k=1}^{\infty }P$(the first success in $k$-the experiment) $+$$P$(no successes)

Derive the probabilities

The first success in the $k$-th experiment occurs when the first $k-1$experiments fail with probability of $\left(1-a_i\right),i\le k,$and the $k$-th experiment succeeds, yielding independence.

$P$(the first success in $k$-th experiment)$=a_k\prod _{i=1}^{k-1}\left(1-a_i\right)$

All tests are independent, and the likelihood that $i$will fail is $\left(1-a_i\right)$, hence the probability that almost all experiments will fail is $\left(1-a_i\right)$.

$P$(no successes) $=\prod _{i=1}^{\infty }\left(1-a_i\right)$

Substituting the values,

$\sum _{k=1}^{\infty }\left[a_k\prod _{i=1}^{k-1}\left(1-a_i\right)\right]+\prod _{i=1}^{\infty }\left(1-a_i\right)=1$