Q. 3.12 - A First Course in Probability

Question

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

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

Step-by-Step Solution

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.

Step1: 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)

step2: 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,

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