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$P_i$ is the probability of one event. As defined below, $a$ is the sum of all probabilities of events (that may or may not be independent), and $c$ is the overall probability that at least one event will happen.

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

$c=(1-\prod_{i=0}^n (1-P_i))$

$0\leq P_i \leq 1$

Take two events with probability of $0.5$ and $0.2$ for example:

$$ a= 0.5 +0.2=0.7~, \qquad c = 1 - 0.5 \cdot 0.8=0.6$$

Is it true $\dfrac{a}{c}$ will increase as $n$, the number of $P_i$, increases?

It seems true to me based on sample results, but I can't think of a formal proof.

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    $\begingroup$ If the events are dependent, then $c$ is not the probability that at least one of the events occurs. $\endgroup$ – grndl Oct 29 '18 at 16:35
  • $\begingroup$ In addition to what @aduh pointed out, note that $c$ could be equal to zero (if all events have probability zero) in which case your ratio $a/c$ becomes infinite - so it is simplest to avoid this case by assuming all events have positive probability. Also, when you say that $a/c$ increases as $n$ increases, I guess you mean that you keep all the existing $P_i$ the same. I implemented these suggestions in my reformulation of your question (and its answer) below. $\endgroup$ – pre-kidney Oct 31 '18 at 8:31
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Yes, the question you meant to ask is true, although (as pointed out already) it is not quite correct as stated. To keep things clear, I will start by restating your question in a more precise form.

Let $n> 1$ be an integer and let $a_1,\ldots,a_n$ be positive real numbers, all between $0$ and $1$. Prove that $$ \frac{\sum_{i=1}^{n-1}a_i}{1-\prod_{i=1}^{n-1}(1-a_i)}\leq \frac{\sum_{i=1}^{n}a_i}{1-\prod_{i=1}^{n}(1-a_i)}. $$

Proof. Let $S=\sum_{i=1}^{n-1}a_i$ and let $P=\prod_{i=1}^{n-1}(1-a_i)$. Since the $a_i$ are positive, both denominators are also positive so we can multiply through and substitute $S$ and $P$ to obtain the equivalent inequality $$ \frac{S}{1-P}\leq \frac{S+a_n}{1-(1-a_n)P}\iff S\bigl(1-(1-a_n)P\bigr)\leq (1-P)(S+a_n). $$ Expanding this out and collecting like terms leads to the equivalent inequality $$ a_nSP \leq a_n(1-P), $$ which (since $a_n$ is positive) is equivalent to $SP\leq 1-P$, which is the same as $P(1+S)\leq 1$.

But this is true, since $$ P(1+S)=\prod_{i=1}^{n-1}(1-a_i) \ \cdot \ \Bigl(1+\sum_{i=1}^{n-1}a_i\Bigr)\leq \prod_{i=1}^{n-1}(1-a_i)\ \cdot \ \prod_{i=1}^{n-1}(1+a_i)=\prod_{i=1}^{n-1}(1-a_i^2)\leq 1. $$ This establishes the desired inequality. $\square$

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  • $\begingroup$ Can you explain the penultimate inequality? $\endgroup$ – grndl Oct 31 '18 at 12:11
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    $\begingroup$ Yes, the reason why $\prod_{i=1}^{n-1}(1+a_i)\geq 1+\sum_{i=1}^{n-1}a_i$ is that if you expand out the product, it is equal to $1+\sum_{i=1}^{n-1}a_i$ plus higher order terms which are all positive. $\endgroup$ – pre-kidney Nov 1 '18 at 4:12

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