# Sum of probabilities of events vs. probability of at least one event

$$P_i$$ 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.

• If the events are dependent, then $c$ is not the probability that at least one of the events occurs. – grndl Oct 29 '18 at 16:35
• 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. – pre-kidney Oct 31 '18 at 8:31

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. $$\square$$
• 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. – pre-kidney Nov 1 '18 at 4:12