A question regarding a special case of the Inclusion–exclusion principle

Question

Allegedly there's a special case in probability theory in which the Inclusion–exclusion principle can be expressed as $P(\displaystyle\bigcup_{i=1}^{n}A_i)=\displaystyle\sum_{i=1}^{n} (-1)^{i-1}\binom{n}{i}a_i$ where for example $\displaystyle\binom{n}{3}a_3 =\sum_{\substack{1≤i<j<k≤n\\}}A_i\cap A_j\cap A_k$. Could someone give me an example where this can be utilized because I'm not quite sure what it means? In addition to that, can this special case be proven with induction without relying on the general case?

I apologize if i'm not giving you much to work with, I've never used $\LaTeX$ before and I don't feel like spending 10 hours writing a question so I kept it short.

Answer 1

$a_3$ is just the sum of all the probabilities for intersections of any three distinct events selected from the sequence, divided by the count of the $\tbinom n 3$ ways to select them.   That is the arithmetic mean.

$$a_3 =\dfrac{\sum\limits_{1\leqslant i<j<k\leqslant n}\mathsf P(A_i\cap A_j\cap A_k)}{\dbinom{n}{3}}$$

And so on for all $a_i$ where $i\in\{1,.., n\}$.

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What you have is not a special case.   It is the principle of inclusion and exclusion.   It is just that you are not normally given the arithmetic means as such.

$$\begin{align}\mathsf P\left(\bigcup_{i=1}^{n}A_i\right) & =\sum_{i=1}^{n} (-1)^{i-1}\binom{n}{i}a_i \\ & = \sum_{i=1}^n (-1)^{i-1} \sum_{1\leqslant k_1<\ldots< k_i\leqslant n} \mathsf P\left(\bigcap_{j\in\{k_1,\ldots,k_i\}}A_j\right) \end{align}$$

What would be a special case would be if the probabilities for any selection of any $i$ distinct events in the sequence were to all be equal to $a_i$.   That is, the intersections of three events were all the same size regardless of which three were selected; et cetera.