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Is this formula right?

$$\renewcommand{\Pr}{\mathbb P}\Pr[\cap_{i=1}^n A_i] = \sum_{i=1}^d (-1)^{i-1} \sum_{|K|=k} \Pr[\cup_{k\in K} A_k],$$ with $K$ a subset (without repetition) of $\{1, ..., n\}$.

I have found a similar formula that expresses $\Pr[\cup_{i=1}^n A_i]$ in terms of $\Pr[\cap_{k\in K} A_k]$, and I'd like to be sure I have derived the other formula in the good way.

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Why don't you write out the formula explicitly for the cases $n=2,3,4$ and see if it is correct? Then use induction to prove it in general. – Dilip Sarwate Apr 5 '12 at 13:36
On the right hand side there seems to be some confusion about the roles of $i$ and $d$. Note that $k$ is not even defined, nor is $d$ specified. The formula, as it stands, therefore is neither right nor wrong: it makes no sense. – whuber♦ Apr 5 '12 at 19:27
Hint: The usual inclusion/exclusion formula, the simplest case of which is $$P(A\cup B) = P(A) + P(B) - P(A\cap B),$$ can be re-written as $$P(A\cap B) = P(A) + P(B) - P(A\cup B)$$ which looks a lot like your result. Is your general result also just a re-arrangement of the standard inclusion/exclusion formula with slightly different notation? – Dilip Sarwate Apr 5 '12 at 19:32

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