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I know that $$P\left(\bigcup_{i=1}^{n} A_i \right)$$

is the sum of of the probabilities of all the sample points that are contained in at least one of the $A_{i}$'s. This is the probability of sample points belonging to exactly 1 event, exactly 2 events, ...,exactly $n$ events. WLOG this can be written as $$P\left(\bigcup_{i=1}^{n} A_i \right) = P(A_1) + P(A_1 \cap A_2) + \cdots + P(A_1 \cap A_2 \cap \cdots \cap A_n)$$

But the $A_i$'s are arbitrary and we have to account for that. So there are $n$ possibilities for the first probability, $\binom{n}{2}$ possibilities for the second probability, ..., and $1$ possibility for the final probability. So we add and subtract these to prevent overcounting?

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    Do you mean $P(A_1) + P(-A_1 \cap A_2) + \cdots + P(-A_1 \cap -A_2 \cap \cdots \cap A_n)$, or are you talking about the Principe of Inclusion and Exclusion (see Wikipedia or some such)?2011-12-29
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    Your assertion about $P(\cup A_i)$ is incorrect. What _would_ be correct is $$P(\cup A_i) = P(A_1) + P(A_1^c\cap A_2) + P(A_1^c\cap A_2^c\cap A_3) + \cdots + P(A_1^c\cap A_2^c \cap \cdots \cap A_{n-1}^c \cap A_n)$$ but then the explanation is all wrong.2011-12-29

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