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I am doing some work that involves combining the probabilities of multiple independent, non-disjoint events into an overall probability of any event occuring. We are using an equation like this:

$P(A_1 \cup A_2 \cup \dots \cup A_N) = 1 - \prod_{i=1}^N{(1 - P(A_i))}$

Is there a name for this equation? I can't seem to find where it comes from in probability theory.

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  • $\begingroup$ @Macro - They are not disjoint events. $\endgroup$ Mar 20, 2013 at 15:01
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    $\begingroup$ I'm not sure this has a name, but I think it follows from the fact that $(\bigcup_{i=1}^{N} A_i)^c = \bigcap_{i=1}^{N} A_{i}^{c}$, which is one of De Morgan's Laws. $\endgroup$
    – Macro
    Mar 20, 2013 at 15:05
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    $\begingroup$ It is called the Principle of Inclusion-Exclusion. It is an immediate consequence of the axioms of probability: a formal proof would use induction on $N$ to reduce this to the case $N=2$. $\endgroup$
    – whuber
    Mar 20, 2013 at 15:20
  • $\begingroup$ @Zen I am absolutely confident that this is what the relationship is commonly called--see the reference--and indeed its proof goes through upon assuming independence, which of course was implicit in the question. (No harm pointing out that independence must be assumed, which you already did in your nice answer, but why keep harping on that point?) $\endgroup$
    – whuber
    Mar 20, 2013 at 17:09
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    $\begingroup$ @whuber While expanding out the right side of the equation and simplifying does give the same result as the one obtained from the Principle of Inclusion-Exclusion and the application of independence of events, I think Macro's point that this is a consequence of DeMorgan's Law (and independence) is a much more useful way of thinking about it, especially when the amount of computation is taken into account. See, for example, this answer of mine on math.SE and the discussion thereafter. $\endgroup$ Mar 20, 2013 at 19:45

2 Answers 2

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If $A_1,\dots,A_n$ are independent events, then $$ \begin{align} P\left(\cup_{i=1}^n A_i\right)&=1-P\left(\left(\cup_{i=1}^n A_i\right)^c\right) \\ &= 1 - P\left(\cap_{i=1}^n A_i^c\right) \qquad &\textrm{(De Morgan)} \\ &= 1 - \prod_{i=1}^n P(A_i^c) \qquad &\textrm{(independence)} \\ &= 1 - \prod_{i=1}^n (1-P(A_i)) \, . \end{align} $$ P.S. Remember that $A$ and $B$ are independent iff $P(A\cap B)=P(A)P(B)$. But then $$P(A^c\cap B^c)=1-P(A\cup B)=1-(P(A)+P(B)-P(A\cap B))=$$ $$1-(P(A)+P(B)-P(A)P(B))$$ $$=(1-P(A))(1-P(B))$$ $$=P(A^c)P(B^c) \, ,$$ so $A^c$ and $B^c$ are also independent.

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  • $\begingroup$ The events are independent, I'll edit the question. $\endgroup$ Mar 20, 2013 at 17:10
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It is called a "Noisy OR". You can read about this and other models that assume Independence of Causal Influence in [1].

[1] Díez, F. J., Druzdzel, M. J., & Francisco, J. D. (2007). Canonical probabilistic models for knowledge engineering (Vol. 9, pp. 1–59).

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