How to combine independent probabilities
Emily Wong 
Given two independent probabilities $P_1 = {1 \over p_1}$ and $P_2 = {1 \over p_2}$, it seems that the probability of either $P_1$ or $P_2$ happening would be:
$$P_1 + P_2 - P_1 P_2 = {1 \over p_1} + {1 \over p_2} - {1 \over p_1 p_2} = {p_1 + p_2 - 1 \over p_1 p_2}$$
What would be the case with three independent probabilities, $P_1$, $P_2$ and $P_3$?
More generally, how do I compute the probability of either one of a given set of probabilities happening?
It is safe to assume that all my probabilities are given in the form $P_k = {1 \over p_k}$, where $p_k \in \mathbb{N}$.
$\endgroup$2 Answers
$\begingroup$What you're looking for is called the Inclusion–exclusion principle.
$\endgroup$ $\begingroup$The easiest way to actually calculate this, especially for a large number of possibilities, is to consider the probability on the double negative that you do not have none of the events happening
So for three you have $$1-(1-P_1)(1-P_2)(1-P_3) = 1 - \left(1-\frac1{p_1}\right)\left(1-\frac1{p_2}\right)\left(1-\frac1{p_3}\right)$$ with the obvious extension to more terms
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