Sorry for being pedantic, but no - these are probabilities not random variates. If the probability of either alone is p then the probability of one or the othe

Sorry for being pedantic, but no - these are probabilities not random variates. If the probability of either alone is p then the probability of one or the other or both is: 1 - (1-p)(1-p) = 1 - (1-2p+p^2) = 2p - p^2 ~ 2p (where the last step is because p^2 << p because p << 1; the basic idea is that it's "certainty minus the probability that neither occur")