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I have a group of 1444 patients that have been included in a group of different trials. The trials found that none (zero) of the patients had a specific complication (independent). The known incidence of the complication in this group is 0.5%-0.6%. (suggesting there should have been about 7-8 events)

I need to know what the probability of finding zero complications given the known event rate of 0.5% in a population size of 1444 (no repeat measures, all independent events).

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  • $\begingroup$ The Poisson approximation with parameter $\lambda = np = 1444 \cdot .005 = 7.22$ gives the probability of zero complications as $\approx e^{-7.22} = 7.31 \cdot 10^{-4}$. $\endgroup$ – Alex Wertheim May 19 '14 at 2:17
  • $\begingroup$ Please do not use the tag (poisson-geometry) for questions related to Poisson processes or Poisson random variables. These are unrelated. $\endgroup$ – Did Mar 7 '17 at 7:44
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The number $X$ of complications in a random sample of $1444$ has Binomial distribution, $p=0.005$, $n=1444$. Thus the probability of $0$ complications is $\binom{1444}{0}(0.005)^0 (0.995)^{1444}$.

But we can do the calculation more simply. The probability that an individual has no complication is $0.995$. so the probability of no complications $1444$ times in a row is $(0.995)^{1444}$.

For problems of this type, namely $p$ small, $n$ large, and $np$ moderate, the Binomial distribution is well-approximated by a Poisson random variable with parameter $\lambda=np$. Thus our probability is well-approximated by $e^{-\lambda}$, where $\lambda=(0.005)(1444)$.

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