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Consider $n$ independent trials, each of which results in one of the outcomes $1,\ldots,k$ with respective probabilities $p_1,\ldots,p_k$, $\sum_{i=1}^{k} p_i =1$. (I interpret this summation as just saying mathematically that definitely one of the outcomes has to occur on each trial). Show that if all the $p_i $are small, then the probability that no trial outcome occurs more than once is a approximately equal to $ \exp\left(−n(n−1)\sum_i \frac{p_i^2}{2}\right). $

So if all the $p_i$are small, in comparison to $n$, then I believe I can approximate this to a Poisson RV (that is where the $\exp$ comes in). (Correct?) So, first I can write the prob that no trial occurs more than once as$ p_1(1−p_1)^{n−1} p_2 (1−p_2)^{n−1}\cdots p_k(1−p_k)^{n−1} \cdot k!=k!\prod_{i=1}^{k} p_i(1−p_i)^{n−1}$ I am not really sure where to go from here. I am guessing at some stage, I will need to take the limit that as $n$ tends to infinity, something tends to $e $ .Thanks.

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    I am only doing an introductory course in Probability and so I haven't heard of 'asymptotics' before. I emailed my professor and he said: 'Consider the probability that a given pair of trials results in the same outcome. This one is easy. Now use the Poisson principle, i.e. use the approximation that all pairs of trials are independent.' So the probability that any two trials are the same is simply $p_i p_i$, right? I think I may have to assume that $n \leq k$. Where to go from here?2012-12-10

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