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So I am a bit confused.

Say $1$ person in $1000$ forget to clean their hands after going to washroom. If $10,000$ go to the washroom, what is the probability of $6$ people forgetting to wash their hands?

So, from my knowledge, I would assume that $\lambda$ is the following: $$\begin{align} 1/1000 &= x/10000 \\ x&=10 \\ \implies& 10/1000 \end{align}$$ Hence, $\lambda$ is $10/10000$, however according to my friend it is $10$ and not $10/10000$. Can someone explain to me which one is correct? I only want to know which $\lambda$ is correct, since I already have a good idea how to solve it once I get this darned lambda thing down.

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    $\begingroup$ Recall that if $X\sim\mathrm{Pois}(\lambda)$, then $\mathbb E[X]=\lambda$. Here the expected number of people forgetting to wash their hands is $10$, so $\lambda=10$. $\endgroup$ – Math1000 Jul 10 '15 at 1:55
  • $\begingroup$ SO you did np? where n is the number and p is the probability. $\endgroup$ – Belphegor Jul 10 '15 at 2:02
  • $\begingroup$ That is for a binomial distribution, but it's pretty much the same concept. You're overthinking it. If we expect $1$ in $1000$ to forget, then we would expect $10$ in $10000$ to forget. $\endgroup$ – Math1000 Jul 10 '15 at 12:03
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Your friend is correct. $\lambda$ is the expected number of successes. So if there are $10,000$ trials each with probability $1/1000$, the expected number is $10$ and that is $\lambda$ The idea is that $\lambda$ will change with more trials- if you had $100,000$ people who visited the washroom you would expect $100$ to forget. That is the $\lambda$ Now the variance is lower.

Incidentally, $1/1000$ is vastly low, but that is not a mathematical problem.

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