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The number of dropped connections per call follows a Poisson distribution. From four calls, the number of dropped connections is $2,\,0,\,3,\,1$.
Obtain the maximum likelihood estimate that the next two calls will be completed without any accidental drops.

I know the maximum likelihood estimate of $\lambda$ is $1.5$. I think I am supposed to use this formula: $$f(x|\lambda) = \frac{\lambda^x e^{-\lambda}}{x!}$$

However, I do not know what $x$ should be. The answer should be $0.0498$.

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You need some assumption about the relative duration of the calls, such as all of them being the same length. – T.. Nov 18 '10 at 2:53
up vote 2 down vote accepted

x is the number of dropped connections, right? So $x=0$ for no dropped connections.

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Then that would be e^-1.5 = 0.223. However, the answer should be 0.0498 – Raptrex Nov 18 '10 at 1:40
Squared (since there are two calls). – Yuval Filmus Nov 18 '10 at 1:41
Your right, thank you – Raptrex Nov 18 '10 at 1:49

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