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## Homework Statement

Suppose that on average, I will receive one phone call every night. Under reasonable assumptions (not given in this question prompt), show that the probability of receiving exactly one phone call, tonight, is e^-1.

## Homework Equations

Possibly the gamma distribution function or exponential distribution function, but I'm not sure, as this problem arises out of material far out of the introductory distribution functions.

That is,

f(y) = [tex]\frac{1}{\Gamma(\alpha)\beta^{\alpha}} * y^{\alpha-1}e^{\frac{-y}{\beta}[/tex]

is a gamma distribution for y>0. An exponential distribution, of course, is just a gamma with alpha = 1.

## The Attempt at a Solution

Well, I thought that if [tex]\beta = 1[/tex], then you simply compute f(1), where y is the number of calls in a day. This, however, seems

*too*easy, considering that this class is Mathematical Statistics 2, and we learned that sort of thing in M.S. 1.

Thanks!