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Exercise
May 14'23
Answer
Key: C
[[math]]
p(k) =\frac{2}{k} p(k-1) = (0 + \frac{2}{k})p(k-1)
[[/math]]
Thus an (a, b, 0) distribution with a = 0, b = 2.
Thus Poisson with [math]\lambda = 2 [/math].
[[math]]
p(4) = \frac{e^{-2}2^4}{4!} = 0.09
[[/math]]
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