<PRE><B>Question 909196</B>
The first thing we will do is assume that our observed frequency of 18 breakdowns over 11 years means we have 18/11 breakdowns every year.

Completing the table 

Let X~Poisson({{{lambda = 18/11}}})

#  times      of            ef
0  1          1/11 = .0909   P[X=0] = .1947
1  5          5/11 = .4545   P[X=1] = .3186
2  3          3/11 = .2727   P[X=2] = .2607
3  1          1/11 = .0909   P[X=3] = .1422
4  1          1/11 = .0909   P[X=4] = .0582
5  0          0/11 = 0       P[X=5] = .0190
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