Question 1170266
would assume a Poisson both because likely to be a large number of such machines and also because there is no way to calculate the probability without being given the sd, unless there is a binomial, Poisson distribution, or a proportion where the two are dependent. 

The mean and variance are both 2, so the sd is sqrt(V)=sqrt(2) or 1.414

P(0)=e^(-2)2^0/0!=0.1353
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at most 2 is 0,1,2
probability of 1 is e^(-2)*2^1/!=0.2707
for 2 it is e^-2)*2^2/2!=0.2707
the probability is 0.6767
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for 3 in 2 weeks it is equivalent of parameter 4 for that time period, since the distribution is proportional to time. e^(-4)^4^3/3!=0.1954