Question 1174138
this is likely a Poisson distribution with parameter proportion to time, discrete variable, and theoretically infinite numbers.
p(3)=e^(-2)*2^3/3!=0.1804
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this is 1- probability (0,1,2)
prob of 0 is e^(-2)=0.1353
p(1) is e^(-2)*2^1/1=0.2707
p(2)=e^(-2)*2^2/2!=0.2707
That probability sum is 0.6767
the answer of p(>2) is 0.3233
also on calculator 1-poissoncdf(2,2) for the left side of the distribution.