A sensible way to model this problem is using the Poisson distribution (which gives a number of events in a certain time frame).
So if the mean number of hurricanes is 0.54, this will be lambda in our poisson distribution formula. So we can find the probability of seeing exactly h hurricanes using:
P( h ; lambda ) = (e-lambda) (lambdah) / h!
We can work all this out by hand, or we can just use a built-in function (e.g. in google sheets use =poisson(h,lambda,0); the final 0 just means you're using the pdf not the cdf, i.e. finding the probability of a single number, not a range).
a) P( h=1 ; lambda=0.54 ) = 0.3147
b) This is either the cdf of the poisson at 1 (in sheets use =poisson(1,0.54,1), or you can find the probability of getting zero hurricanes (=poisson(0,0.54,0)) and add it to the probability of getting 1, above. Either way:
P( h<=1 ; lambda=0.54 ) = 0.8974
c) This is just the inverse of the previous answer, right? So subtract 0.8974 from 1:
P( h>1 ; lambda=0.54 ) = 1 - P( h<=1 ; lambda=0.54 ) = 1 - 0.8974 = 0.1026
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