Question 321126
This appears to be a Poisson probability.

It is given by {{{P(X)=(L)^X*e^(-L)/X!}}}. Where L represents lambda and the mean. In this case, 1.8 or 9/5.

Since it says more than 5, we sum up the probabilities from 0 to 5 and subtract from 1.

{{{1-(sum((9/5)^X*e^(-9/5)/X!,X=0,5))=1-(374179*e^(-9/5)/62500)=.01038}}}