If I am thinking about this correctly, we can use a Poisson probability.
The mean number of plates broken on a given day is \(\displaystyle {\lambda}=\frac{8}{5}\).
What is the probability that there are no plates broken on any given day. Say, Saturday.
\(\displaystyle P(x)=\frac{{\lambda}^{n}e^{-\lambda}}{n!}\)
\(\displaystyle P(0)=\frac{(\frac{8}{5})^{0}e^{-\frac{8}{5}}}{0!}=e^{\frac{-8}{5}}\approx .2019\)
