Because all 10 of the excited states in this exercise have the same probability, we just calculate that probability once and multiply by 10.

Z = e^{0} + 10e^{-ε/(kT)} = 1 + 10(0.1445) = 2.445

Now you can plug that in to find that the probability of finding the system in its ground state is:

So the probability is 41%. Just to check, we can calculate the combined probability of being in any of the excited states: