Answer:
[tex](\frac{m-1}{m})^n[/tex]
Step-by-step explanation:
Given a basket, the probability of a ball to end there is [tex]\frac{1}{m}[/tex] (because there are m baskets).
Then, the probability of a ball to end in other basket is [tex]\frac{m-1}{m}[/tex].
Finally, the probability of the basket to remain empty after n throws is
[tex](\frac{m-1}{m})^n[/tex]
This last is because, given n independets events with probability p, the probability for all of them to happen is [tex]p^n[/tex]. In this case [tex]p=\frac{m-1}{m}[/tex]