$m$ distinct boxes and $n$ distinct balls probability problem

Question

I have looked at similar questions, however I could not find an answer to my specific problem. The problem is the following: If I have $m$ distinct boxes and $n$ distinct balls, what is the probability of finding all the balls in the same box.

I thought that I can do this via $\frac{number of successes}{number of arrangements}$, where the number of arrangements should be $m^n$?

Answer 1

As the outcomes are equally likely (balls are distinct), the answer can be obtained as follows by counting:

$$ \frac{ \binom{m}{1} \times 1^n}{m^n}=\frac{1}{m^{n-1}}$$

Note that if the balls are not distinct, the counting method cannot be used. Instead, the following method can be used for both cases:

$$ \binom{m}{1} \left ( \frac{1}{m} \right) ^n=\frac{1}{m^{n-1}}.$$

Answer 2

We can calculate the probability that all the balls go in a marked box the probability of that happening is (1/M)^N the marked box can be chosen in M ways so the answer is M * (1/M)^N .