Probability of throwing balls into bins
Solution 1:
It is helpful to imagine the balls are distinguishable. That makes no difference to the probability.
There are $m^n$ equally likely ways to distribute the $n$ balls among the $n$ bins. There are $(m-1)^n$ ways to distribute the balls among the last $m-1$ bins. Thus the probability the first bin is empty is $\frac{(m-1)^n}{m^n}$.
Or else when we throw a ball, the probability it misses Bin 1 is $\frac{m-1}{m}$. The probability of missing Bin 1 $n$ times in a row is $\left(\frac{m-1}{m}\right)^n$.
The other problems are solved using similar reasoning.