8. The Lonely Ball (from 8th grade, 2 points). In 100 boxes, exactly 100 balls are placed randomly. What is the probability that the last box will contain only one ball?
Answer: approximately $0.370$.

Solution. Consider the placement of balls into boxes as a series of $n$ independent Bernoulli trials. A success will be considered as a ball landing in the last box. The probability of exactly one success is

$$
C_{n}^{1}\left(\frac{n-1}{n}\right)^{n-1} \cdot \frac{1}{n}=\left(\frac{n-1}{n}\right)^{n-1}
$$

Substituting $n=100$, we get $0.99^{99} \approx 0.3697$.

Remark. Those familiar with the number $e$ and know that $e=\lim _{x \rightarrow 0}(1+x)^{x}$, can verify that as $n$ increases, the number $\left(\frac{n-1}{n}\right)^{n-1}$ approaches $\frac{1}{e} \approx 0.368$.