Assume that there are $n$ participants in an exam. Each participant with a probability of $\frac{1}{4}$ does not need any paper because he/she has brought some paper with them, with a probability of $\frac{1}{2}$ someone may need $1$ piece of paper and with a probability of $\frac{1}{4}$ someone may need $2$ pieces of paper. At least how many papers should the teacher buy to make sure that with the probability of $99\%$, we have enough pieces of paper?
I thought maybe we can solve it with the central limit theorem but I got confused.
The central limit theorem will work here for an approximation. Let $X_i$ be the number of pieces of paper required by the $i$th student, $i=1,\dots,n$. Then these are IID random variables, and you can compute their mean $\mu$ and variance $\sigma^2$. By the central limit theorem, $$\sqrt{n}\left(\frac{\frac{1}{n}\sum_{i=1}^n X_i-\mu}{\sigma} \right)\overset{d}{\to}N(0,1) \quad \text{ as } n \to \infty.$$ Now using a standard $z$-score table you can find a value $z$ such that if $Z \sim N(0,1)$, then $P(Z \leq z)\geq 0.99$. For large $n$, we can make the approximation that $\sqrt{n}\left(\frac{\frac{1}{n}\sum_{i=1}^n X_i-\mu}{\sigma} \right)$ has the distribution $N(0,1)$, so then we have: $$P\left(\sqrt{n}\left(\frac{\frac{1}{n}\sum_{i=1}^n X_i-\mu}{\sigma} \right)\leq z\right)\geq 0.99.$$ Now rearrange the inequality inside the probability until you get something that looks like $P(\sum_{i=1}^n X_i \leq \text{blah}) \geq 0.99$, and $\text{blah}$ will be your answer.