There are $10$ students and $16$ classes. Those $10$ students will attend all of those $16$ classes.Teachers in each classes can select any student to ask question. For example, the teacher in class $1$ can select no students or can also select $3$ students to ask question. This process will repeat in each classes. Then, what is the probability that at least two students will be selected in each class?
My solution: $$\frac{\sum_{k=0}^{16}(-1)^k\binom{16}{k}\bigg[\binom{10}{0}+\binom{10}{1}\bigg]^k\bigg[\binom{10}{2}+\binom{10}{3}...+\binom{10}{10}\bigg]^{16-k}}{\bigg[\binom{10}{0}+\binom{10}{1}...+\binom{10}{10}\bigg]^{16}}$$
$$\frac{\sum_{k=0}^{16}(-1)^k\binom{16}{k}(11)^k(2^{10}-11)^{16-k}}{2^{160}}$$
Am I right ?
Assume that in each class, the probability of $0$ or $1$ student being selected is the same probability $p$. If I understand your problem statement correctly, then the probability of at least $2$ students being selected in every class is simply $(1-p)^{16}$.
Since you did not specify any information that allows us to determine $p$, there is no unique answer. However, if we assume that in each class, each student has $\frac{1}{2}$ chance of being selected, then $p=(\frac{1}{2})^{10}+10\cdot (\frac{1}{2})^{10}$. The final answer would be $(\frac{1013}{1024})^{16}$.
I don't see why you would apply the inclusion-exclusion principle. I assume it has to do with the unspecified mechanism of determining $p$.