# What is the probability of the occurrence of three independent events?

I'm trying to solve this probability problem with my son:

In a classroom with 24 students, three teachers randomly ask some students each day to check their notebooks. The first teacher asks 4 students for their notebooks. The second teacher asks 6 students. And the third teacher asks 8 students. Find the probability that, for a given student on a given day, none ask for his/her notebook. In other words, no teacher will ask for the notebook.

I think the solution is:

$$\frac{4}{24}.\frac{6}{24}.\frac{8}{24}=\frac{1}{72}$$

But I'm not sure.

The probability that teacher 1 doesn't ask for their notebook is $$1 - \frac{4}{24} = \frac{20}{24}$$. Similarly for teacher 2 and 3, it is $$\frac{18}{24}$$ and $$\frac{16}{24}$$. So the answer should be:
$$\frac{20\cdot18\cdot16}{24\cdot24\cdot24}$$
The teacher that checks only $$4$$ notebooks can do it in $${24\choose 4}$$ ways. If the the teacher doesn't check the notebook of the given student then there is $${23 \choose 4}$$ ways to check the others. so for the first teacher there is a $$\frac{23 \choose 4}{24\choose 4}$$ chance that the teacher doesn't check the given student notebook. We can apply the same logic for the other teachers and get the final answer of $$\frac{23 \choose 4}{24\choose 4} \times \frac{23 \choose 6}{24\choose 6} \times \frac{23 \choose 8}{24\choose 8} = \frac{5}{12}.$$
• imagine you have $n$ distinct balls and you want to choose $r$ of them. the number of ways you can choose these $r$ balls is $n\choose r$ and is equal to $\frac{n!}{r!(n - r)!}$. here $a! = a \times (a - 1) \times (a - 2) \times \dots \times 2 \times 1$. for example $4! = 4 \times 3 \times 2 \times 1 = 24$ and ${7\choose 3} = \frac{7!}{3!4!} = \frac{5040}{6 \times 24} = 35$. Jun 5, 2023 at 21:31