At school, each person gets put in a class at the start of the year, so I was just thinking how likely you were to get with a certain number of your friends in every period. I formulated this question:

You are at school with 6 periods and each period has 5 teachers teaching that one period in separate classes and each class has 30 students. You have 11 friends. What is probability that you have at least 2 friends in your class per period?

Now my original approach was to look at each friend as having a choice of which class to choose from. Since there are 5 teachers to chose from per period, you could think of it that way and then sort out the rest of the students' spots. I'm not really sure where to go from here.


One possible way to approach this problem is to first note the symmetry of the problem, in that the probabilities of this happening is the same for each period. It is then apparent that we just have to find the probability of getting at least 2 friends in your class for one period, then raise that probability to the power of 6 to account for each period.

The next thing that comes in mind is complementary counting. Notice the term "at least $2$". If we consider using complementary counting, that will become "$0$ or $1$", which(in my opinion) is way easier to count.

We are then to split it into two cases:

Case 1: You share a class with exactly $0$ friends

Consider it from "your" perspective. You have $5$ teachers to choose from. Then note that the placement of your friends doesn't matter at this point, as long as none of them are placed in your class. We then have that there are $\binom{150-12}{30-1}=\binom{138}{29}$ ways to choose the rest of the students in "your" class, noting that none of your friends can be placed in your class. It then follows that the number of ways to choose the rest of the students in the class is $\prod_{n=0}^3\binom{120-30n}{30}$. This comes from the fact that after every successive class, there are $30$ less students to choose from.

So the total for case 1 is $5\cdot\binom{138}{29}\cdot\prod_{n=0}^3\binom{120-30n}{30}$

Case 2: You share a class with exactly $1$ friend

Like the last case, you have $5$ teachers to choose from. You then have $11$ friends that you can choose from that can be in your class. You then have $\binom{150-12}{30-2}=\binom{138}{28}$ ways to choose the rest of "your" class. Like the previous case, there are $\prod_{n=0}^3\binom{120-30n}{30}$ ways to choose the rest of the $4$ classes.

So the total for case 2 is $5\cdot11\cdot\binom{138}{28}\cdot\prod_{n=0}^3\binom{120-30n}{30}$

We then have that the total amount of ways is the sum of those two cases. We now have to find the total ways that the 5 classes can be chosen. This is easy to calculate: $\prod_{n=0}^4\binom{150-30n}{30}$

So we have that the probability after considering complementary counting and the other 5 periods is:$$\left(1-\frac{5\cdot\binom{138}{29}\cdot\prod_{n=0}^3\binom{120-30n}{30}+5\cdot11\cdot\binom{138}{28}\cdot\prod_{n=0}^3\binom{120-30n}{30}}{\prod_{n=0}^4\binom{150-30n}{30}}\right)^6\approx\boxed{9.915\%}$$

If there are any errors, please let me know! I'm kind of a noob at math so I most likely made some error somewhere.


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