60 students sitting at 5 tables randomly

Question

There are $60$ students from 12 different grade years, $5$ students from each year.

Also, there are $5$ tables A,B,C,D,E. Tables A-D have $14$ seats and E has $4$ . Students sit randomly to the Tables.

What is the probability that all students from a specific grade year(5-students) sit at table A.

I was thinking it is a classical probability with maybe hyper-geometric form. something like: $\frac{\binom{5}{5}*\binom{55}{9}}{\binom{60}{14}}$ but I am not sure if i have to consider the other tables as well, since students sit randomly?

Any advice appreciated!

Answer 1

Let us label the grade years with $i=1,\dots12$ and let $E_i$ denote the event that the students of grade $i$ are all seated at table A.

Then to be found is: $$P\left(\bigcup_{i=1}^{12}E_i\right)$$

For this we can use the principle of inclusion/exclusion together with symmetry based on the fact that the events $E_i$ are evidently equiprobable.

This results in:$$P\left(\bigcup_{i=1}^{12}E_i\right)=\binom{12}1P(E_1)-\binom{12}2P(E_1\cap E_2)=\frac{\binom{12}1\binom{14}5}{\binom{60}5}-\frac{\binom{12}2\binom{14}{10}}{\binom{60}{10}}$$

Answer 2

Just counting friend:

Total number of different seat combination: X = 60!

Total number of combination, which only one specific grade sits at A: $Y= \binom{14}5 * 55! - 11 * \binom{9}5*50!$

Total number of combination, which only two specific grades sit at A: $Z=\binom{12}2 * \binom{14}{10}*50!$

Probability = (Y+Z)/X