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!

  • 1
    $\begingroup$ You don't have to consider other tables for this question. Think: is there any difference if there are no extra tables, one extra table, or two extra tables? No, all the other students will not be sitting at the first table. $\endgroup$ – Quang Hoang May 21 at 11:54
  • $\begingroup$ Since you have determined table A, so your respond is correct. If you want to change sitting problem, you must add more details, like : probability of sitting all student from a grade on one table, or probability of sitting all student from at least one grade on one table ,... $\endgroup$ – BarzanHayati May 21 at 11:57

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.