60 students are going to be split into two rooms, whats the probability that students A,B,C,D,E (which are super friends) end up toghether? As the title says, 
60 students are getting split into two rooms of $30$. 
What is the probability that Ann, Belle, Clarice, Diego and Evelyn end up in the same room? 
I can't understand if the number $60$ influences the probability that they end up in the same room. 
because if not, then it would be $\left(\frac{1}{2}\right)^5$. 
But I do believe we have to take into account the other $55$ students and that the divisions have a limit of $30$ per each.
 A: Yes, the number 60 influences this ... as an extreme example, what if there were only 6 students to be divided into two rooms of 3? So you are right: given that there is a maximum number that goes in the room has an effect.
Also, it doesn't matter what room Ann is in ... as long as the other four go in that same room ... so the probability will be closer to $(\frac{1}{2})^4$ than to $(\frac{1}{2})^5$
But it is not going to be exactly $(\frac{1}{2})^4$ either:
Given that Ann is in one of the rooms, the chance of Belle being in the same room is $\frac{29}{59}$. For Clarice it then is $\frac{28}{58}$, etc.
So, the probability of all 5 ending up in the same room is:
$$\frac{29}{59} * \frac{28}{58} *\frac{27}{57} *\frac{26}{56}$$
And now you see why the number 60 matters ... change this number slightly, and the probability will change slightly as well.  The higher the number, the more closely the probability will get to $\frac{1}{2}$, but for smaller numbers, the probability will get smaller and smaller. Indeed, once we get to two rooms of 4 or smaller, the probability of the 'fab five' ending up in the same room has become $0$.
A: Anne is going to be in one from the two rooms.   Who cares which.
What is the probability for Ann's roommates being a selection of all her $4$ friends and another $25$ from the $55$ remaining students, when selecting $29$ from the $59$ other students without bias?
