A group of 60 second graders is to be randomly assigned to two classes of 30 each. Five of the second graders, Marcelle, Sarah, Michelle, Katy, and Camerin, are close friends.
(a) What is the probability that they will all be in the same class?
Note: This question was answered before: Probability that 4 friends will be in the same class
But I have some issues:
1) I don't see how that is the answer
2) What is wrong with the approach I was attempting?
My attempt at a)
So first I found the sample space which I felt was $$\Omega = \binom{60}{30}$$
Then with that being the case I asked the question of "given that the 5 friends need to be together, in how many ways can I get the other 25 students for their group?" I wanted to get some arrangement of the form: $$(55)(54)….(32)(31)$$ But If I do that then I have assumed that the 5 kids are in a specific group, which is not the case. But multiplying by $2$ will also no work because the other group is to have 30 kids in it…….
How could I (if possible) reconcile my answer with the other one?
Best Answer
Suppose the chairs in the two classes are numbered $1-30$ and $31-60$. When we assign the students to classes, the five friends end up with random numbered chairs. The question is then whether the random set of five chairs we come up with are either all $\le 30$ or all $>30$.
Now there are $\binom{60}{5}$ sets of five possible chairs the friends could end up with, and $\binom{30}{5}+\binom{30}{5}$ of those keep the friends together ...