PROBABILITY
posted by Anonymous .
A group of 90 students is to be split at random into 3 classes of equal size. All partitions are equally likely. Joe and Jane are members of the 90student group. Find the probability that Joe and Jane end up in the same class.
My idea:
the total combination: 90C30
since 2 have to be together then I only have 88 students to arrange in groups fo 303028, and It'll be:
88!/(30! 30! 28!)
But then, should I multiply by 3 or how should I add the two that must remain together.
P = [(??) * 88!/(30! 30! 28!)] / 90C30
Or maybe I'm doing this wrong.

PROBABILITY 
Count Iblis
Method 1:
Assign a different number to each student ranging from 1 to 90, numbers 1 to 30 go in group 1, 31 to 60 go to group 2 61 to 90 go to group 3.
All possible partitions are obtained with equal probablity by a random assignment if these numbers, it doesn't matter with which students we start, so we are free to start by assinging a random number to Joe and then we assign a random number to Jane. After Joe has been assigned a random mumber there are 89 random numbers available for Jane and 29 of these will land her in the same group as Joe. Therefore the probability is 29/89.
Method 2:
We can consider the total number of ways of partitioning N objects into groups of n1, n2,...nr groups. There are
N!/(n1! n2!...nr!) ways to do that. But when two groups become of equal size we need to consider if we consider a permutation that has the effect of interchanging these groups as a different partition or the same partition. If we don't count that as diferent, then we need to divide by additional symmetry factors.
If we choose to do this, then the number of ways to make a partitioning where Joe and Jane end up in the same group is:
88!/(30!^2 2! 28! )
The total number of partitions is:
90!/(30!^3 3!)
Dividing the two numbers gives:
30!^3 3! 88!/(30!^2 2! 28! 90!) =
29/89
Method 3
The same as above, but now without symmetry factors for permuting groups. This amounts to labeling the 3 groups and it then matters in which group Joe and Jane end up in as these are counted as distinct possibilities.
The number of ways to make a partitioning where Joe and Jane end up in the same group is now:
3 88!/(30!^2 28!)
The total number of partitionings is now 90!/30!^3. Dividing the two numbers gives:
3 30!^3 88!/(30!^2 28! 90!) =
29/89
Method 4:
Similar as method 1 we just count all ways of placing the students in the 3 groups without taking into account that permuting the students in one group leads to the same partitioning.
There are then 3 ways to choose in which group Joe and Jane will go. And given that choice there are 30*29 ways to assing them a position in that chosen group. The total number of ways to divide all the students with Joe and Jane in the same group is thus:
3*30*29*88!
The total number of ways of partitioning the students is 90!, the ratio is:
3*30*29*88!/90! =
29/89
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