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Imagine a school with 100 students from nine different countries (a to i): A 50 students B 20 students C 15 students D 10 students E-I, 1 student from each

For any randomly selected student Sitting in class, what is the probability that both her immediate left AND right neighbors are not from the same country as her?

I'm stuck on the 'no replacement ' concept when looking at the right neighbor. My gut feeling is to minus 1 from somewhere and recalculate and sum or aggregate the probabilities for each country. Obviously if the left neighbor is from a country with only one student, the probability that the right neighbor is also from that country should be nil. The same for our initial random student. So it's pick 3, no replacement, I think...

Edit: For simplicity's sake, they're all seated in a giant circle, completely at random and have both left and right neighbors.

To calculate the probability for the left student being from a different country from our first student, I took the weighted average of the probabilities by nationality, i.e. 50prob(a) + 20prob(b) [....] /100

then subtracted the result from 1 for the prob that they are from different countries:

1-0.323 = 0.677

But now having taken the initial student plus her left neighbour out of the mix, how do I go about calcuating the probablity of the right hand neighbour also being from a different country, there are now only 98 students.

Jimbo78
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  • So...just go by cases, according to the which country the selected student is from. Note that it's not clear how the students are arranged. Are they in a circle? If not, then some of the students will only have neighbors on one side or the other, right? Also, I assume that each student is seated uniformly at random (so any given student is equally likely to be in any given seat), but this ought to be stated. – lulu Jun 17 '22 at 13:48
  • Thanks, good points, I've edited the question. In reality, students sit in rows in small classrooms, but I'm looking for a fairly general measure of diversity to improve on 'number of countries represented', so the completely random circle should suffice. The selected student is also hypothetical and could come from any country. – Jimbo78 Jun 17 '22 at 14:06
  • Given that they are in a circle, the method sketched in my first comment should go through without difficulty. – lulu Jun 17 '22 at 14:08
  • Thank you, this is exactly what I was looking for. If you post as a separate answer I can check it as answered – Jimbo78 Jun 17 '22 at 15:04
  • Done.................. – user2661923 Jun 17 '22 at 16:47

1 Answers1

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Response transferred from the comments, per OP's (i.e. original poster's) request.

Per the comment of lulu, which I agree with, I advise against trying to find any elegant computations. Simply plow through the cases one at a time.

Indexing the nationalities by the elements in $\{1,2,\cdots,9\}$, let $f(k)$ denote the probability that the randomly selected student is from country $(k)$.

Then, let $g(k)$ denote the probability that neither neighbor is from country $(k)$, given that the one in the middle is from country $(k)$. Then, the desired computation is

$$\sum_{k=1}^9 \left[f(k) \times g(k)\right].$$

If you have $S$ people from country $(k)$ and $T$ people not from country $(k)$, and you assume that one of the people from country $(k)$ is in the middle, then there are only $(S-1)$ people from country $(k)$ remaining, who might end up as neighbors.

Then, the probability that neither neighbor is from country $(k)$ will be

$$\frac{\binom{T}{2}}{\binom{T + [S-1]}{2}}.$$

user2661923
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