# probability question

Ok, probablilty enthusiasts , a couple of questions please:

1. Assuming there are 136 children in a school (all in the same grade level). Each year, the children are divided into 8 classes of 17 children each. The groupings are random except that each child must be in a class with at least one other child she was in class with the year before. What are the odds that a child is in a class with only one other child who was in her class in either of the prior 2 years?

2. Same school (136 children, 8 classes of 17 children each). If there are 116 boys and 20 girls, what are the chances that a girl is not in a class with any other girls?

Many thanks! Let me know if I’m missing any material facts.

In both questions, do you mean a particular specified child/girl, or at least one child/girl?

A specified child. Thanks

Question 2 is fairly simple. The quantity of interest is hypergeometrically distributed, which means that the probability is given by C(19, 0)C(116, 16)/C(135, 16), where C(n, k) is a binomial coefficient. Matlab tells me that this is roughly equal to .066.

The requirement that each student be put in a class with at least one other student they were in class with last year makes question 1 considerably more difficult.

Thanks for the reply. Yeah, the first question is certainly more complicated because of that requirement. What about if I eliminated the requirement and asked what the chances are that the child was in class with either 1 or 0 students they were in class with for either of the prior 2 years?

I don’t believe there is enough information to answer this question. Presumably random means something like a deck of cards with student names is shuffled and dealt into 8 “hands”. The problem is with the restriction. What happens when the hands are revealed and it is not met?

One answer is the entire deal is thrown out and a complete new deal is made. That would have a specific answer.

Another way would be to deal face up. Each time you put a child into a class which does not have a child from her last year class you search through the deck until you find the next child who was from her class and immediately put him/her into the same class. (Though there are cases here when you might run into problems as well and need to start over.) This would have a different answer.

I suspect from your wording that you are actually referring to a real-life situation involving a girl you know and what to know if she was treated unfairly and the school is misinforming you as to how classes were decided upon. If this is real life, I suspect the random process used is some kind of ad hoc process. Kids are assigned at random and the classes are examined. If somebody is “alone”, then a child from her class from last year who is in the class with a lot of other kids from her last year class is swapped into the class. This will have a third solution and not be answerable until we know the swapping algorithm.

If it’s real world, I also suspect things are not quite random. Considerations like, a teacher says two children did not get along at all last year so make sure they are not in the same class will arise.

This depends on how many kids total she shared a class with in the past two years. We know there were 16 kids last year and 16 two years ago, but presumably this is not 32 different kids. Suppose there were N such kids (N will be in the range 16 to 32. Then there are 135-N “strangers” (excluding herself of course),

There are C(135,16) possible different classes she could be in. Of these there are C(135-N,16) classes with only strangers. There are N possible single matches and for each of these there are C(135-N,15) sets of 15 strangers she could be matched with.

Depending on N I get the following answers for no match, exactly one match, 0 or 1 match

N none one 0 or 1
16 11.7% 28.7% 40.4%
17 10.1% 26.7% 36.8%
18 8.7% 24.6% 33.4%
19 7.5% 22.7% 30.2%
20 6.5% 20.8% 27.3%
21 5.6% 19.0% 24.6%
22 4.8% 17.3% 22.1%
23 4.1% 15.7% 19.8%
24 3.5% 14.1% 17.7%
25 3.0% 12.7% 15.8%
26 2.6% 11.4% 14.0%
27 2.2% 10.2% 12.5%
28 1.9% 9.2% 11.0%
29 1.6% 8.2% 9.7%
30 1.4% 7.2% 8.6%
31 1.2% 6.4% 7.6%
32 1.0% 5.7% 6.6%

so it’s not wildly improbable even in the worst case.

(Sorry I can’t make the table pretty)

Use [noparse]

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[/noparse] to get something like this:

``````

N	none	one	0 or 1
16	11.7%	28.7%	40.4%
17	10.1%	26.7%	36.8%
18	8.7%	24.6%	33.4%
19	7.5%	22.7%	30.2%
20	6.5%	20.8%	27.3%
21	5.6%	19.0%	24.6%
22	4.8%	17.3%	22.1%
23	4.1%	15.7%	19.8%
24	3.5%	14.1%	17.7%
25	3.0%	12.7%	15.8%
26	2.6%	11.4%	14.0%
27	2.2%	10.2%	12.5%
28	1.9%	9.2%	11.0%
29	1.6%	8.2%	9.7%
30	1.4%	7.2%	8.6%
31	1.2%	6.4%	7.6%
32	1.0%	5.7%	6.6%

``````

Thank you for your responses! Yes, it is a real life situation, and I was in fact trying to prove that it is very improbable for a student to be only matched 1 person that it shared a class with over the prior 2 years.