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"Six brothers sign up to take a class. The class has 90 registered students including the six. There are three sections, A, B, and C, of 30 students each by random assignment. What is the probability that each class section has exactly 2 brothers?"

What I've tried: I know there are 15 different ways to pair the brothers, from 6 factorial divided by (3 factorial times (2 factorial) cubed)). But I don't know how to set up the probability of any given pairing scheme seeing pairs ending up together, for all three classes.

I figured we could let the first brother go anywhere, then his partner had a 29/89 chance of ending up in his section, then the next brother had to go to either remaining section (60/88), his partner had to follow (29/87), then the last two take up the last section (30/86 * 29/85).

Am I overcounting or undercounting due to ordering? What is the above calculation finding the probability of, if it is not the probability of one pairing scheme ending up in separate classes?

Thanks.

Dragonborn Torinn
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2 Answers2

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If you intended multiplying the fractions by $15$ (this wasn't very clear to me)
i.e. $15\times\frac{29}{89}\frac{60}{88}\frac{29}{87}\frac{30}{86}\frac{29}{85},$ you are dead right !

I have just added an alternative method which may be of interest.

An alternative way

If you prefer to use combinations, while retaining the slot concept,
here too, you can make the simplification of considering only the $6$ brothers

$Pr = \dfrac{\dbinom{30}{2}^3}{\dbinom{90}{6}}$

true blue anil
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  • How is this different from the way I described? Your first way also has the 15 different pairing schemes, just split into a 5 x 3. I get a probability of (189225 / 1431298) using my method or either of these two methods. – Dragonborn Torinn Sep 29 '16 at 13:00
  • I think I may have found a loophole in second method. Doesn't having (90 choose 6) in the denominator imply that all spots are equally likely to be picked? Since each class section is capped at 30 people, the probabilities are not independent. This means that, for example, 6 slots in section A being picked is less likely that 3 slots in B and 3 slots in C being picked. – Dragonborn Torinn Sep 29 '16 at 13:24
  • If you meant to multiply by $15$, (this was not clear to me) your method and answer are correct. 2) There is no loophole in the second method. We aren't worried about the probabilities of each allocation of brothers, we are interested in the ratio of favorable allocations to total possible allocations of the brothers, which is what the numerator and denominator respectively count, and you will note that the answer it gives is identical, viz. 189225 / 1431298. I will add a note re your first method in my answer.
    • – true blue anil Sep 29 '16 at 13:58
  • I have modified the answer to give credit for your correct answer. You may like to have a look at http://math.stackexchange.com/questions/1345413/is-there-an-alternative-intuition-for-solving-the-probability-of-having-one-ace, and see how favorably what I call the "slot method" compares to more complicated methods. – true blue anil Sep 29 '16 at 14:38
  • Thanks a ton for your help! It's finally clicked in my head. – Dragonborn Torinn Sep 29 '16 at 17:47
  • You're welcome ! – true blue anil Sep 30 '16 at 03:29