N people sit down at random a classroom containing n+p seats

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SUMMARY

The discussion focuses on calculating the probability that all red seats will be occupied when n people randomly sit in a classroom with n+p seats, including m red seats (where m ≤ n). The correct formula for this probability is derived as P = \frac{C^{n+p}_{m} C^{n+p-m}_{n-m}}{C^{n+p}_{n}}. The participants clarify that after allocating m people to the red seats, only n+p-m seats remain for the remaining n-m people, which is crucial for accurate probability calculation.

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n people sit down at random a classroom containing n+p seats. There are m red seats (m<=n) in the classroom, what is the probability that all red seats will be occupied?

I know the bottom should be n+p choose n but I'm not sure what the numerator should be, any ideas would be great.

I was thinking n+p choose m since that will give the different ways that the red seats could be chosen, times n+p choose n-m which gives the choices that the non-red seats could be chosen.

Or: \frac{ C^{n+p} _{m} C^{n+p} _{n-m} } { C^{n+p} _{n} }

Does this logic make sense?
 
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indigojoker said:
I was thinking n+p choose m since that will give the different ways that the red seats could be chosen, times n+p choose n-m which gives the choices that the non-red seats could be chosen.
Close, but after having allocated m people amongst the n+p seats there are now only n+p-m seats to allocate the remaining n-m people.
 


then:
<br /> \frac{ C^{n+p} _{m} C^{n+p-m} _{n-m} } { C^{n+p} _{n} }?
 


indigojoker said:
then:
<br /> \frac{ C^{n+p} _{m} C^{n+p-m} _{n-m} } { C^{n+p} _{n} }?
EDIT: my error; see the other post.
 
Last edited:

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