Probability of All Red Seats Occupied in a Classroom

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In summary: I don't understand how this relates to the original question.In summary, the probability that all red seats will be occupied is 1/3.
  • #1
indigojoker
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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?

It is asking for a probability. The denominator is easy, I think it
should be (n+p) choose m since we are looking for the number of ways
to choose the m specified seats from all seats.

I'm not sure what to write for the numerator, I was thinking n+p
choose m since that will give the different ways that the red seats
could be chosen, times (n+p)-m choose n-m which gives the choices that
the non-red seats could be chosen.

Any ideas would be very appreciated.
 
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  • #2
Sorry I made a mistake, never mind.
 
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  • #3
I'm not sure I completely follow what you are saying. What exactly do you mean by a relation between n+p and m?
 
  • #4
indigojoker said:
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?

It is asking for a probability. The denominator is easy, I think it
should be (n+p) choose m since we are looking for the number of ways
to choose the m specified seats from all seats.

I'm not sure what to write for the numerator, I was thinking n+p
choose m since that will give the different ways that the red seats
could be chosen, times (n+p)-m choose n-m which gives the choices that
the non-red seats could be chosen.

Any ideas would be very appreciated.
This is a repeat post: https://www.physicsforums.com/showthread.php?t=286955
 
  • #5
I don't think it's [tex]

\frac{ C^{n+p} _{m} C^{n+p-m} _{n-m} } { C^{n+p} _{n} } [/tex]

The reason why is, I was told that in order for a relation like this to hold, you must have [tex] \frac{ C^{A} _{B} C^{D} _{F} } { C^{E} _{G} } [/tex]

A+D=E
B+F=G

Also, I believe the denominator should be n+p choose m, since we are looking for the total ways the red seats can be filled.
 
  • #6
No the denominator is right you're looking for how many people could of chose any of the seats, in proportion to how many chose red. So the chances are over all possible choices.
 
  • #7
indigojoker said:
I don't think it's [tex]

\frac{ C^{n+p} _{m} C^{n+p-m} _{n-m} } { C^{n+p} _{n} } [/tex]
You can think about it like this (I thought you had):

P(n people sit down in the n+p seats such that all m red seats are taken) = (number of ways to sit n people down in the n+p seats such that all m red seats are taken) / (number of ways to sit n people down in the n+p seats).

The right-hand side is precisely the expression you have above.
 
  • #8
It can be helpful to try out your formula on some easy cases. Like suppose you had 2 people, 3 chairs, 1 red. Or 2 people, 3 chairs, 2 red? Etc. I don't think you'll find your proposed formula works very well. You shouldn't be thinking of how the red seats are chosen or who sits in them. For example, I would start with the denominator C(n+p,n). That's the number of ways to chose a subset of n chairs from n+p (indifferent to whether they are red or not).
 
  • #9
Ok, so the probability, I believe, to chose the red chairs should be:

[tex] \frac{ C^{n+p}_{m} }{C^{n+p}_{n} }[/tex]

But I believe after we know how likely it is to choose the red seats, do we then multiply it by the number of people? Since there are n people choosing seats, then each would have an equal prob. of choosing a red seat unless it is taken up.. this is becoming confusing.
 
  • #10
Ok, 2 people, 3 chairs, 1 red. That's n=2, p=1, m=1. C(3,1)/C(3,2)=1. That's not right. The odds pretty clearly ought to be 2/3. Think again about the numerator. In the numerator you ought to be assuming that the m red chairs are already occupied. Now you just have to seat the people who aren't in red chairs. Yes, it can be confusing.
 
  • #11
I thought I needed to find the probability of the red chairs being taken up?

If the red chairs are already taken up, then the numerator should be [tex] C^{n+p-m} _{n-m} }[/tex]

But I'm not sure what this is getting at? The possible ways that the non-red chairs can be filled?
 
  • #12
Yes, in the numerator assume the red chairs are already occupied. Just seat the non red chairs. In the denominator you seat regardless of chair color.
 
  • #13
You do but your students have the choice of any chair. so it's the probability of all red chairs being taken in relation to the number of students, and the number of other chairs that could be taken in relation to the number of students.

You just need to denote the number of red chairs taken so as they are all filled, the number of remaining chairs taken if any, all over the total number of chairs.
 
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  • #14
The possible ways to seat those who are not in red chairs should be:

[tex]
\frac{ C^{n+p-m} _{n-m} }{C^{n+p}_{n} }[/tex]
 
  • #15
indigojoker said:
The possible ways to seat those who are not in red chairs should be:

[tex]
\frac{ C^{n+p-m} _{n-m} }{C^{n+p}_{n} }[/tex]

That looks like the correct probability to me. Try it with some easy cases if you lack confidence.
 
  • #16
Dick said:
That looks like the correct probability to me. Try it with some easy cases if you lack confidence.
... and I agree. Apologies to Indigo, Dick and all for my poor response.
 
  • #17
I'm still confused. If the above gives me the probability to seat those who are not in red chairs, then what happened to the probability to seat those who are IN red chairs? Does the question ask what is the probability to seat the red chairs INSTEAD of the probability to seat those who are NOT in red chairs? Clarification on this method would be great.
 
  • #18
Unco said:
... and I agree. Apologies to Indigo, Dick and all for my poor response.

I wrote down the same thing first off. As Indigo knows, it can be a confusing problem. Apologies unnecessary.
 
  • #19
indigojoker said:
I'm still confused. If the above gives me the probability to seat those who are not in red chairs, then what happened to the probability to seat those who are IN red chairs? Does the question ask what is the probability to seat the red chairs INSTEAD of the probability to seat those who are NOT in red chairs? Clarification on this method would be great.

The 'above', C(n+p-m,n-m)/C(n+p,n) gives you the probability that all of the red seats are occupied. It's what the problem asked for. In the numerator is arrangements assuming the red seats are all occupied. In the denominator is all arrangements. That's all.
 

Related to Probability of All Red Seats Occupied in a Classroom

What is the "Probability of All Red Seats Occupied in a Classroom"?

The "Probability of All Red Seats Occupied in a Classroom" refers to the likelihood that all the red seats in a classroom will be occupied by students. This is a concept in probability, which is the branch of mathematics that deals with the likelihood of events occurring.

Why is the probability of all red seats occupied in a classroom important?

The probability of all red seats occupied in a classroom is important because it can help us make predictions about the likelihood of certain events occurring. In a classroom setting, this can be used to determine the likelihood of all students being present for a class or the chances of a seating arrangement being filled.

How is the probability of all red seats occupied in a classroom calculated?

The probability of all red seats occupied in a classroom can be calculated by dividing the number of ways all the red seats can be occupied by the total number of possible seating arrangements in the classroom. This can be expressed as a fraction, decimal, or percentage.

What factors can affect the probability of all red seats occupied in a classroom?

The probability of all red seats occupied in a classroom can be affected by various factors such as the number of red seats in the classroom, the total number of seats, and the number of students present. It can also be influenced by the seating preferences of the students and their likelihood of showing up for class.

How can the probability of all red seats occupied in a classroom be used in real-life situations?

The probability of all red seats occupied in a classroom can be used in real-life situations to make predictions about the likelihood of certain events occurring, such as all students being present for a class or the chances of a seating arrangement being filled. It can also be used in business or marketing to make decisions about the likelihood of a product being purchased by all customers or the chances of a certain event happening during a campaign or promotion.

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