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Probability question

  1. Jan 24, 2009 #1
    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.
     
  2. jcsd
  3. Jan 24, 2009 #2
    Sorry I made a mistake, never mind.
     
    Last edited: Jan 24, 2009
  4. Jan 24, 2009 #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?
     
  5. Jan 24, 2009 #4
    This is a repeat post: https://www.physicsforums.com/showthread.php?t=286955
     
  6. Jan 24, 2009 #5
    I dont 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.
     
  7. Jan 24, 2009 #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.
     
  8. Jan 24, 2009 #7
    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.
     
  9. Jan 24, 2009 #8

    Dick

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    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).
     
  10. Jan 24, 2009 #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.
     
  11. Jan 24, 2009 #10

    Dick

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    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.
     
  12. Jan 24, 2009 #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?
     
  13. Jan 24, 2009 #12

    Dick

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    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.
     
  14. Jan 24, 2009 #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.
     
    Last edited: Jan 24, 2009
  15. Jan 24, 2009 #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]
     
  16. Jan 24, 2009 #15

    Dick

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    That looks like the correct probability to me. Try it with some easy cases if you lack confidence.
     
  17. Jan 24, 2009 #16
    ... and I agree. Apologies to Indigo, Dick and all for my poor response.
     
  18. Jan 24, 2009 #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.
     
  19. Jan 24, 2009 #18

    Dick

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    I wrote down the same thing first off. As Indigo knows, it can be a confusing problem. Apologies unnecessary.
     
  20. Jan 24, 2009 #19

    Dick

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    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.
     
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