1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question on permutation.

  1. Feb 5, 2006 #1
    A group of 5 boys and 10 girls is lined up in random order-that is, each of the 15! permutaions is assumed to be equally likely.

    a) What is the probability that the person in the 4th position is a boy?
    b) What is the probability that a particular boy is in the 3rd position?

    i tried to list out 1 possible combination but it gets pretty ugly since there are a total of 15 choices.. can someone please help?
     
  2. jcsd
  3. Feb 5, 2006 #2

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    I'm not sure about this, so I'll take a guess at what I think.

    For a), wouldn't you just find the number of possible ways to organize the line such that the 4th position is a boy? Once you got that, you can just compare it with the total permutation, which should give you the probability of that happening.

    For b), it's something extremely similiar, but you have to remember that you cannot change position 3.

    Is your problem with the idea of finding the number of ways to organize the line with some conditions?

    If that's the case, try googling some sample exercises about permutations. After a couple, you'll understand enough to apply it to this situation. Practice is what makes perfect.
     
    Last edited: Feb 5, 2006
  4. Feb 5, 2006 #3
    Thanks for replying.

    for part a), If I just look at the first 3 picks, i can either have 0,1,2,3 boys and the 4th postion have to be a boy. I calculated the probabilities of each of them individually and they added up to be 1/3. For which i multiplied it by 11! (5-15th selection) and divided by 15! (total permutation), but i ended up with 1.0175x10^-5 which seems way to small to be a correct answer..
    what did i do wrong in my reasoning??
     
  5. Feb 6, 2006 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The precise position isn't relevant. What may be confusing you is that "permutations" are not at all relevant. Suppose, out of 5 boys and 10 girls, you choose one person at random (each equally likely to be picked). What is the probability that the person is a boy? (That should be easy.)

    "What is the probability that a particular boy is in the third position."
    Given that the person in the third position is a boy?? Or given that the person in the fourth position was a boy? If neither of those is intended: there are 15 persons who are all equally likely to be chosen. What is the probability that a specific one of them will be chosen?

    If it is intended that the person in the third position is a boy, then there are 5 persons all equally likely to be chosen.
     
  6. Feb 9, 2006 #5
    Halls of Ivy: The precise position isn't relevant.

    I remember an old Mutt and Jeff cartoon. As they are traveling on a train, Jeff asks Mutt, "How many sheep do you think are in that herd we just passed?"

    Mutt replies, "876."

    "That's amazing! You mean you counted every one as the train sped past?"

    "No, not really. I just counted the legs and divided by four."
     
    Last edited: Feb 9, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Question on permutation.
  1. Permutation Question. (Replies: 2)

  2. Permutation Question (Replies: 5)

  3. Permutations question (Replies: 7)

  4. Permutation question (Replies: 3)

Loading...