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Permutation homework

  1. Jan 9, 2008 #1
    This is problem 4 from section 2.3 of Fundamentals of Probability by Saeed Ghahramani.

    1. The problem statement, all variables and given/known data

    Robert has eight guests, two of whom are Jim and John. If the guests will arrive in a random order, what is the probability that John will not arrive right after Jim?

    2. Relevant equations

    ...

    3. The attempt at a solution

    There are 8! ways for the guests to arrive. There are 7 ways for John to arrive right after Jim. The probability of John arriving right after Jim is then 7/8!, and the probability of John not arriving right after Jim is 1-7/8!=0.9998.

    I am pretty sure it is right, but not 99.98% sure.
     
  2. jcsd
  3. Jan 9, 2008 #2

    Dick

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    That's wrong. There aren't 7 ways for John to arrive right after Jim. There are 7 times the number of ways for all of the other guests to arrive.
     
  4. Jan 9, 2008 #3
    True. So,

    1 - (7*6!)/8! = .875
     
  5. Jan 9, 2008 #4

    Dick

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    Much better. I'm 99.98% sure that's right.
     
  6. Jan 17, 2012 #5
    Re: Permutation

    There are only 7 ways for John to arrive right after Jim, apart from this detail being insufficient for the solution, or?
     
  7. Jan 17, 2012 #6

    Dick

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    Re: Permutation

    So you didn't read the rest of the thread? Please do it.
     
  8. Jan 18, 2012 #7

    Dick

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    Re: Permutation

    I'm not sure what your question is. acme37 thought that of the 8! orders that the guests could arrive there were 7 in which John arrives right after Jim. That's wrong. The right answer is perfectly sufficient for a solution.
     
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