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Math Help - Basic probability question with permutations

  1. #1
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    Basic probability question with permutations

    What is the probability of these events when we randomly select a permutation of 1,2,....,n where n>=4

    a) 1 precedes 2
    I have no idea how to do this one

    b) 1 immediately precedes 2
    This one is simple I think. I just considered 1 and 2 as one unit to be permuted with the rest of the numbers and I get (n-1)!/n!= 1/n

    I could really use some advice, thank you so much!!
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  2. #2
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    Quote Originally Posted by mulaosmanovicben View Post
    What is the probability of these events when we randomly select a permutation of 1,2,....,n where n>=4
    a) 1 precedes 2
    I have no idea how to do this one

    b) 1 immediately precedes 2
    This one is simple I think. I just considered 1 and 2 as one unit to be permuted with the rest of the numbers and I get (n-1)!/n!= 1/n
    Your part (b) is correct.

    Part (a) is so easy!
    Either 1 comes before 2 or not.
    Now what is the answer?
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  3. #3
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    I am guessing answer to part a then is 1/2 for n>=4
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  4. #4
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    Quote Originally Posted by mulaosmanovicben View Post
    I am guessing answer to part a then is 1/2 for n>=4
    Say n=2: 12; 21
    Say n=3: 123; 132; 213; 231; 312; 321.

    Does it matter if n\ge 4?
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  5. #5
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    no it does not matter what n is. thank you i understand now. I am really having some difficulty with some of these problems.

    how about this one:

    n precedes 1 and n precedes 2

    should this be 1/4 then? (either n precedes 1 or it doesnt and either n precedes 2 or it doesnt)
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  6. #6
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    Quote Originally Posted by mulaosmanovicben View Post
    about this one:
    n precedes 1 and n precedes 2
    should this be 1/4 then? (either n precedes 1 or it doesnt and either n precedes 2 or it doesnt)
    No it is \frac{1}{3}.

    Look at 12n; 1n2; 21n; 2n1; n12; n21.
    That is two out of six.
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