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Homework Help: Probability question

  1. Jan 8, 2006 #1
    There are n stations between two cities X and Y. At train is to stop at three of these n stations. Find the probability that no two of these three stations are consecutive.

    This is what I did:

    Total number of possibilities nc3 .
    Now suppose that the stations train stops at is such that exactly two of these are consecutive.
    (1,2) then it may stop at (4,5,6…..n) and no. of possibilities are n-3
    (2,3) then it may stop at (5,6,7…..n) and no. of possibilities are n-4
    …………………………………….
    (n-3,n-2) then it may stop only at n and no. of possibilities are 1

    So the number of ways in which the train may stop at exactly two consecutive stations is
    (n-3) + (n-4) + ……… + 1 n-3 terms

    using formula of A.P. [n/2 (a+l)] n is no. of terms, a is first term, l last ]
    possibilities are (n-3)(n-2)/2

    Now if exactly 3 stations are consecutive then (1,2,3) (2,3,4)….. (n-2,n-2,n)
    Hence here the no. of possibilities are n-2

    So the things I have to exclude are (n-3)(n-2)/2 + (n-2)
    Which is equal to (n-1)(n-2)/2

    So we have Pr{event} = 1 – (n-1)(n-2)/2*nC3 = (n-3)/n

    But I am not getting the right answer. Any help where my analysis is wrong?
     
  2. jcsd
  3. Jan 10, 2006 #2

    EnumaElish

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    Let's see... Suppose the train were to make 2 stops. You could make an n-by-n table, where a row shows the 1st stop and a column shows the 2nd stop. (E.g. if n=7 then the table will have 49 cells.) Since 1st stop < 2nd stop, you need to consider the cells above the diagonal only. The consecutive stops are of the form (k, k+1); they are the cells just above the diagonal. There are n-1 such cells.

    Now you could apply this logic to your case (with 3 stops).
     
  4. Jan 12, 2006 #3
    Gaganpreet Singh, nice to see another sardar here. :wink: WJKK WJKF! _/|\_
     
  5. Jan 12, 2006 #4

    EnumaElish

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    Do you have something to contribute to the discussion here?
     
  6. Jan 13, 2006 #5

    ehild

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    The first stop can be anything betveen 1 and n-4. Let it be "j".
    The second stop "k" can be any number from j+2 to n-2.
    The third stop "l" can be any number from k+2 to n.
    The number of possible arrangement of stops is
    [tex]M = \sum_{1}^{n-4} \sum_{j+2}^{n-2} \sum_{k+2}^{n}1=\frac{(n-2)(n-3)(n-4)}{6} [/tex]

    ehild
     
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