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20 chickens

  1. Jun 3, 2006 #1
    Hi, i some time ago my physics teacher asked me this problem: There are 20 chickens coming out of an hen-coop. 15 are white, 5 are black. What's the probability of at least 2 black chickens come out consecutively.

    so basicly i tried to calc P = 1 - (Probability of no black chicken come out consecutively )

    So i thought P = 1 - D(20,5)/C(20,5)

    being D = Ways to arrange the 5 chickens in 20 "boxes" so that the 5 black never come out consecutivelly

    after some thinking i figured out that
    D(n,p) = D(n-2,p-1) + D(n-1,p)

    which is somewhat similar to C(n,p) = C(n-1,p-1) + C(n-1,p)

    So i started making some sort of pascal triangle for the D function which lead me to this complicated sum:

    [tex] D(n,p) = \sum_{i=1}^{n-2p+2} C(p-3+i,p-2) \times (k-i+1) [/tex]

    hope i got this right (first time using latex ^^)

    well, i came to the same result as my teacher but he had a much more simple formula, to which he arrived empirically:

    P = 1 - C(20-5+1,5)*15!*5!/20!

    So, D(n,p)=C(n-p+1,p) i tried to deduce this formula from what i previously had but with no success. Can some1 plz tell me how i could have arrived to it.

    Ty

    PS: By the way, i'm only in highschool so try to keep it simple.:biggrin:
     
  2. jcsd
  3. Jun 3, 2006 #2

    NateTG

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    Here's one way:

    Saying that there are no two black chickens in a row is the same as saying that every black chicken is directly in front of a white chicken, or directly in front of the end of the row.

    So we can think of this as 16 spots (1 in front of each white chicken and 1 at the end) and choosing 5.

    You should be able to generalize from there.
     
  4. Jun 3, 2006 #3

    Gokul43201

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    How many ways are there of putting k black objects in the n+1 spaces formed by a string of n white objects, with no more than 1 black object per space ?

    Edit : This seems to be exactly what Nate was saying. The spots threw me! I guess I'm a space person.
     
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