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Math Help - Hyper geometric Distribution

  1. #1
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    Hyper geometric Distribution

    There is a box of 12 flares, four are selected at random. If the box contains four flares that don't work:

    a- what is the probability that all four will not work?

    b- What is the probability that, at most two will not work?

    MY ANSWERS:

    a = p(x=4)= C(4,4) x C(8,0)
    /divided C(12,4)

    24x1/19958400
    0.0000012025 = 0.00012%

    (that seems like a very low number for me)

    b= x=equal or less 2
    = C(4,2) x C(8,2) C(12,4) + C(4,1) x C(8,3) + (4,0) x (8,4)
    /divided C(12,4)

    = 403392/19958400 = 0.02021164 = 2%

    That seems off to me as well...where am I going wrong????
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  2. #2
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    Re: Hyper geometric Distribution

    Quote Originally Posted by staceylynnxx View Post
    There is a box of 12 flares, four are selected at random. If the box contains four flares that don't work:
    a- what is the probability that all four will not work?
    b- What is the probability that, at most two will not work?

    MY ANSWERS:
    a = p(x=4)= C(4,4) x C(8,0)
    /divided C(12,4)
    24x1/19958400
    0.0000012025 = 0.00012%
    (that seems like a very low number for me)
    You method in a) is correct, but your arithmetic is lousy.
    \frac{1}{C(12,4)}=0.002020202020202.

    For part b) \sum\limits_{k = 0}^2 {\frac{C(4,k)\cdot C(8,4-k)}{C(12,4)}}~.
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  3. #3
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    Re: Hyper geometric Distribution

    Thank you again.

    I don't know how to do the math signs on the forum.
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  4. #4
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    Re: Hyper geometric Distribution

    Quote Originally Posted by staceylynnxx View Post
    I don't know how to do the math signs on the forum.
    Use LaTeX tags.
    [TEX]\frac{1}{C(12,4)}=0.002020202020202[/TEX] gives \frac{1}{C(12,4)}=0.002020202020202

    [TEX]\sum\limits_{k = 0}^2 {\frac{C(4,k)\cdot C(8,4-k)}{C(12,4)}} [/TEX] gives \sum\limits_{k = 0}^2 {\frac{C(4,k)\cdot C(8,4-k)}{C(12,4)}}
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