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Math Help - Probability - sampling without replacement.

  1. #1
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    Probability - sampling without replacement.

    I have a feeling it's not as easy as I originally thought!

    "There are 20 solar cells, 12 of which are flat and the rest are concentrated. If you were to pick six random cells, what is the probability you'd get three of each?"

    So... 60% of all the cells are flat, meaning that the remaining 40% are concentrated and those would also be the probabilities of grabbing all six cells of the same description.

    would this formula apply here: (p^x)(1-p)^(x-n)

    where p is the probability, x being the description and n being the total cells?

    (0.6^12)(1-0.6)^(12-20) = (0.002177) (0.000655) = 3.324 ??

    A little lost, as you can surely see!

    Thank you!
    Last edited by mr fantastic; May 17th 2011 at 03:47 PM. Reason: Re-titled.
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  2. #2
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    Quote Originally Posted by Gretchy View Post
    I have a feeling it's not as easy as I originally thought!
    "There are 20 solar cells, 12 of which are flat and the rest are concentrated. If you were to pick six random cells, what is the probability you'd get three of each?"
    There are \binom{12}{3}\binom{8}{3} ways to select three of each.

    How many ways to select twelve altogether?

    What do you do with that?
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  3. #3
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    Quote Originally Posted by Plato View Post
    There are \binom{12}{3}\binom{8}{3} ways to select three of each.

    How many ways to select twelve altogether?

    What do you do with that?
    Hmm.

    I know that \binom{n}{x}p^x(1-p)^{n-x} where \binom{n}{x} = \frac{n!}{x!(n-x)!}, then

    \binom{12}{3} = \frac{12!}{3!(9!)} = 220

    Plug it in and...

    220\cdot [(0.6)^{3}\cdot(1-0.6)^{12-3}]

    = 220\cdot [(0.6)^{3}\cdot(1-0.6)^{9}]
    = 220\cdot [(0.6)^{3}\cdot(0.4)^{9}]
    = 220\cdot [(0.216)\cdot(0.000262)]
    = 220 \cdot (0.000056592)
    = 0.01245024

    Similarly,
    \binom{8}{3} = \frac{8!}{3!(5!)} = 56

    56\cdot [(0.4)^{3}\cdot(1-0.4)^{8-3}]

    = 56\cdot [(0.4)^{3}\cdot(0.6)^{5}]
    = 56\cdot [(0.064)\cdot(0.7776)]
    = 56\cdot(0.0497664)
    = 2.7869184

    So... I add them? (3 + 3 = 6)

    2.7869184 + 0.01245024 = 2.79936864

    Yes? No?

    (PS: LaTex is NOT easy...!]
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  4. #4
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    Quote Originally Posted by Gretchy View Post
    Hmm.

    I know that \binom{n}{x}p^x(1-p)^{n-x} where \binom{n}{x} = \frac{n!}{x!(n-x)!}, then

    \binom{12}{3} = \frac{12!}{3!(9!)} = 220

    Plug it in and...

    220\cdot [(0.6)^{3}\cdot(1-0.6)^{12-3}]

    = 220\cdot [(0.6)^{3}\cdot(1-0.6)^{9}]
    = 220\cdot [(0.6)^{3}\cdot(0.4)^{9}]
    = 220\cdot [(0.216)\cdot(0.000262)]
    = 220 \cdot (0.000056592)
    = 0.01245024

    Similarly,
    \binom{8}{3} = \frac{8!}{3!(5!)} = 56

    56\cdot [(0.4)^{3}\cdot(1-0.4)^{8-3}]

    = 56\cdot [(0.4)^{3}\cdot(0.6)^{5}]
    = 56\cdot [(0.064)\cdot(0.7776)]
    = 56\cdot(0.0497664)
    = 2.7869184

    So... I add them? (3 + 3 = 6)

    2.7869184 + 0.01245024 = 2.79936864
    Yes? No?
    SORRY, BUT NO.
    All of that is a serious misunderstanding of this problem.
    You see this as a binomial distribution. It is not.
    In binomial distributions the trials must be independent. These are not.
    This sampling without replacement, so not independent.
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  5. #5
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    thats my fault i told him it was binomial in my (removed) post. sorry gretchy!
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