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Thread: Please help!! Basic Probability.

  1. #16
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    for the first question my answer was 0.006 is that correct by your reckoning?

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    Quote Originally Posted by el1 View Post
    and the bottom part of the formula says


    X!(n-x)! which if x=0 then


    o*10*9*8.... is that right?
    for X!*(n-X)!

    if X=0

    then we have

    0!*10!

    0! is defined to be 1

    0!=1

    so

    0!*10!=1*10*9*8*7*6*5*4*3*2*1=3628800

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    Quote Originally Posted by el1 View Post
    for the first question my answer was 0.006 is that correct by your reckoning?
    after rounding...yes

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    i think i've got it!! Thanks alot dude!

    Thank you for your time and patience.

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    Would you be so kind as to enlighten me about the third question ? Any ideas?

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    Quote Originally Posted by el1 View Post
    Would you be so kind as to enlighten me about the third question ? Any ideas?
    the assumptions for the binomial distribution are independent trials with the same probability of success

    so for your problem ....

    each caller needs to be independent from the others and they all have to have the same probability of opting for an operator.

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    Quote Originally Posted by el1 View Post
    Would you be so kind as to enlighten me about the third question ? Any ideas?
    http://childrensmercy.org/stats/defi...s/bin_dist.htm

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