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Math Help - binomial distribution

  1. #1
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    binomial distribution

    please refer to the second line of solution, since we only concerned about the probability of getting number (5) , then why cant I just just say P=(5/6)^5 , why should I times =(5/6)^5 with (1/6)^2 ?binomial distribution-img_20140605_112229-1-.jpg
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  2. #2
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    Re: binomial distribution

    Quote Originally Posted by delso View Post
    please refer to the second line of solution, since we only concerned about the probability of getting number (5) , then why cant I just just say P=(5/6)^5 , why should I times =(5/6)^5 with (1/6)^2 ?Click image for larger version. 

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    We have strings of length seven made up it numbers on the faces of a die.
    The questions is: "how many of those strings contain exactly two 5's"
    There $^7C_2$ places for the two 5's. The probability of two 5's is $\left(\frac{1}{6}\right)^2$
    The probability of five non-5's is $\left(\frac{5}{6}\right)^5$.
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    Re: binomial distribution

    what do u mean by We have strings of length seven made up it numbers on the faces of a die.? do you mean we have length of 7 unit consists of 7 dice which each of it with length 1unit?
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    Re: binomial distribution

    Quote Originally Posted by delso View Post
    what do u mean by We have strings of length seven made up it numbers on the faces of a die.? do you mean we have length of 7 unit consists of 7 dice which each of it with length 1unit?
    I just don't think you understand any of this.
    The outcome $6~3~5~6~4~4~5$ is one favorable case.
    ANY outcome $X~X~5~X~X~X~5$ is also a favorable case if $X\ne 5$
    There are $\dbinom{7}{2}$ places for the two 5's.

    Hence $\dbinom{7}{2}{\left( {\dfrac{1}{6}} \right)^2}{\left( {\dfrac{5}{6}} \right)^5}$
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