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Thread: Drawing balls randomly with replacement

  1. #1
    Senior Member I-Think's Avatar
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    Drawing balls randomly with replacement

    An urn contains $\displaystyle n$ balls numbered $\displaystyle 1$ through $\displaystyle n$. If you withdraw $\displaystyle m$ balls randomly in sequence, each time replacing the ball selected previously, find $\displaystyle P[X=k], k=1,2,...,m$, where $\displaystyle X$ is the maximum of the $\displaystyle m$ chosen numbers

    I'm not even sure what this question is asking me to find. Can someone help me clarify the question please?
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  2. #2
    Grand Panjandrum
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    Re: Drawing balls randomly with replacement

    Quote Originally Posted by I-Think View Post
    An urn contains $\displaystyle n$ balls numbered $\displaystyle 1$ through $\displaystyle n$. If you withdraw $\displaystyle m$ balls randomly in sequence, each time replacing the ball selected previously, find $\displaystyle P[X=k], k=1,2,...,m$, where $\displaystyle X$ is the maximum of the $\displaystyle m$ chosen numbers

    I'm not even sure what this question is asking me to find. Can someone help me clarify the question please?
    What is the probability that the maximum of m balls drawn with replacement is k?

    CB
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  3. #3
    MHF Contributor

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    Re: Drawing balls randomly with replacement

    Quote Originally Posted by I-Think View Post
    An urn contains $\displaystyle n$ balls numbered $\displaystyle 1$ through $\displaystyle n$. If you withdraw $\displaystyle m$ balls randomly in sequence, each time replacing the ball selected previously, find $\displaystyle P[X=k], k=1,2,...,m$, where $\displaystyle X$ is the maximum of the $\displaystyle m$ chosen numbers
    I will discuss one case. You can generalize.
    Suppose $\displaystyle \{1,2,3,4,5,6\}$ is the set of numbered balls and $\displaystyle m=4.$

    Letís find $\displaystyle P(X=3)$. There $\displaystyle 3^4$ ways to have 4-tuples containing only 1, 2, or 3.
    There $\displaystyle 2^4$ ways to have 4-tuples containing only 1 or 2.
    Therefore there $\displaystyle 3^4-2^4$ ways to have 4-tuples made of 1, 2, or 3 with at least one 3.

    Then $\displaystyle P(X=3)=\frac{3^4-2^4}{6^4}$
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