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Math Help - A urn contain N white

  1. #1
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    A urn contain N white

    Question : A urn contain N white and M black balls. Balls are randomly selected, one at a time , until a black one is obtained . It we assume that each selected ball is replaced before the next one is drawn, what is the probability that
    i) exactly n draws are needed
    ii) atleast k draws are needed
    ----------------------------------------------------------------

    Should i be using geometric distribution
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  2. #2
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    Quote Originally Posted by zorro View Post
    Question : A urn contain N white and M black balls. Balls are randomly selected, one at a time , until a black one is obtained . It we assume that each selected ball is replaced before the next one is drawn, what is the probability that
    i) exactly n draws are needed
    ii) atleast k draws are needed
    ----------------------------------------------------------------

    Should i be using geometric distribution
    Yes. p = M/(M + N). Now go and thoroughly review that topic.
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  3. #3
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    Hello, zorro!

    Who wrote this problem?
    What a confusing set of variables!


    An urn contains {\color{red}W} white and {\color{red}B} black balls.
    Balls are randomly selected, one at a time , until a black one is obtained.
    Each selected ball is replaced before the next one is drawn.

    Find the probability that:

    a) exactly n draws are needed.

    b) at least k draws are needed.

    Should i be using geometric distribution? . . . . no
    Let W + B \,=\,T, the total number of balls.

    Since the balls are replaced, the probabilities remain constant.

    . . \begin{array}{ccc}P(W) &=& \dfrac{W}{T} \\ \\[-3mm] P(B) &=& \dfrac{B}{T} \end{array}



    (a) A black ball is drawn on the n^{th} draw.
    . . .The first n\!-\!1 balls are White, the n^{th} is Black.

    . . . P(n\text{ draws}) \;=\;\left(\frac{W}{T}\right)^{n-1}\!\!\left(\frac{B}{T}\right)



    (b) At least k draws.

    We have this list of possible outcomes:

    . . \begin{array}{ccc}P(k\text{ draws})  &=& \left(\frac{W}{T}\right)^{k-1}\left(\frac{B}{T}\right) \\ \\[-3mm]<br />
P(k\!+\!1\text{ draws}) &=& \left(\frac{W}{T}\right)^k\left(\frac{B}{T}\right) \\ \\[-3mm]<br />
P(k\!+\!2\text{ draws}) &=& \left(\frac{W}{T}\right)^{k+1}\left(\frac{B}{T}\ri  ght) \\ \\[-3mm]<br />
\vdots && \vdots \\\end{array}


    Our probability is the sum of these probabilities:

    P(\text{at least }k\text{ draws}) \;=\;\left(\frac{W}{T}\right)^{k-1}\!\!\left(\frac{B}{T}\right) \;+\; \left(\frac{W}{T}\right)^k\!\!\left(\frac{B}{T}\ri  ght) \;+\; \left(\frac{W}{T}\right)^{k+1}\!\!\left(\frac{B}{T  }\right) \;+\; \hdots

    . . . . . . . . . . . . =\;\left(\frac{W}{T}\right)^{k-1}\!\!\left(\frac{B}{T}\right)\underbrace{\bigg[1 + \frac{W}{T} + \left(\frac{W}{T}\right)^2 + \hdots \bigg]}_{\text{geometric series}} .[1]


    \text{The sum of the geometric series is: }\:\frac{1}{1-\frac{W}{T}} \:=\:\underbrace{\frac{T}{T-W}}_{\text{This is }B} \;=\;\frac{T}{B}


    Substitute into [1]:

    . . P(\text{at least }k\text{ draws}) \;\;=\;\;\left(\frac{W}{T}\right)^{k-1}\!\!\left(\frac{B}{T}\right)\!\left(\frac{T}{B}\  right) \;\;=\;\;\left(\frac{W}{T}\right)^{k-1}

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  4. #4
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    Thanks mite

    Thanks mite
    Happy and prosperous NEW YEAR
    Cheers
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