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Math Help - win/lose probability

  1. #1
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    win/lose probability

    An urn contains five balls, one marked win and four marked lose. You and another player take turns selecting a ball from the urn , one at a time. The first person to select the WIN ball is the winner. If you draw first, find the probability that you will win if the sampling is done

    a. with replacement
    b. without replacement

    My work:
    a. (1/5)+(4/5)(1/5)+(4/5)(4/5)(1/5)=5/9
    b. (1/5)+(4/5)(3/4)(1/3)+(4/5)(3/4)(2/3)(1/2)=3/5
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  2. #2
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    Quote Originally Posted by antman View Post
    An urn contains five balls, one marked win and four marked lose. You and another player take turns selecting a ball from the urn , one at a time. The first person to select the WIN ball is the winner. If you draw first, find the probability that you will win if the sampling is done

    a. with replacement
    b. without replacement

    My work:
    a. (1/5)+(4/5)(1/5)+(4/5)(4/5)(1/5)=5/9
    b. (1/5)+(4/5)(3/4)(1/3)+(4/5)(3/4)(2/3)(1/2)=3/5
    Hi

    I agree with your results (5/9 and 3/5) even if I do not understand how you manage to find 5/9 for a.

    \frac15 + \frac{4}{5}\:\frac{4}{5}\:\frac{1}{5}+\frac{4}{5}\  :\frac{4}{5}\:\frac{4}{5}\:\frac{4}{5}\:\frac{1}{5  } \cdots = \frac15\:\left(1+\frac{16}{25}+\left(\frac{16}{25}  \right)^2+\cdots \right)= \frac15\:\frac{1}{1-\frac{16}{25}}=\frac59
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  3. #3
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    Hello, antman!

    I too agree with your answers.
    and I think I know how you got \tfrac{5}{9}.


    An urn contains five balls, one marked WIN and four marked LOSE.
    You and another player take turns selecting a ball from the urn , one at a time.
    The first person to select the WIN ball is the winner.
    If you draw first, find the probability that you will win if the sampling is done

    . . a. with replacement . . b. without replacement

    My work:

    a. (1/5) + (4/5)(1/5) + (4/5)(4/5)(1/5) = 5/9 .??

    b. (1/5) + (4/5)(3/4)(1/3) +(4/5)(3/4)(2/3)(1/2) = 3/5

    For (a), I believe you meant to type: . P \;=\;\frac{1}{5} + \left(\frac{4}{5}\right)^2\frac{1}{5} + \left(\frac{4}{5}\right)^4\frac{1}{5} + \hdots

    \text{Then: }\;P \;=\;\frac{1}{5}\underbrace{\bigg[1 + \left(\frac{4}{5}\right)^2 +  \left(\frac{4}{5}\right)^4 + \hdots \bigg] }_{\text{geometric series}}

    The geometric series has sum: . \frac{1}{1-\frac{16}{25}} \:=\:\frac{1}{\frac{9}{25}} \:=\:\frac{25}{9}

    . . Therefore: . P \;=\;\frac{1}{5}\cdot\frac{25}{9} \;=\;\frac{5}{9}


    If that is the case, you did terrific work on both parts.

    . . Nice going!

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