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Math Help - Two Urns Problem

  1. #1
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    Two Urns Problem

    How can 20 balls, 10 white and 10 black, be put into two urns so as to maximize the probability of drawing a white ball if an urn is selected at random and a ball is drawn at random from it?
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  2. #2
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    Hello, novas51791!

    This is a classic (very old) puzzle.


    How can 20 balls, 10 white and 10 black, be put into two urns
    so as to maximize the probability of drawing a white ball
    if an urn is selected at random and a ball is drawn at random from it?

    Solution: . \begin{array}{c}\text{Put 1 white ball in Urn A} \\ \text{put the rest in urn B}\end{array}


    We have: . \boxed{\begin{array}{c} \text{Urn A} \\\text{1 W} \\  \\ \end{array}} \qquad \boxed{\begin{array}{c} \text{Urn B} \\\text{9 W} \\ \text{10 B} \end{array}}


    P(\text{urn A}) \,=\,\frac{1}{2}

    . . P(\text{W from urn A}) \,=\,1
    . . Hence: . P(\text{urn A }\wedge\text{ W}) \:=\:\left(\frac{1}{2}\right)(1) \:=\:\frac{1}{2}


    P(\text{Urn B}) \:=\:\frac{1}{2}
    . . P(\text{W from urn B}) \:=\:\frac{9}{19}
    . . Hence: . P(\text{urn B }\wedge\text{ W}) \:=\:\left(\frac{1}{2}\right)\left(\frac{9}{19}\ri  ght) \:=\:\frac{9}{38}


    Therefore: . P(\text{White}) \;=\;\frac{1}{2} + \frac{9}{38} \;=\;\frac{14}{19} \;\approx\; 73.7\%

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  3. #3
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    Thank you so much! That's the answer I got as well, from guessing and checking.

    Is it just a matter of guessing and checking, or is there some mathematical way to obtain that answer? Thanks!
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