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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- Feb 4th 2010, 09:12 AMnovas51791Two 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?

- Feb 4th 2010, 10:12 AMSoroban
Hello, novas51791!

This is a classic (very old) puzzle.

Quote:

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

We have: .$\displaystyle \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}} $

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

. . $\displaystyle P(\text{W from urn A}) \,=\,1$

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

$\displaystyle P(\text{Urn B}) \:=\:\frac{1}{2}$

. . $\displaystyle P(\text{W from urn B}) \:=\:\frac{9}{19}$

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

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

- Feb 4th 2010, 10:32 AMnovas51791
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!