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?
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: .$\displaystyle \begin{array}{c}\text{Put 1 white ball in Urn A} \\ \text{put the rest in urn B}\end{array}$
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: .$\displaystyle P(\text{White}) \;=\;\frac{1}{2} + \frac{9}{38} \;=\;\frac{14}{19} \;\approx\; 73.7\%$