Question

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?

Short answer

The white ball from each urn is the maximizes possibility of drawing is one white ball from urn$1$and nine white ball and ten black ones from urn$2.$

Step-by-step solution

Step: 1 Events:

The probability formula is

$P\left(\text{White}\right)=P(\text{White}\mid \text{Urn}1)P\left(\text{Urn}1\right)+P(\text{White}\mid \text{Urn}2)P\left(\text{Urn}2\right)$

Urn randomly as

role="math" localid="1649501479994" $$\begin{gathered} P\left(Urn1\right)=\frac{1}{2} \\ P\left(Urn2\right)=\frac{1}{2} \\ P\left(\text{White}\right)=\frac{1}{2}\left[P\right(\text{White}\mid \text{Urn}1)+P(\text{White}\mid \text{Urn}2\left)\right] \end{gathered}$$

Since,P(white) is greater than P( white/Urn1) and P(white/Urn 2).

Step: 2 Upper bounds:

May be both urns are same as many white and black balls as

$$\begin{gathered} P(\text{White}\mid Urn1)=P(\text{White}\mid Urn2)=\frac{1}{2} \\ P\left(\text{White}\right)=\frac{1}{2}\left[\frac{1}{2}+\frac{1}{2}\right] \\ P\left(\text{White}\right)=\frac{1}{2}. \end{gathered}$$

One urn less than half white balls.so it's urn $2.$

$P(\text{White}\mid Urn2)<\frac{1}{2}.$

Step: 3 Getting probability:

In total $20$balls,the nearest probabality can get $1/2$is $9/19.$So it's upper limit.

Probabilities are less are equal to one.

$P(\text{White}\mid \text{Urn}1)\le 1P(\text{White}\mid \text{Urn}2)\le \frac{9}{19}$

It's theoretically maximum and the distribution satisfies maximum.

Urns are not differentiated.if not reverse will be a solution.