Q. 3.3 - A First Course in Probability

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?

Step-by-Step Solution

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: 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

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

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.