Chapter 3: Q. 3.3 (page 109)

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

Expert verified

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

$P (White) = P (White ∣ Urn 1) P (Urn 1) + P (White ∣ Urn 2) P (Urn 2)$

Urn randomly as

role="math" localid="1649501479994" $P (U r n 1) = \frac{1}{2} P (U r n 2) = \frac{1}{2} P (White) = \frac{1}{2} [P (White ∣ Urn 1) + P (White ∣ Urn 2)]$

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

$P (White ∣ U r n 1) = P (White ∣ U r n 2) = \frac{1}{2} P (White) = \frac{1}{2} [\frac{1}{2} + \frac{1}{2}] P (White) = \frac{1}{2} .$

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

$P (White ∣ U r n 2) < \frac{1}{2} .$

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.

$P (White ∣ Urn 1) \leq 1 P (White ∣ Urn 2) \leq \frac{9}{19}$

It's theoretically maximum and the distribution satisfies maximum.

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

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

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

Step by step solution

Step: 1 Events:

Step: 2 Upper bounds:

Step: 3 Getting probability:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Applied Mathematics

Calculus

Pure Maths

Decision Maths

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

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

Step by step solution

Step: 1 Events:

Step: 2 Upper bounds:

Step: 3 Getting probability:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Applied Mathematics

Calculus

Pure Maths

Decision Maths

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

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?