Q. 2.35 Seven balls are randomly withdra... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 2: Q. 2.35 (page 51)

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that
(a) 3 red, 2 blue, and 2 green balls are withdrawn;
(b) at least 2 red balls are withdrawn;
(c) all withdrawn balls are the same color;
(d) either exactly 3 red balls or exactly 3 blue balls are withdrawn.

Short Answer

Expert verified

The respective probabilities are

(a) $P (3 R + 2 B + 2 G) = \frac{3060}{40549}$

(b) $P (R 鈮� 2) = \frac{363707}{608235}$

(d) $P (3 R or 3 B) = \frac{13}{892078}$

Step by step solution

Given information

An urn that contains 12 red(R), 16 blue(B), and 18 green(G) balls.

Part(a)

$P (3 R + 2 B + 2 G) = \frac{(\begin{matrix} 12 \\ 3 \end{matrix}) (\begin{matrix} 16 \\ 2 \end{matrix}) (\begin{matrix} 18 \\ 2 \end{matrix})}{(\begin{matrix} 46 \\ 7 \end{matrix})} = \frac{3060}{40549}$

Part(b)

P(atleast 2 red balls are drawn) $= P (R 鈮� 2)$

$P (R 鈮� 2) = 1 - P (R 鈮� 1) = 1 - P (R = 0) - P (R = 1) = 1 - \frac{(\begin{matrix} 34 \\ 7 \end{matrix})}{(\begin{matrix} 46 \\ 7 \end{matrix})} - \frac{(\begin{matrix} 12 \\ 1 \end{matrix}) (\begin{matrix} 34 \\ 6 \end{matrix})}{(\begin{matrix} 46 \\ 7 \end{matrix})} 鈬� P (R 鈮� 2) = \frac{363707}{608235}$

Part(c)

P(all balls are of same color) = P(all R) + P(all B) + P(all G) = $p$ (say)

p = \frac{(\begin{matrix} 12 \\ 7 \end{matrix})}{(\begin{matrix} 46 \\ 7 \end{matrix})} + \frac{(\begin{matrix} 16 \\ 7 \end{matrix})}{(\begin{matrix} 46 \\ 7 \end{matrix})} + \frac{(\begin{matrix} 18 \\ 7 \end{matrix})}{(\begin{matrix} 46 \\ 7 \end{matrix})} p = \frac{5507}{6690585}

Part(d)

P(either exactly 3 red balls or exactly 3 blue balls are withdrawn) = P(3R or 3B)
$P (3 R or 3 B) = P (3 R) + 鈥� P (3 B) P (3 R or 3 B) = \frac{(\begin{matrix} 12 \\ 3 \end{matrix})}{(\begin{matrix} 46 \\ 3 \end{matrix})} + \frac{(\begin{matrix} 16 \\ 3 \end{matrix})}{(\begin{matrix} 46 \\ 3 \end{matrix})} 鈬� P (3 R or 3 B) = \frac{13}{892078}$

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Short Answer

Step by step solution

Given information

Part(a)

Part(b)

Part(c)

Part(d)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Mechanics Maths

Geometry

Calculus

Pure Maths

Logic and Functions

Study anywhere. Anytime. Across all devices.