Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 11 - Counting Methods and Probability Theory - 11.3 Combinations - Exercise Set 11.3 - Page 709: 80

Answer

1365 samples

Work Step by Step

A sample implies that the order of selecting its contents in unimportant. We work with combinations. The sample with no defective phones has been chosen in a way that: (1) 4 phones were chosen from the 15 working phones (2) 0 phones were chosen from the 5 defective phones Total: ${}_{15}C_{4}\displaystyle \cdot {}_{5}C_{0}=\frac{15!}{(15-4)!4!}\times\frac{5!}{(5-0)!0!}$ ... by definition, $0!=1$... $=\displaystyle \frac{15\times 14\times 13\times 12}{1\times 2\times 3\times 4}\cdot 1$ $=1365$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.