Elementary Statistics (12th Edition)

Published by Pearson
ISBN 10: 0321836960
ISBN 13: 978-0-32183-696-0

Chapter 4 - Probability - Review - Review Exercises: 14

Answer

Probably no, $0.0211$.

Work Step by Step

By using the complement rule: $P(at \ least \ one \ positive)=1-P(no \ positive)=1-(1-\frac{213}{100,000})^{10}=1-(\frac{99,787}{100,000})^{10}\approx0.0211$. This is very close to 0, therefore the combined test samples probably won't be positive.
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.