Elementary Statistics (12th Edition)

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

Chapter 4 - Probability - Review - Review Exercises - Page 187: 14


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.
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