Elementary Statistics (12th Edition)

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

Chapter 4 - Probability - 4-3 Addition Rule - Basic Skills and Concepts - Page 156: 38

Answer

0.91, complement of the result from Exercise 37.

Work Step by Step

We know that $probability=\frac{number \ of \ good \ outcomes}{number\ of\ all\ outcomes}$, and also $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Let A=true positive, B=true negative, then $P(A\cup B)=\frac{119}{300}+\frac{154}{300}-\frac{0}{300}=\frac{273}{300}=0.91.$, which is the complement of the result from Exercise 37.
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