Thinking Mathematically (6th Edition)

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

Chapter 11 - Counting Methods and Probability Theory - 11.6 Events Involving Not and Or; Odds - Exercise Set 11.6 - Page 736: 79

Answer

$\frac{9}{10}$.

Work Step by Step

The total number of outcomes in this sample: $38$, the number of outcomes satisfying the requirements in this sample: $18$, thus the odds for: $\frac{18}{38-18}=\frac{18}{20}=\frac{9}{10}$.
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