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: 51

Answer

$\frac{65}{71}$

Work Step by Step

The total number of outcomes in this sample: $1420$ (thousand), the number of outcomes satisfying the requirements in this sample: $270+490+80+190+270=1300$ (thousand), thus the probability: $\frac{1300}{1420}=\frac{65}{71}$
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