Thinking Mathematically (6th Edition)

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

Chapter 11 - Counting Methods and Probability Theory - 11.7 Events Involving And; Conditional Probability - Exercise Set 11.7 - Page 748: 89

Answer

$\frac{11}{36}$

Work Step by Step

The number of all possible outcomes is $6\cdot6=36$. The outcomes satisfying the requirements: $\{(1,5),(1,6),(3,5),(3,6),(5,1),(5,3),(5,5),(5,6)(6,1),(6,3),(6,5)\}$ Hence here the probability: $\frac{11}{36}$
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