Thinking Mathematically (6th Edition)

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

Chapter 11 - Counting Methods and Probability Theory - 11.4 Fundamentals of Probability - Exercise Set 11.4 - Page 716: 37

Answer

$\frac{1}{9}$

Work Step by Step

P(E) = $\frac{no.\ of\ outcomes\ in\ favor}{Total\ no.\ of\ outcomes}$ A single die is rolled twice. The 36 equally likely outcomes are shown in the attached picture. E: two numbers whose sum is 5. Favorable outcomes= {(1,4),(2,3),(3,2),(4,1)} P(E) = $\frac{4}{36}$ = $\frac{1}{9}$
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