Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 11 - 11.7 - Probability - 11.7 Exercises - Page 834: 16

Answer

$P(E)=\frac{N(E)}{N(S)}=\frac{3}{8}$

Work Step by Step

H -> head T -> tail All the possible outcomes (sample space): $S=[(H,H,H),(H,H,T),(H,T,H),(H,T,T),(T,H,H),(T,H,T),(T,T,H),(T,T,T)]$ We have that the total number of possible outcomes in the sample space is: $N(S)=8$ We want exactly two tails (Event): $E=[(H,T,T),(T,H,T),(T,T,H)]$ $N(E)=3$ $P(E)=\frac{N(E)}{N(S)}=\frac{3}{8}$
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