Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 12 - Data Analysis and Probability - 12-8 Probability of Compound Events - Practice and Problem-Solving Exercises - Page 769: 29

Answer

1/36

Work Step by Step

You do not replace the coins, so the events are dependent: P(penny)=$\frac{2}{9}$ -2 of the 9 coins are pennies- P(quarter after penny)=$\frac{1}{8}$ -1 of the 8 remaining coins are quarters- Probability: P( penny then quarter)=P(quarter) $\times$ P( quarter after penny) P( penny then quarter)=$\frac{2}{9}$ $\times$ $\frac{1}{8}$ P( penny then quarter)=$\frac{1}{36}$
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