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: 27

Answer

1/36

Work Step by Step

You do not replace the coins, so the events are dependent: P(quarter)=$\frac{1}{9}$ -1 of the 9 coins is a quarter- P(penny after quarter)=$\frac{2}{8}$=$\frac{1}{4}$ -2 of the 8 remaining coins are pennies- Probability: P(quarter then penny)=P(quarter) $\times$ P(penny after quarter- P(quarter then penny=$\frac{1}{9}$ $\times$ $\frac{1}{4}$ P(quarter then penny)=$\frac{1}{36}$
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