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: 32

Answer

1/12

Work Step by Step

You do not replace the coins, so the events are dependent: P(dime)=$\frac{3}{9}$=$\frac{1}{3}$ -3 of the 9 coins are dimes- P(dime after dime)=$\frac{2}{8}$=$\frac{1}{4}$ -2 of the 8 remaining coins are dimes- robability: P(dime then dime)=P(dime) $\times$ P(dime after dime) P(dime then dime)=$\frac{1}{3}$ $\times$ $\frac{1}{4}$ P(dime then dime)=$\frac{1}{12}$
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