# Chapter 12 - Data Analysis and Probability - 12-8 Probability of Compound Events - Practice and Problem-Solving Exercises - Page 781: 31

$\frac{1}{12}$

#### Work Step by Step

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

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