## Algebra 1: Common Core (15th Edition)

$\frac{1}{12}$
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 dimes: P(dime after dime)=$\frac{2}{8}$=$\frac{1}{4}$ 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)=$\frac{1}{12}$