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

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