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

$\frac{1}{9}$
Because the tiles are replaced, the events are independent: 4 of the 12 tiles are Y's: P(Y)=$\frac{4}{12}$=$\frac{1}{3}$ 4 of the 12 tiles are Y's: P(Y after Y)=$\frac{4}{12}$=$\frac{1}{3}$ P(Y then Y)=P(Y) $\times$ P(Y after Y) P(Y then Y)=$\frac{1}{3}$ $\times$ $\frac{1}{3}$= $\frac{1}{9}$