## Algebra and Trigonometry 10th Edition

$\frac{11}{15}$
From Exercises 31 and 32, $P(\text{both red})=\frac{1}{5}$, $P(\text{both yellow})=\frac{1}{15}$, thus $P(\text{both same colour})=\frac{1}{5}+\frac{1}{15}=\frac{3}{15}+\frac{1}{15}=\frac{4}{15}$ $P(\text{different colour})=1-P(\text{both same colour})=1-\frac{4}{15}=\frac{11}{15}$