## Thinking Mathematically (6th Edition)

$\frac{1}{9}$
P(E) = $\frac{no.\ of\ outcomes\ in\ favor}{Total\ no.\ of\ outcomes}$ A single die is rolled twice. The 36 equally likely outcomes are shown in the attached picture. E: two numbers whose sum is 5. Favorable outcomes= {(1,4),(2,3),(3,2),(4,1)} P(E) = $\frac{4}{36}$ = $\frac{1}{9}$