## Thinking Mathematically (6th Edition)

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 exceeds 12 Favorable outcomes= {⌀} P(E) = $\frac{0}{36}$ =0