#### Answer

0

#### Work Step by Step

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