## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$0.3$.
We know that probability of the event E is $P(E)=\frac{\text{number of favourable outcomes}}{\text{number of all outcomes}}=\frac{n(E)}{n(S)}$. Since here we have $n(E)=3$ and $n(S)=10$, then $P(E)=\dfrac{3}{10}=0.3$.