Solution
= Solution
An excess of thirty seconds is $T>1/2$ in minute units. Therefore
$$
\boxed{\mathbb P(T>1/2)=\int_{1/2}^1(1-t)\,dt=\frac18}.
$$
This is the shaded triangular <area> under the <probability density function>.
= Solution
An excess of thirty seconds is $T>1/2$ in minute units. Therefore
$$
\boxed{\mathbb P(T>1/2)=\int_{1/2}^1(1-t)\,dt=\frac18}.
$$
This is the shaded triangular <area> under the <probability density function>.