Answer
$\frac{1}{6}$.
Work Step by Step
$n(S)=4\cdot3=12$, because the first spinner can have $4$ outcomes and the second spinner can have $3$ outcomes.
$n(E)=2$, because either a $2$ followed by a red or a $4$ followed by a red are the required outcomes.
We know that $P(E)=\frac{n(E)}{n(S)}$, hence $P=\frac{2}{12}=\frac{1}{6}$.
