#### Answer

$\dfrac{1}{6}$

#### Work Step by Step

The experiment involves spinning $\text{Spinner 1}$ followed by $\text{Spinner 2}$.
Thus, the sample space $S$ is:
$S=\left\{1\text{Green}, 1\text{Yellow}, 1\text{Red}, 2\text{Green}, 2\text{Yellow}, 2\text{Red}, 3\text{Green}, 3\text{Yellow}, 3\text{Red},\\
\space \space \space \space \space \space \space \space \space 4\text{Green}, 4\text{Yellow}, 4\text{Red}\right\}$
Note that the sample space has $12$ equally-likely outcomes so $n(S)=12$.
Let $E_1$ be the event that the outcome is $2$ followed by red.
Then, $P(E_1)=\dfrac{1}{12}$.
Let $E_2$ be the event that the outcome is $4$ followed by red.
Then, $P(E_2)=\dfrac{1}{12}$.
Let $E$ = event that a $2$ or a$4$ comes out folllowed by a red.
Then,
$P(E)=P(E_1)+P(E_2)=\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{2}{12}=\dfrac{1}{6}$