Answer
$\frac{1}{12}$.
Work Step by Step
$n(S)=4\cdot4\cdot3=48$, because the first spinner can have $4$ outcomes (but we spin it twice) and the second spinner can have $3$ outcomes.
$n(E)=2$, because either a $2$ followed by a $2$ and a red or a $2$ followed by a $4$ and a red or a $2$ followed by a $2$ and a green or a $2$ followed by a $4$ and a green are the required outcomes.
We know that $P(E)=\frac{n(E)}{n(S)}$, hence $P=\frac{4}{48}=\frac{1}{12}$.
