Answer
$\frac{1}{8}$.
Work Step by Step
$n(S)=2\cdot4\cdot4=32$, because the third spinner can have $2$ outcomes and the first spinner can have $4$ outcomes (but we spin it twice) .
$n(E)=4$, because either a forward followed by a $1$ and $2$ or a forward followed by a $1$ and a $4$ or a forward followed by a $3$ and a $2$ or a forward followed by a $3$ and a $4$ are the required outcomes.
We know that $P(E)=\frac{n(E)}{n(S)}$, hence $P=\frac{4}{32}=\frac{1}{8}$.
