Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

The probability is $\dfrac{1}{12}$. Refer to the step-by-step part bellow for the model.
The experiment involves spinning $\text{Spinner I}$ followed by $\text{Spinner II}$ then finally by $\text{ Spinner III}$. Thus, the sample space $S$ is: $\left\{\text{1 Yellow Forward }, \text{1 Yellow Backward}, \text{1 Green Forward }, \text{1 Green Backward}, \text{1 Red Forward }, \text{1 Red Backward}, \text{2 Yellow Forward }, \text{2 Yellow Backward}, \text{2 Green Forward }, \text{2 Green Backward}, \text{2 Red Forward }, \text{2 Red Backward}, \text{3 Yellow Forward }, \text{3 Yellow Backward}, \text{3 Green Forward }, \text{3 Green Backward}, \text{3 Red Forward }, \text{3 Red Backward}, \text{4 Yellow Forward }, \text{4 Yellow Backward}, \text{4 Green Forward }, \text{4 Green Backward}, \text{4 Red Forward }, \text{4 Red Backward}, \right\}$ Note the sample space has $24$ equally-likely outcomes so $n(S)=24$. Let $E_1$ be the event that the outcome is $1$ followed by a Red. then by Backward. Then, from the sample space, we have $P(E_1)=\dfrac{1}{24}$. Let $E_2$ be the event that the outcome is $1$ followed by a Green, then by a Backward. Then, $P(E_2)=\dfrac{1}{24}$. Let $E$ = event that a $1$ comes out followed by a Red or a Green, then by a Backward. Then, $P(E)=P(E_1)+P(E_2)=\dfrac{1}{24}+\dfrac{1}{24}=\dfrac{2}{24}=\dfrac{1}{12}$