Question

In Problems 17–22, use the following spinners to construct a probability model for each experiment.

Spin spinner II, then I, then III. What is the probability of getting Yellow, followed by a 2 or a 4, followed by Forward?

Short answer

The possible outcome is

$S=\{\mathrm{Y}1\mathrm{B},\mathrm{Y}1\mathrm{F},\mathrm{Y}2\mathrm{B},\mathrm{Y}2\mathrm{F},\mathrm{Y}3\mathrm{B},\mathrm{Y}3\mathrm{F},\mathrm{Y}4\mathrm{B},\mathrm{Y}4\mathrm{F},\mathrm{R}1\mathrm{B},\mathrm{R}1\mathrm{F},\mathrm{R}2\mathrm{B},\mathrm{R}2\mathrm{F},\mathrm{R}3\mathrm{B},\mathrm{R}3\mathrm{F},\mathrm{R}4\mathrm{B},\mathrm{R}4\mathrm{F},\mathrm{G}1\mathrm{B},\mathrm{G}1\mathrm{F},\mathrm{G}2\mathrm{B},\mathrm{G}2\mathrm{F},\mathrm{G}3\mathrm{B},\mathrm{G}3\mathrm{F},\mathrm{G}4\mathrm{B},\mathrm{G}4\mathrm{F}\}$

The probability is$\frac{1}{12}$.

Step-by-step solution

Step 1. Given information.

Spin spinner II, then I, then III. What is the probability of getting Yellow, followed by a 2 or a 4, followed by Forward?

PY2F+PY4FStep 2. Construct the probability model.  

Letting R stands for Red , Y for Yellow , G for Green , B for Backward and F for Forward , we are asked to look for

$P\left(\mathrm{Y}2\mathrm{F}\right)+P\left(\mathrm{Y}4\mathrm{F}\right)$

We added them since they are mutually exclusive events. Listing the possible outcomes from Spinner I, II then III, we have

$S=\{\mathrm{Y}1\mathrm{B},\mathrm{Y}1\mathrm{F},\mathrm{Y}2\mathrm{B},\mathrm{Y}2\mathrm{F},\mathrm{Y}3\mathrm{B},\mathrm{Y}3\mathrm{F},\mathrm{Y}4\mathrm{B},\mathrm{Y}4\mathrm{F},\mathrm{R}1\mathrm{B},\mathrm{R}1\mathrm{F},\mathrm{R}2\mathrm{B},\mathrm{R}2\mathrm{F},\mathrm{R}3\mathrm{B},\mathrm{R}3\mathrm{F},\mathrm{R}4\mathrm{B},\mathrm{R}4\mathrm{F},\mathrm{G}1\mathrm{B},\mathrm{G}1\mathrm{F},\mathrm{G}2\mathrm{B},\mathrm{G}2\mathrm{F},\mathrm{G}3\mathrm{B},\mathrm{G}3\mathrm{F},\mathrm{G}4\mathrm{B},\mathrm{G}4\mathrm{F}\}$

From this, we were able to get the following

$$\begin{gathered} P\left(Y2F\right)=\frac{1}{24} \\ P\left(Y4F\right)=\frac{1}{24} \end{gathered}$$

Since they just appeared once in the sample space with a total of 24 possible outcomes .

Thus,$P\left(Y2F\right)+P\left(Y4F\right)=\frac{1}{24}+\frac{1}{24}=\frac{2}{24}=\frac{1}{12}$.

Step 3. Conclusion. 

The possible outcome is

$S=\{\mathrm{Y}1\mathrm{B},\mathrm{Y}1\mathrm{F},\mathrm{Y}2\mathrm{B},\mathrm{Y}2\mathrm{F},\mathrm{Y}3\mathrm{B},\mathrm{Y}3\mathrm{F},\mathrm{Y}4\mathrm{B},\mathrm{Y}4\mathrm{F},\mathrm{R}1\mathrm{B},\mathrm{R}1\mathrm{F},\mathrm{R}2\mathrm{B},\mathrm{R}2\mathrm{F},\mathrm{R}3\mathrm{B},\mathrm{R}3\mathrm{F},\mathrm{R}4\mathrm{B},\mathrm{R}4\mathrm{F},\mathrm{G}1\mathrm{B},\mathrm{G}1\mathrm{F},\mathrm{G}2\mathrm{B},\mathrm{G}2\mathrm{F},\mathrm{G}3\mathrm{B},\mathrm{G}3\mathrm{F},\mathrm{G}4\mathrm{B},\mathrm{G}4\mathrm{F}\}$
The probability is$\frac{1}{12}$