Question

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

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

Short answer

The possible outcome is $S=\left\{1\mathrm{Yellow},1\mathrm{Red},1\mathrm{Green},2\mathrm{Yellow},2\mathrm{Red},2\mathrm{Green},3\mathrm{Yellow},3\mathrm{Red},3\mathrm{Green},4\mathrm{Yellow},4\mathrm{Red},4\mathrm{Green}\right\}$

The probability is$P\left(2\mathrm{Red}\right)+P\left(4\mathrm{Red}\right)=\frac{1}{6}$.

Step-by-step solution

Step 1. Given information.

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

Step 2. Construct the probability model.

The probability of getting 2 or 4 from spinner I followed by spinner II, that is

$P\left(2\mathrm{Red}\right)orP\left(4\mathrm{Red}\right)$

Thus the formula will be $P\left(2\mathrm{Red}\right)+P\left(4\mathrm{Red}\right)$

Since they are mutually exclusive events.

Listing the possible outcome from spinner I then II, we have

$S=\left\{1\mathrm{Yellow},1\mathrm{Red},1\mathrm{Green},2\mathrm{Yellow},2\mathrm{Red},2\mathrm{Green},3\mathrm{Yellow},3\mathrm{Red},3\mathrm{Green},4\mathrm{Yellow},4\mathrm{Red},4\mathrm{Green}\right\}$

Each outcome has probability of $\frac{1}{12}$.

So, role="math" localid="1647172300119" $$\begin{gathered} P\left(2\mathrm{Red}\right)=\frac{1}{12} \\ P\left(4\mathrm{Red}\right)=\frac{1}{12} \end{gathered}$$

Since they just appeared once in the sample space with a total of 12 possible outcomes.

Thus,$P\left(2\mathrm{Red}\right)+P\left(4\mathrm{Red}\right)=\frac{1}{12}+\frac{1}{12}=\frac{2}{12}=\frac{1}{6}$

Step 3. Conclusion.

The possible outcome is $S=\left\{1\mathrm{Yellow},1\mathrm{Red},1\mathrm{Green},2\mathrm{Yellow},2\mathrm{Red},2\mathrm{Green},3\mathrm{Yellow},3\mathrm{Red},3\mathrm{Green},4\mathrm{Yellow},4\mathrm{Red},4\mathrm{Green}\right\}$

The probability is$P\left(2\mathrm{Red}\right)+P\left(4\mathrm{Red}\right)=\frac{1}{6}$.