Question

Weird dice Nonstandard dice can produce interesting distributions of outcomes. Suppose you have two balanced, six-sided dice. Die A has faces with $2,2,2,2,6$, and$6$ spots. Die B has three faces with $5$ spots and three faces with 1 spot. Imagine that you roll both dice at the same time.

(a) Find a probability model for the difference (Die A − Die B) in the total number of spots on the up-faces.

(b) Which die is more likely to roll a higher number? Justify your answer.

• Use basic probability rules, including the complement rule and the addition rule for mutually exclusive events.

Short answer

Part (a) Probability distribution is$$\begin{gathered} DProbability \\ -3\frac{1}{3} \\ 1\frac{1}{2} \\ 5\frac{1}{6} \end{gathered}$$

Part (b) Die A can take higher values in comparison of the die B.

Step-by-step solution

Part (a) Step 1. Given Information

Die A takes the values$2$ and $6$, whereas Die B takes the values $5$ and $1$

Part (a) Step 2. Concept

$Probability=\frac{Favorableoutcomes}{Totaloutcomes}$

Part (a) Step 3. Calculation

The difference can be determined using the following formula:

$$\begin{gathered} 2-5=-3 \\ 2-1=6-5=1 \\ 6-1=5 \end{gathered}$$

The probabilities could be estimated in the following way:

$$\begin{gathered} P(2,5)=\frac{4}{6}\times \frac{3}{6}=\frac{1}{3} \\ P(2,1)=\frac{4}{6}\times \frac{3}{6}=\frac{1}{3} \\ P(6,5)=\frac{2}{6}\times \frac{3}{6}=\frac{1}{6} \\ P(6,1)=\frac{2}{6}\times \frac{3}{6}=\frac{1}{6} \end{gathered}$$

As a result, the probability distribution is as follows:

$$\begin{gathered} DProbability \\ -3\frac{1}{3} \\ 1\frac{1}{2} \\ 5\frac{1}{6} \end{gathered}$$

Part (b) Step 3. Explanation

Die A could take the values $2$ and $6$based on the information supplied, whereas Die B may take the values $5$ and 1. As a result, it's possible to say that die A can take higher values than die B.