Chapter 5: Q R5.4. (page 334)

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

Expert verified

Part (a) Probability distribution is $D P r o b a b i l i t y - 3 \frac{1}{3} 1 \frac{1}{2} 5 \frac{1}{6}$

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

$P r o b a b i l i t y = \frac{F a v o r a b l e o u t c o m e s}{T o t a l o u t c o m e s}$

Part (a) Step 3. Calculation

The difference can be determined using the following formula:

2 - 5 = - 3 2 - 1 = 6 - 5 = 1 6 - 1 = 5

The probabilities could be estimated in the following way:

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}

As a result, the probability distribution is as follows:

$D P r o b a b i l i t y - 3 \frac{1}{3} 1 \frac{1}{2} 5 \frac{1}{6}$

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.

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Short Answer

Step by step solution

Part (a) Step 1. Given Information

Part (a) Step 2. Concept

Part (a) Step 3. Calculation

Part (b) Step 3. Explanation

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