Chapter 8: Problem 8
Two dice are rolled. One is fair, but the other is loaded: It shows the face with six spots half the time and the remaining five faces with equal frequencies. (a) Describe the experiment in terms of a cross product sample space. (b) Define a probability density on the cross product space. (c) Verify by direct computation that the probability density found in part (b) is legitimate. (d) Does it matter in what order the dice are considered? Explain your answer.
Short Answer
Step by step solution
Understand the Problem
Define the Sample Space
Define Probability Density for Fair Die
Define Probability Density for Loaded Die
Defining Probability on Cross Product Space
Verify Probability Density
Determine Order Relevance
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cross Product Sample Space
For rolled dice, if we have two dice, each die has six faces, making their individual sample spaces: for the fair die, it's \( \{1, 2, 3, 4, 5, 6\} \) and likewise for the loaded die. To form the cross product sample space when rolling these dice, you create ordered pairs \((x, y)\), where \(x\) is an outcome from the fair die and \(y\) from the loaded die. This results in a total of 36 ordered pairs like \((1,1), (1,2), \,\ldots\,, (6,6) \).
This sample space gives a comprehensive view of all possible outcomes from the experiment where both dice are rolled, capturing the various combinations of results.
Probability Density
Let's start with the fair die. When you roll a fair die, each of the six outcomes \(1, 2, 3, 4, 5, \) and \(6\) has an equal chance of occurring, giving a probability density of \( \frac{1}{6} \) for each face.
Now, let's consider the loaded die. This die is biased, showing "6" half the time. Thus, it assigns a probability of \( \frac{1}{2} \) to the "6" face. The remaining five faces share the remaining probability equally, giving each a probability density of \( \frac{1}{10} \). In the cross product sample space, we determine the probability of any outcome \((x, y)\) by multiplying the independent probabilities: \( P((x,y)) = P(X = x) \times P(Y = y) \). This provides individual outcome likelihoods within the larger experiment context.
Loaded Die
This change significantly impacts the probabilities for the die. While a regular die would give each face a chance of \( \frac{1}{6} \), the loaded die's approach skews these chances. In this scenario, only the face featuring "6" has increased probability, specifically \( \frac{1}{2} \). The rest of the faces (\(1, 2, 3, 4, \) and \(5\)) share the leftover probability, each with \( \frac{1}{10} \), so that the total probability across all outcomes sums up to 1. Understanding this differentiation allows analysts to calculate expected outcomes accurately when dealing with non-standard dice.
Fair Die
This fairness translates directly into a straightforward probability density function for any fair die roll. With six faces and equal likelihood for each, the resulting uniform distribution is a basic starting point for calculating probabilities in dice-related experiments.
The fair die, in contrast to a loaded die, provides a benchmark for understanding the principle of equally likely outcomes. It is often used in theoretical scenarios to help students and analysts comprehend the foundational elements of random experiments and the behavior expected from perfectly unbiased, unaltered dice.