Question

Cast a die two independent times and let \(X\) equal the absolute value of the difference of the two resulting values (the numbers on the up sides). Find the pmf of \(X .\) Hint: It is not necessary to find a formula for the pmf.

Short answer

Hence, the pmf of \(X\) is \(P(X=0) = 1/6\), \(P(X=1) = 5/18\), \(P(X=2) = 2/9\), \(P(X=3) = 1/6\), \( P(X=4) = 1/9\), \(P(X=5) = 1/18\).

Step-by-step solution

Identify the possible outcomes of the roll

The two dice are rolled independently, which can result in \(6 \times 6 = 36\) possible outcomes. The absolute difference between the rolled numbers could be in the range from 0 to 5. The task is to find the count of situations where the absolute difference between rolls is each of these values.

Count the occurrences

Now the task is to compute how often each absolute difference occurs in the 36 possible outcomes. \n\nFor absolute difference 0, we have outcomes (1,1), (2,2), (3,3), (4,4), (5,5), (6,6); that is a total of 6 outcomes. \n\nFor absolute difference 1, we have outcomes (1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4), (5,6), (6,5); a total of 10 outcomes. \n\nContinuing similarly, absolute difference 2 has 8 outcomes, 3 has 6 outcomes, 4 has 4 outcomes, and 5 has 2 outcomes. So, we have a total of 36 outcomes.

Calculate the probabilities

Since the two dice are fair, each of the 36 outcomes has an equal probability of occurrence, which is \(1/36\). The probability of each absolute difference can, therefore, be computed as count of that difference divide by total outcomes. These are as follows: \n\n\(P(X=0) = 6/36 = 1/6 \n P(X=1) = 10/36 = 5/18 \n P(X=2) = 8/36 = 2/9 \n P(X=3) = 6/36 = 1/6 \n P(X=4) = 4/36 = 1/9 \n P(X=5) = 2/36 = 1/18\).

Key concepts

Random Variables

When we talk about random variables, we are discussing a fundamental idea in probability and statistics. A random variable is essentially a numerical outcome of a random process. In the context of our exercise, we're looking at a scenario where we roll a die twice, which is a random experiment. The result of these rolls can be captured by a random variable.

• Random variables can be discrete or continuous, depending on the values they can take. In our example, the random variable is discrete.
• For the problem we have, the random variable, denoted by \(X\), represents the absolute difference between two independent die rolls.
• Understanding random variables helps in assigning probabilities to different possible outcomes of these random processes.

Grasping the concept of a random variable is crucial for tackling problems that involve predicting the likelihood of various outcomes based on random experiments.

Discrete Probability

In probability theory, the discrete probability concept is key when dealing with scenarios where possible outcomes are countable. The exercise with the die is a good example. Here, our interest lies in finding the probability of obtaining different values for \(X\), the absolute difference between two die rolls.

• The possible outcomes in a single die roll are finite and known: numbers 1 to 6. When we roll two dice, this results in a total of 36 possible combinations for the outcome of \(X\).
• Each possible outcome has a probability since we're dealing with fair dice. Hence, the probability of each specific \((x_1, x_2)\) pair occurring is \(\frac{1}{36}\).
• By counting how often each absolute difference occurs among these 36 pairs, we can determine the probability for each value of \(X\).

This approach is characteristic of discrete probability, where we can associate a probability mass function (PMF) with the discrete random variable, \(X\). The PMF outputs probabilities directly related to the random variable's possible values.

Independent Events

Understanding the concept of independent events is crucial when analyzing scenarios involving multiple random processes. In our exercise, we have two dice rolls that are independent of each other.

• Independence implies that the outcome of the first die roll does not affect the outcome of the second die roll.
• This is an essential fact because it allows us to calculate the total number of outcomes as the product of outcomes of individual events (i.e., \(6 \times 6 = 36\)).
• When dealing with independent events in probability, the probabilities of combined outcomes can be determined by multiplying the probabilities of each independent event.

For our exercise, the independence of the die rolls justifies why each pair of outcomes is equally likely, simplifying the calculation of the PMF for \(X\). This foundational concept makes it easier to apply probability rules faithfully and accurately.