Question

Cast a fair die and let $X=0$ if 1,2, or 3 spots appear, let $X=1$ if 4 or 5 spots appear, and let $X=2$ if 6 spots appear. Do this two independent times, obtaining $X_1$ and $X_2.$ Calculate $P(|X_1-X_2|=1)$.

Short answer

The absolute difference between $X_1$ and $X_2$ equals one with a probability of $\frac{2}{3}$.

Step-by-step solution

Identify the Possibilities

The absolute difference between $X_1-X_2$ can be only 1 when (1) $X_1=0$ and $X_2=1$, or (2) $X_1=1$ and $X_2=0$, or (3) $X_1=1$ and $X_2=2$, or (4) $X_1=2$ and $X_2=1$.

Calculate Each Probability

Calculate the probability for each of these possibilities. The possibilities have the same probability because the throws are independent and symmetric. For example, for possibility (1), the probability $P(X_1=0)$ is $\frac{1}{2}$ because 1,2, or 3 spots can appear, and the probability $P(X_2=1)$ is $\frac{1}{3}$ because 4 or 5 spots can appear. Therefore, $P(|X_1-X_2|=1)=P(X_1=0,X_2=1)=\text{(}\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$. Perform similar calculations for all other possibilities.

Sum All Probabilities

Finally, sum up the probabilities of all four possibilities. Let's denote the probability $P(|X_1-X_2|=1)$ as $P_{abs}$. Then, $P_{abs}=4\times \frac{1}{6}=\frac{2}{3}$. This is because these are mutually exclusive events, and hence the probability of their union is equal to the sum of their individual probabilities.

Key concepts

Independent Events

When we talk about independent events in probability theory, we refer to two or more events where the occurrence of one does not affect the occurrence of another. In this exercise, the rolls of the die represent independent events. This is because the outcome of the first roll does not change the probabilities of the outcomes of the second roll. Each roll is a separate event with its own outcome possibilities.

In this particular problem, when we roll a die twice, getting a specific number in the first roll has no bearing on what we might get in the second roll. Thus, each roll is independent. This independence allows us to calculate the probability of compound events (like the combined result of two die rolls) by multiplying their individual probabilities. Remember, for events to be truly independent, the probability of their intersection is the product of their probabilities, i.e., $$P(A\text{ and }B)=P(A)\times P(B)$$ where each event is independent of the other.

Random Variables

Random variables are a fundamental concept in probability and statistics. They are variables that can take on different values based on the result of a random event. In simpler terms, a random variable is a way to map outcomes of a random process to numerical values.

In this exercise, the random variable $X$ is assigned a value based on the result of a die roll. If the die lands on 1, 2, or 3, $X$ is 0; if it lands on 4 or 5, $X$ is 1; and if it lands on 6, $X$ is 2. This mapping helps in analyzing and interpreting the results of the die rolls mathematically.

• For $X=0$, the outcomes are \{1, 2, 3\}
• For $X=1$, the outcomes are \{4, 5\}
• For $X=2$, the outcome is \{6\}

Each roll of the die produces a random variable value $X$, and since the die is fair, the probability of each set of outcomes can be calculated easily.

Absolute Difference

The concept of absolute difference is a way to measure how different two numbers are without considering their order. In mathematical terms, the absolute difference between two values $a$ and $b$ is $$|a-b|$$.

For this exercise, we need to find the probability that the absolute difference between the two random variables $X_1$ and $X_2$ is 1: $$|X_1-X_2|=1$$.

This means that $X_1$ and $X_2$ differ by exactly one unit. The possible scenarios are:

• $X_1=0$ and $X_2=1$
• $X_1=1$ and $X_2=0$
• $X_1=1$ and $X_2=2$
• $X_1=2$ and $X_2=1$

Each scenario leads to an absolute difference of 1. By considering the probability of each of these events—$\frac{1}{6}$ per scenario—and summing them up (since they are mutually exclusive), we achieve the overall probability $\frac{2}{3}$ that the absolute difference is indeed one.