Problem 17 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Two dice are rolled. Let the random variable \(X\) denote the number that falls uppermost on the first die, and let \(Y\) denote the number that falls uppermost on the second die. a. Find the probability distributions of \(X\) and \(Y\). b. Find the probability distribution of \(X+Y\).

Step-by-Step Solution

Answer

The probability distributions of \(X\) and \(Y\) are P(X=k) = P(Y=k) = \(\frac{1}{6}\) for k = 1, 2, 3, 4, 5, 6. The probability distribution of the sum \(Z = X + Y\) is: P(Z=k)= \[\frac{1}{36}\] for k = 2, 12;\[\frac{2}{36}\] for k = 3, 11;\[\frac{3}{36}\] for k=4, 10;\[\frac{4}{36}\] for k=5, 9;\[\frac{5}{36}\] for k=6, 8;\[\frac{6}{36}\] for k=7.

Step 1: Probability distributions of X and Y

As the two dice are fair and identical, the probability distributions of \(X\) and \(Y\) will be the same. There are 6 possible outcomes for each die (1, 2, 3, 4, 5, and 6), each with an equal probability of occurrence. So, the probability of each outcome is \(\frac{1}{6}\). This gives us the probability distribution for both X and Y: P(X=k) = P(Y=k) = \(\frac{1}{6}\) for k = 1, 2, 3, 4, 5, 6

Step 2: Probability distribution of X+Y

Now, let's find the probability distribution of the sum \(Z = X + Y\). Since X and Y can each take values from 1 to 6, the minimum value of Z is 2 (when X = 1 and Y = 1) and the maximum value of Z is 12 (when X = 6 and Y = 6). To find the probability of each possible value of Z, we need to count the number of ways to get that sum and divide it by the total number of possible outcomes (which is 36 since there are 6 possible outcomes for each die). We can use a table to enumerate the possibilities: | X=1 | X=2 | X=3 | X=4 | X=5 | X=6 ---|-----|-----|-----|-----|-----|---- Y=1| 2 | 3 | 4 | 5 | 6 | 7 Y=2| 3 | 4 | 5 | 6 | 7 | 8 Y=3| 4 | 5 | 6 | 7 | 8 | 9 Y=4| 5 | 6 | 7 | 8 | 9 |10 Y=5| 6 | 7 | 8 | 9 |10 |11 Y=6| 7 | 8 | 9 |10 |11 |12 In the table, the sum Z is obtained by adding the value in row Y and the value in column X. By counting the occurrences of each possible value of Z, we find the following probabilities: P(Z=2) = 1/36 P(Z=3) = 2/36 P(Z=4) = 3/36 P(Z=5) = 4/36 P(Z=6) = 5/36 P(Z=7) = 6/36 P(Z=8) = 5/36 P(Z=9) = 4/36 P(Z=10) = 3/36 P(Z=11) = 2/36 P(Z=12) = 1/36 So the probability distribution of the sum Z = X + Y is as follows: P(Z=k)= \[\frac{1}{36}\] for k = 2, 12;\[\frac{2}{36}\] for k = 3, 11;\[\frac{3}{36}\] for k=4, 10;\[\frac{4}{36}\] for k=5, 9;\[\frac{5}{36}\] for k=6, 8;\[\frac{6}{36}\] for k=7.

Key Concepts

Discrete Random Variables

Discrete random variables are foundational in the study of probability, often representing countable outcomes of an experiment or process. For instance, when rolling a fair die, the outcome can only be one of the six integers: 1, 2, 3, 4, 5, or 6, with no possibility of a fraction or decimal value. This perfectly encapsulates what a discrete random variable is—a variable that takes on a finite or countably infinite number of distinct values.

In the provided exercise, the variable \(X\) representing the outcome of the first die, and the variable \(Y\) representing the outcome of the second die are both discrete random variables. They can only assume the values {1, 2, 3, 4, 5, 6}, making their behavior easy to predict and map out using a probability distribution. Each outcome has an equal chance, representing a uniform probability distribution that is crucial when calculating the likelihood of individual outcomes.

To aid clarity for students, pictorial representations such as tables or bar graphs can visually communicate the probability distribution of discrete random variables. Such visual tools support comprehension, especially when digesting the abstract concepts of probability theory.

Sum of Independent Random Variables

When we consider the sum of two independent random variables, we explore a fundamental concept in probability theory that reveals the combined behavior of multiple stochastic processes. In this case, the roll of one die does not influence the outcome of the other—hence, \(X\) and \(Y\) are independent.

Calculating the probability distribution of the sum, \(Z = X + Y\), involves a bit more complexity. You can imagine each roll as a separate event and that the eventual sum, ranging from 2 to 12, has various ways of occurring. The independence of \(X\) and \(Y\) is crucial here because it allows for the simplification of the problem. If the variables were dependent, the addition of their outcomes would require a different approach.

The visualization of this addition is commonly undertaken with a two-dimensional grid (as used in the steps provided), where rows and columns represent possible outcomes for \(X\) and \(Y\) respectively. This visual approach helps students to enumerate all possible sums and their probabilities systematically. Moreover, when it comes to analyzing the sum of random variables, it's helpful to reiterate to students that the resulting distribution can—and often will—be different from the distributions of the individual variables.

Uniform Probability

Uniform probability refers to the idea that every event in a sample space has an equal chance of occurring. In the context of rolling a fair six-sided die, this means each number (1 through 6) has an identical probability of landing face up, which is \(\frac{1}{6}\). This uniformity is an assumption that underpins many probability problems and is easily understood by considering physical symmetries or the design of 'fair' games.

The practical implication of a uniform probability distribution is that it simplifies calculations, as no single outcome is more or less likely than any other. This forms the basis of many textbook probability exercises, like the dice roll scenario presented in the exercise.

To deepen students' understanding, it is beneficial to contrast uniform probability with non-uniform distributions, where some outcomes are more likely than others. It's also productive to discuss real-world scenarios where uniform probability applies and where it does not. Such discussions can promote critical thinking and better grasp the nuances of probability theory.