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Two fair dice are rolled. Find the joint probability mass function of \(X\) and \(Y\) when (a) \(X\) is the largest value obtained on any die and \(Y\) is the sum of the values; (b) \(X\) is the value on the first die and \(Y\) is the larger of the two values; (c) \(X\) is the smallest and \(Y\) is the largest value obtained on the dice.

Short Answer

Expert verified
(a) Joint pmf for $X$ (largest value) and $Y$ (sum) is: \[P_{XY}(x,y) = \frac{\text{number of combinations with }(x,y)}{36}\] (b) Joint pmf for $X$ (value on first die) and $Y$ (larger value) is: \[P_{XY}(x,y) = \frac{\text{number of combinations with }(x,y)}{36}\] (c) Joint pmf for $X$ (smallest value) and $Y$ (largest value) is: \[P_{XY}(x,y) = \frac{\text{number of combinations with }(x,y)}{36}\]

Step by step solution

01

List all possible outcomes

List all possible outcomes when rolling two dice, as ordered pairs. There are a total of 36 combinations, i.e. (1,1), (1,2), (1,3), ... ,(6,6).
02

Identify X and Y in each outcome

For each combination, determine the largest value obtained (X) and the sum of the values (Y). For example, for the combination (2, 4), X = 4 and Y = 2+4 = 6.
03

Compute Joint Probabilities

Count how many times each pair (X,Y) occurs in the 36 outcomes. Then, divide the counts by 36 to find the probability for each pair. This gives you the joint probability mass function for case (a). #Case (b)#
04

Use all possible outcomes

Since we already have the list of all possible outcomes from Case (a), use the same list for this case.
05

Identify X and Y in each outcome

For each combination, determine the value of the first die (X) and the larger of the two values (Y). For example, for the combination (2, 4), X = 2 and Y = 4.
06

Compute Joint Probabilities

Count how many times each pair (X,Y) occurs in the 36 outcomes. Then, divide the counts by 36 to find the probability for each pair. This gives you the joint probability mass function for case (b). #Case (c)#
07

Use all possible outcomes

Since we already have the list of all possible outcomes from Case (a), use the same list for this case.
08

Identify X and Y in each outcome

For each combination, determine the smallest value (X) and the largest value (Y). This is different from Case (a) as the values of X & Y are symmetrical here. For example, for the combination (2, 4), X = 2 and Y = 4.
09

Compute Joint Probabilities

Count how many times each pair (X,Y) occurs in the 36 outcomes. Then, divide the counts by 36 to find the probability for each pair. This gives you the joint probability mass function for case (c).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Theory
Probability theory is a branch of mathematics that deals with the likelihood of events occurring. It provides us with a systematic way to make predictions and inform decisions under uncertainty. In the context of rolling two dice, probability theory comes into play when we try to determine the likelihood of various outcomes such as the roll sums or comparing the values on the dice. The foundation of probability begins with the idea that the likelihood of a single event can be calculated by taking the number of favorable outcomes and dividing it by the total number of possible outcomes.

When we consider two dice, the outcomes are represented as pairs, reflecting the results on each die. For instance, (3,5) would mean the first die showed a 3 and the second a 5. As we can have various ways to look at the dice, for example focusing on the largest value or the sum, the joint probability mass function (PMF) comes into play. It represents the probability of each pair occurring, considering the specific conditions or rules we are interested in, such as 'the sum of the dice' or 'the largest value rolled'. We calculate the joint PMF by counting how many times each outcome meets our condition within all possible combinations. This count is then divided by the total number of outcomes, which is 36 for two six-sided dice.

The joint PMF is particularly valuable because it allows us to analyze the relationship between two discrete random variables, like the highest number rolled and the sum of both dice, and is fundamental in disciplines ranging from statistics to decision sciences. Understanding how to compute and interpret a joint PMF is essential for anyone studying probability theory.
Combinatorial Analysis
Combinatorial analysis is a field of mathematics that focuses on counting, arranging, and grouping items without actually having to enumerate every possibility. This technique is particularly helpful in probability theory, especially when dealing with discrete random variables like rolling dice, where each outcome must be accounted for. In the dice example, we use combinatorial analysis to understand that there are 36 unique outcomes when rolling two six-sided dice.

By listing all the possible outcomes, combinatorial methods guide us in creating a systematic approach to manage all the pairs of dice rolls and then help in recognizing the different events for which we want to calculate probabilities. This serves as a base for more advanced analysis, such as calculating joint probability mass function (PMF). Moreover, combinatorial analysis is not just about counting but also about making sure that the counts represent meaningful combinations that cater to the rules of our probability experiment, such as identifying which die shows the largest number, or determining the sum of both dice.

Understanding combinatorial principles is crucial for interpreting results correctly and ensuring that all possible combinations have been considered in our probability calculations. In cases where the counting becomes complex, combinatorial formulas like the permutation and combination equations come into play to simplify the work. However, in the case of rolling two dice, the manual listing is still manageable and serves as a clear example of how combinatorial analysis supports probability theory.
Discrete Random Variables
Discrete random variables are a staple of probability theory, where they represent variables that can take on a countable number of distinct values. Taking the dice-rolling scenario as an example, the numbers that appear on the dice after a roll are discrete random variables because they can only be whole numbers from 1 to 6. Each of these numbers has a calculable probability associated with it, representing the likelihood of the dice landing on that particular number.

When we talk about joint probability mass functions, we are actually referring to the probabilities associated with pairs of discrete random variables, such as the largest dice number (X) and their sum (Y). These pairs have distinct probabilities which we calculate separately for each outcome. For example, in case (a), X and Y are dependent on each other because X influences the sum that Y can take, while in case (b) and case (c), the relationships between X and Y change according to the new definitions.

It's essential to understand that discrete random variables are often dependent on the context of the scenario and its constraints. For instance, when the scenario changes, as with the different cases (a), (b), and (c), the interpretation and values of X and Y change correspondingly. A solid grasp of discrete random variables is not only important for solving textbook problems but also for real-world applications where decision-making often relies on the analysis of discrete outcomes.

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Most popular questions from this chapter

A rectangular array of \(m n\) numbers arranged in \(n\) rows, each consisting of \(m\) columns, is said to contain a saddlepoint if there is a number that is both the minimum of its row and the maximum of its column. For instance, in the array $$ \begin{array}{rr} 1 & 3 & 2 \\ 0 & -2 & 6 \\ .5 & 12 & 3 \end{array} $$ the number 1 in the first row, first column is a saddlepoint. The existence of a saddlepoint is of significance in the theory of games. Consider a rectangular array of numbers as described above and suppose that there are two individuals \(-A\) and \(B\) - that are playing the following game: \(A\) is to choose one of the numbers \(1,2, \ldots, n\) and \(B\) one of the numbers \(1,2, \ldots, m\). These choices are announced simultaneously, and if \(A\) chose \(i\) and \(B\) chose \(j\), then \(A\) wins from \(B\) the amount specified by the number in the ith row, \(j\) th column of the array. Now suppose that the array contains a saddlepoint-say the number in the row \(r\) and column \(k-\) call this number \(x_{r k}\). Now if player \(A\) chooses row \(r\), then that player can guarantee herself a win at least \(x_{r k}\) (since \(x_{r k}\) is the minimum number in the row \(r\) ). On the other hand, if player \(B\) chooses column \(k\), then he can guarantee that he will lose no more than \(x_{r k}\) (since \(x_{r k}\). is the maximum number in the column \(k\) ). Hence, as \(A\) has a way of playing that guarantees her a win of \(x_{r k}\) and as \(B\) has a way of playing that guarantees he will lose no more than \(x_{r k}\), it seems reasonable to take these two strategies as being optimal and declare that the value of the game to player \(A\) is \(x_{r k}\). If the \(n m\) numbers in the rectangular array described above are independently chosen from an arbitrary continuous distribution, what is the probability that the resulting array will contain a saddlepoint?

Suppose that \(X, Y\), and \(Z\) are independent random variables that are each equally likely to be either 1 - or 2 . Find the probability mass function of (a) \(X Y Z\), (b) \(X Y+X Z+Y Z\), and (c) \(X^{2}+Y Z\).

If \(X\) and \(Y\) are independent standard normal random variables, determine the joint density function of $$ U=X \quad V=\frac{X}{Y} $$ Then use your result to show that \(X / Y\) has a Cauchy distribution.

An ambulance travels back and forth, at a constant speed, along a road of length \(L\). At a certain moment of time an accident occurs at a point uniformly distributed on the road. [That is, its distance from one of the fixed ends of the road is uniformly distributed over \((0, L)\).] Assuming that the ambulance's location at the moment of the accident is also uniformly distributed, compute, assuming independence, the distribution of its distance from the accident.

The joint probability density function of \(X\) and \(Y\) is given by $$ f(x, y)=\frac{6}{7}\left(x^{2}+\frac{x y}{2}\right) \quad 0Y\\}\). Chapter 6 Jointly Distributed Random Variables (d) Find \(P\left\\{Y>\frac{1}{2} \mid X<\frac{1}{2}\right\\}\). (e) Find \(E[X]\).

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