Problem 168
Question
Two fair dice are tossed. Let \(X\) denote the number appearing on the first die and \(Y\) the number on the second. Show that \(X\) and \(Y\) are independent.
Step-by-Step Solution
Verified Answer
Yes, \(X\) and \(Y\) are independent because their joint probability distribution equals the product of their individualprobability distributions.
1Step 1: Define individual probabilities
Each die has 6 outcomes, and each outcome is equally likely. Therefore, the probability of any given outcome on one die is \( \frac{1}{6} \). Hence, \( P(X = x) = P(Y = y) = \frac{1}{6} \) for \( x, y \) in {1, 2, 3, 4, 5, 6}.
2Step 2: Define joint probability
The joint probability of two events \(X\) and \(Y\) is defined as \( P(X = x, Y = y) \). Since we have two dice and each die has 6 outcomes, there are a total of \(6 * 6 = 36\) possible outcomes. Therefore, the joint probability \( P(X = x, Y = y) = \frac{1}{36} \) for \( x, y \) in {1, 2, 3, 4, 5, 6}.
3Step 3: Prove Independence
We need to show that \( P(X = x, Y = y) = P(X = x) * P(Y = y) \). Substituting the known probabilities, we get \( \frac{1}{36} = \frac{1}{6} * \frac{1}{6} \), which simplifies to \( \frac{1}{36} = \frac{1}{36} \). Thus, we can conclude that \(X\) and \(Y\) are independent.
Key Concepts
Independent EventsJoint ProbabilityFair DiceRandom Variables
Independent Events
In probability theory, independent events are events whose outcomes do not affect each other. Imagine two fair dice being thrown. The result of one die does not change or influence the outcome of the other die. This makes them independent.
For two events to be independent, the probability of both occurring together must equal the product of their individual probabilities. Mathematically, two events \( A \) and \( B \) are independent if:
For two events to be independent, the probability of both occurring together must equal the product of their individual probabilities. Mathematically, two events \( A \) and \( B \) are independent if:
- \( P(A \cap B) = P(A) \times P(B) \)
Joint Probability
Joint probability refers to the probability of two events occurring at the same time. In our exercise, it's the probability that the first die rolls a certain number \( X \), and the second die rolls a certain number \( Y \).
When dealing with dice, each outcome is equally likely. So the joint probability of any two specific outcomes is determined by the product of individual probabilities. For two dice, both having 6 outcomes, the joint probability is given by:
When dealing with dice, each outcome is equally likely. So the joint probability of any two specific outcomes is determined by the product of individual probabilities. For two dice, both having 6 outcomes, the joint probability is given by:
- \( P(X = x, Y = y) = \frac{1}{36} \)
Fair Dice
A fair die is a theoretical dice that has an equal chance of landing on any of its faces. This means each side (or outcome) is equally probable. For a standard six-sided die, each face has a probability of:
In scenarios involving fair dice, like our exercise, knowing this fact allows us to determine the probabilities of combined events, given that each possible result is equally likely.
- \( \frac{1}{6} \)
In scenarios involving fair dice, like our exercise, knowing this fact allows us to determine the probabilities of combined events, given that each possible result is equally likely.
Random Variables
In probability theory, a random variable represents a numerical outcome of a random event. It is a variable whose possible values are numerical outcomes of a random phenomenon. In our exercise, \( X \) and \( Y \) are random variables representing the numbers on two dice.
Random variables can be discrete, taking on a finite set of values (like a die roll), or continuous, taking on a range of values. In the case of dice:
Random variables can be discrete, taking on a finite set of values (like a die roll), or continuous, taking on a range of values. In the case of dice:
- \( X \) and \( Y \) are discrete random variables.
- Each can take on any integer value from 1 to 6.
Other exercises in this chapter
Problem 166
Suppose that the random variables \(X, Y\), and \(Z\) have the multivariate pdf $$ f_{X, Y, Z}(x, y, z)=(x+y) e^{-z} $$ for \(00\). Find (a) \(f_{X, Y}(x, y)\),
View solution Problem 167
The four random variables \(W, X, Y\), and \(Z\) have the multivariate pdf $$ f_{W, X, Y, Z}(w, x, y, z)=16 w x y z $$ for \(0 \leq w \leq 1,0 \leq x \leq 1,0 \
View solution Problem 169
Let \(f_{X, Y}(x, y)=\lambda^{2} e^{-\lambda(x+y)}, 0 \leq x, 0 \leq y .\) Show that \(X\) and \(Y\) are independent. What are the marginal pdfs in this case?
View solution Problem 170
Suppose that each of two urns has four chips, numbered 1 through 4 . A chip is drawn from the first urn and bears the number \(X\). That chip is added to the se
View solution