Answer :
Certainly! Let's go through each part of the question step by step.
### a) Joint probability mass function of X and Y
When you roll a pair of four-sided dice (one red and one black), each combination of outcomes (X, Y) is equally likely. Since each die has 4 faces, there are [tex]\(4 \times 4 = 16\)[/tex] possible outcomes.
The probability of any specific outcome such as (1, 1) or (3, 4) is given by:
[tex]\[ P(X = x, Y = y) = \frac{1}{16} \][/tex]
Thus, the joint probability mass function can be organized in a 4x4 matrix where each entry represents the probability of a particular pair (X, Y):
[tex]\[ P(X, Y) = \begin{array}{c|cccc} & Y=1 & Y=2 & Y=3 & Y=4 \\ \hline X=1 & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} \\ X=2 & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} \\ X=3 & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} \\ X=4 & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} \\ \end{array} \][/tex]
Or more concisely:
[tex]\[ \text{Joint PMF} = \begin{bmatrix} 0.0625 & 0.0625 & 0.0625 & 0.0625 \\ 0.0625 & 0.0625 & 0.0625 & 0.0625 \\ 0.0625 & 0.0625 & 0.0625 & 0.0625 \\ 0.0625 & 0.0625 & 0.0625 & 0.0625 \end{bmatrix} \][/tex]
### b) Marginal probability mass function of X
The marginal PMF of X, [tex]\(P(X)\)[/tex], is found by summing the joint PMF over all possible values of Y.
For each value of X (1 through 4):
[tex]\[ P(X=x_i) = \sum_{j=1}^4 P(X=x_i, Y=y_j) = \sum_{j=1}^4 \frac{1}{16} = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4} = 0.25 \][/tex]
So the marginal PMF of X is:
[tex]\[ P(X) = \begin{bmatrix} 0.25 & 0.25 & 0.25 & 0.25 \end{bmatrix} \][/tex]
### c) Marginal probability mass function of Y
Similarly, the marginal PMF of Y, [tex]\(P(Y)\)[/tex], is found by summing the joint PMF over all possible values of X.
For each value of Y (1 through 4):
[tex]\[ P(Y=y_j) = \sum_{i=1}^4 P(X=x_i, Y=y_j) = \sum_{i=1}^4 \frac{1}{16} = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4} = 0.25 \][/tex]
So the marginal PMF of Y is:
[tex]\[ P(Y) = \begin{bmatrix} 0.25 & 0.25 & 0.25 & 0.25 \end{bmatrix} \][/tex]
### d) Are X and Y independent?
Two variables X and Y are considered independent if the joint probability [tex]\(P(X=x, Y=y)\)[/tex] is equal to the product of their marginal probabilities [tex]\(P(X=x) \cdot P(Y=y)\)[/tex] for all x and y.
Given the joint PMF:
[tex]\[ P(X=x, Y=y) = \frac{1}{16} \][/tex]
And the marginal PMFs:
[tex]\[ P(X=x) = 0.25 \quad \text{and} \quad P(Y=y) = 0.25 \][/tex]
We calculate the product:
[tex]\[ P(X=x) \cdot P(Y=y) = 0.25 \times 0.25 = 0.0625 \][/tex]
Since [tex]\(P(X=x, Y=y) = P(X=x) \cdot P(Y=y)\)[/tex] holds for all combinations of x and y, X and Y are independent.
### e) Calculate [tex]\(E(X)\)[/tex], [tex]\(E(Y)\)[/tex], [tex]\(\operatorname{Var}(X)\)[/tex], [tex]\(\operatorname{Var}(Y)\)[/tex]
#### Expected values:
The expected value [tex]\(E(X)\)[/tex] is calculated as:
[tex]\[ E(X) = \sum_{x=1}^{4} x \cdot P(X=x) = 1 \cdot 0.25 + 2 \cdot 0.25 + 3 \cdot 0.25 + 4 \cdot 0.25 = 0.25 \cdot (1+2+3+4) = 0.25 \cdot 10 = 2.5 \][/tex]
Similarly, the expected value [tex]\(E(Y)\)[/tex] is:
[tex]\[ E(Y) = \sum_{y=1}^{4} y \cdot P(Y=y) = 0.25 \cdot (1+2+3+4) = 2.5 \][/tex]
#### Variances:
The variance [tex]\(\operatorname{Var}(X)\)[/tex] is calculated as:
[tex]\[ \operatorname{Var}(X) = \sum_{x=1}^{4} (x - E(X))^2 \cdot P(X=x) = (1-2.5)^2 \cdot 0.25 + (2-2.5)^2 \cdot 0.25 + (3-2.5)^2 \cdot 0.25 + (4-2.5)^2 \cdot 0.25 \][/tex]
Breaking it down:
[tex]\[ \operatorname{Var}(X) = 0.25 \cdot [(1.5)^2 + (0.5)^2 + (0.5)^2 + (1.5)^2] = 0.25 \cdot [2.25 + 0.25 + 0.25 + 2.25] = 0.25 \cdot 5 = 1.25 \][/tex]
Similarly, the variance [tex]\(\operatorname{Var}(Y)\)[/tex] is:
[tex]\[ \operatorname{Var}(Y) = \sum_{y=1}^{4} (y - E(Y))^2 \cdot P(Y=y) = 1.25 \][/tex]
In summary:
[tex]\[ E(X) = E(Y) = 2.5 \][/tex]
[tex]\[ \operatorname{Var}(X) = \operatorname{Var}(Y) = 1.25 \][/tex]
### a) Joint probability mass function of X and Y
When you roll a pair of four-sided dice (one red and one black), each combination of outcomes (X, Y) is equally likely. Since each die has 4 faces, there are [tex]\(4 \times 4 = 16\)[/tex] possible outcomes.
The probability of any specific outcome such as (1, 1) or (3, 4) is given by:
[tex]\[ P(X = x, Y = y) = \frac{1}{16} \][/tex]
Thus, the joint probability mass function can be organized in a 4x4 matrix where each entry represents the probability of a particular pair (X, Y):
[tex]\[ P(X, Y) = \begin{array}{c|cccc} & Y=1 & Y=2 & Y=3 & Y=4 \\ \hline X=1 & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} \\ X=2 & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} \\ X=3 & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} \\ X=4 & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} & \frac{1}{16} \\ \end{array} \][/tex]
Or more concisely:
[tex]\[ \text{Joint PMF} = \begin{bmatrix} 0.0625 & 0.0625 & 0.0625 & 0.0625 \\ 0.0625 & 0.0625 & 0.0625 & 0.0625 \\ 0.0625 & 0.0625 & 0.0625 & 0.0625 \\ 0.0625 & 0.0625 & 0.0625 & 0.0625 \end{bmatrix} \][/tex]
### b) Marginal probability mass function of X
The marginal PMF of X, [tex]\(P(X)\)[/tex], is found by summing the joint PMF over all possible values of Y.
For each value of X (1 through 4):
[tex]\[ P(X=x_i) = \sum_{j=1}^4 P(X=x_i, Y=y_j) = \sum_{j=1}^4 \frac{1}{16} = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4} = 0.25 \][/tex]
So the marginal PMF of X is:
[tex]\[ P(X) = \begin{bmatrix} 0.25 & 0.25 & 0.25 & 0.25 \end{bmatrix} \][/tex]
### c) Marginal probability mass function of Y
Similarly, the marginal PMF of Y, [tex]\(P(Y)\)[/tex], is found by summing the joint PMF over all possible values of X.
For each value of Y (1 through 4):
[tex]\[ P(Y=y_j) = \sum_{i=1}^4 P(X=x_i, Y=y_j) = \sum_{i=1}^4 \frac{1}{16} = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4} = 0.25 \][/tex]
So the marginal PMF of Y is:
[tex]\[ P(Y) = \begin{bmatrix} 0.25 & 0.25 & 0.25 & 0.25 \end{bmatrix} \][/tex]
### d) Are X and Y independent?
Two variables X and Y are considered independent if the joint probability [tex]\(P(X=x, Y=y)\)[/tex] is equal to the product of their marginal probabilities [tex]\(P(X=x) \cdot P(Y=y)\)[/tex] for all x and y.
Given the joint PMF:
[tex]\[ P(X=x, Y=y) = \frac{1}{16} \][/tex]
And the marginal PMFs:
[tex]\[ P(X=x) = 0.25 \quad \text{and} \quad P(Y=y) = 0.25 \][/tex]
We calculate the product:
[tex]\[ P(X=x) \cdot P(Y=y) = 0.25 \times 0.25 = 0.0625 \][/tex]
Since [tex]\(P(X=x, Y=y) = P(X=x) \cdot P(Y=y)\)[/tex] holds for all combinations of x and y, X and Y are independent.
### e) Calculate [tex]\(E(X)\)[/tex], [tex]\(E(Y)\)[/tex], [tex]\(\operatorname{Var}(X)\)[/tex], [tex]\(\operatorname{Var}(Y)\)[/tex]
#### Expected values:
The expected value [tex]\(E(X)\)[/tex] is calculated as:
[tex]\[ E(X) = \sum_{x=1}^{4} x \cdot P(X=x) = 1 \cdot 0.25 + 2 \cdot 0.25 + 3 \cdot 0.25 + 4 \cdot 0.25 = 0.25 \cdot (1+2+3+4) = 0.25 \cdot 10 = 2.5 \][/tex]
Similarly, the expected value [tex]\(E(Y)\)[/tex] is:
[tex]\[ E(Y) = \sum_{y=1}^{4} y \cdot P(Y=y) = 0.25 \cdot (1+2+3+4) = 2.5 \][/tex]
#### Variances:
The variance [tex]\(\operatorname{Var}(X)\)[/tex] is calculated as:
[tex]\[ \operatorname{Var}(X) = \sum_{x=1}^{4} (x - E(X))^2 \cdot P(X=x) = (1-2.5)^2 \cdot 0.25 + (2-2.5)^2 \cdot 0.25 + (3-2.5)^2 \cdot 0.25 + (4-2.5)^2 \cdot 0.25 \][/tex]
Breaking it down:
[tex]\[ \operatorname{Var}(X) = 0.25 \cdot [(1.5)^2 + (0.5)^2 + (0.5)^2 + (1.5)^2] = 0.25 \cdot [2.25 + 0.25 + 0.25 + 2.25] = 0.25 \cdot 5 = 1.25 \][/tex]
Similarly, the variance [tex]\(\operatorname{Var}(Y)\)[/tex] is:
[tex]\[ \operatorname{Var}(Y) = \sum_{y=1}^{4} (y - E(Y))^2 \cdot P(Y=y) = 1.25 \][/tex]
In summary:
[tex]\[ E(X) = E(Y) = 2.5 \][/tex]
[tex]\[ \operatorname{Var}(X) = \operatorname{Var}(Y) = 1.25 \][/tex]
