Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
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]
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
