Provided that X and Y are independent and discreet random variables. Fill in the table below with missing values. Show calculations. This table represents a joint probability distribution of two discrete random variables, X and Y. The entries in the table correspond to the joint probabilities \( P(X = x, Y = y) \).### Table Explanation:- **Rows and Columns** - The rows represent different values of \( Y \): \( Y = 0, 1, \) and \( 2 \). - The columns represent different values of \( X \): \( X = 4, 5, \) and \( 6 \).- **Joint Probabilities** - The intersection of each row and column gives the probability \( P(X = x, Y = y) \). - For example, \( P(X = 4, Y = 0) = \frac{1}{5} \).- **Marginal Probabilities** - The last column labeled \( P_Y(y) \) provides the marginal probabilities of \( Y \). - \( P_Y(Y = 0) = \frac{3}{5} \) - \( P_Y(Y = 1) = \) (value not given) - \( P_Y(Y = 2) = \frac{3}{20} \) - The last row labeled \( P_X(x) \) provides the marginal probabilities of \( X \). - \( P_X(X = 4) = \frac{1}{3} \) - \( P_X(X = 5) = \frac{2}{5} \) - \( P_X(X = 6) = \) (value not given but can be computed)- **Total Probability** - The sum of the marginal probabilities \( P_Y(y) \) and \( P_X(x) \) across their respective rows and columns should equal 1, indicating a complete probability distribution.This table is used to understand the relationship and individual distributions of the random variables \( X \) and \( Y \).

Given that, and are independent discrete random variables with joint PMF is as follows: 1

Answered: Y = 0 Y = 1 Y = 2 Px(x) X = 4 1/5 1/12…

Math

Probability

Y = 0 Y = 1 Y = 2 Px(x) X = 4 1/5 1/12 1/3 X = 5 1/10 2/5 X = 6 Py(y) 3/5 3/20 1

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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