### Question 5Suppose the random variables $ X $ and $ Y $ have a probability density function (pdf) given by\[f(x, y) = x + y \quad \text{on} \quad 0 < x < 1, \, 0 < y < 1\]**a. Find $ P(X > Y) $.**\[\text{prob} = \boxed{\phantom{answer}}\]**b. Find $ F\left(\frac{1}{4}, \frac{1}{2}\right) $.**\[\text{ans} = \boxed{\phantom{answer}}\]

Answered: Suppose the random variables X and Y…

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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Introduction

Given that,

$X$ and $Y$ are two random variables

The joint pdf of them is given by ,

$f (x, y) = \{\begin{cases} x + y & 0 < x < 1, 0 < y < 1 \\ 0 & otherwise \end{cases}$

Step 1

(a) $P (X > Y)$ is to be obtained.

It can be written as $P (X > Y) = 1 - P (X < Y)$

Marginal pdf of $X$ is given by

$\begin{array}{rcl} \int_{0}^{1} (x + y) d y \\ = & \int_{0}^{1} x d y + \int_{0}^{1} y d y \\ = & x {(y)}_{0}^{1} + {(\frac{y^{2}}{2})}_{0}^{1} \\ = & x + \frac{1}{2} \end{array}$

Now,

$\begin{array}{rcl} P (X < Y) & = & \int_{0}^{y} x + \frac{1}{2} d x \\ = & {(\frac{x^{2}}{2})}_{0}^{y} + {(\frac{x}{2})}_{0}^{y} \\ = & \frac{y^{2}}{2} + \frac{y}{2} \end{array}$

Therefore, $P (X > Y) = 1 - \begin{array}{rcl} \frac{y^{2}}{2} \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} \frac{y}{2} \end{array}$

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