A privately owned business operates both a drive in facility and a walk -in facility on a random selected day, X and Y respectively, be the proportions of the time that the drive - in and the walk - in facilities are in use, and suppose that the joint density function of these random variable is F(x,y) = { 2/5 ( 2x + 3x ) , 0<= X <= 1 ,0<=Y<=1 0 , Otherwise i) Verify condition F(x,y) is joint density function ii) Find P[(x,y) € A ] whereas A= {(x,y) | 0

Let's break down the problem step by step.

### Part i: Verifying the Condition that F(x, y) is a Joint Density Function

A function \( F(x, y) \) is a valid joint probability density function if the following conditions are satisfied:

1. **Non-negativity**: \( F(x, y) \geq 0 \) for all \( x \) and \( y \).
2. **Normalization**: The total probability must integrate to 1 over the entire range of \( x \) and \( y \):
   \[
   \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} F(x, y) \, dx \, dy = 1
   \]

Given \( F(x, y) = \frac{2}{5} (2x + 3y) \) for \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \), and \( F(x, y) = 0 \) otherwise, we need to check these two conditions.

#### 1. Non-Negativity
For \( F(x, y) \) to be non-negative, we need to verify that:
\[
F(x, y) = \frac{2}{5} (2x + 3y) \geq 0
\]
Since \( 2x + 3y \geq 0 \) for \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \), and \( \frac{2}{5} \) is a positive constant, the non-negativity condition is satisfied.

#### 2. Normalization
We need to integrate \( F(x, y) \) over the entire range of \( x \) and \( y \):
\[
\int_{0}^{1} \int_{0}^{1} \frac{2}{5} (2x + 3y) \, dx \, dy
\]
Let's compute this integral.

First, integrate with respect to \( x \) from 0 to 1:
\[
\int_{0}^{1} \frac{2}{5} (2x + 3y) \, dx = \frac{2}{5} \int_{0}^{1} (2x + 3y) \, dx
\]

\[
\frac{2}{5} \left[ x^2 + 3yx \right]_{0}^{1} = \frac{2}{5} \left[ 1 + 3y \right]
\]

This simplifies to:
\[
\frac{2}{5} (1 + 3y)
\]

Now, integrate with respect to \( y \) from 0 to 1:
\[
\int_{0}^{1} \frac{2}{5} (1 + 3y) \, dy = \frac{2}{5} \left[ y + \frac{3}{2}y^2 \right]_{0}^{1} 
\]
\[
= \frac{2}{5} \left( 1 + \frac{3}{2} \right) = \frac{2}{5} \times \frac{5}{2} = 1
\]
Since the total integral equals 1, the normalization condition is satisfied.

### Part ii: Finding \( P[(x, y) \in A] \) for a Region \( A \)

The problem asks for \( P[(x, y) \in A] \) where \( A = \{(x, y) | 0 \leq x \leq 1, 0 \leq y \leq 1\} \).

Since \( A \) is the entire region over which \( F(x, y) \) is defined, \( P[(x, y) \in A] \) is simply the total probability, which we already computed to be 1.

Thus, \( P[(x, y) \in A] = 1 \).

Would you like further details or have any questions about these steps?

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Here are 5 related questions:

1. What would happen if the region \( A \) were different from the entire domain?
2. How would the joint density function change if the coefficients of \( x \) and \( y \) were different?
3. Can you find the marginal density functions for \( X \) and \( Y \) from this joint density function?
4. How would you find the expected values \( E[X] \) and \( E[Y] \) using the joint density function?
5. What is the significance of the joint density function in probability and statistics?

**Tip:** When dealing with joint probability density functions, always check for non-negativity and proper normalization to ensure that the function is valid.

Explore the verification process of a joint density function F(x, y) = \frac{2}{5} (2x + 3y) over a specified domain, and calculate the probability P[(x, y) \in A] where A = {(x, y) | 0 \leq x \leq 1, 0 \leq y \leq 1}. Detailed step-by-step solution included.

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