3. Let (X,Y) have the following joint density function: J ?y? 0
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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![### Joint Probability Density Function
**Problem 3: Joint Density Function**
Let \((X, Y)\) have the following joint density function:
\[
f(x, y) =
\begin{cases}
\frac{9}{64} x^2 y^2 & \text{if } 0 \leq x \leq 2, \, 0 \leq y \leq 2 \\
0 & \text{otherwise}
\end{cases}
\]
**Tasks:**
a. What is \(P(X = 0.6)\)?
b. Compute \(P(X + Y \leq 1)\).
### Explanation
The joint density function \(f(x, y)\) is defined for values of \(x\) and \(y\) between 0 and 2. For values outside this range, the function is zero. We need to determine the probabilities for specific conditions based on this function:
- **\(P(X = 0.6)\):**
The probability of a continuous random variable taking any specific value (like \(X = 0.6\)) in a continuous distribution is always zero.
- **\(P(X + Y \leq 1)\):**
To find this probability, integrate the joint density function over the region where \(X + Y \leq 1\). This would involve calculating the integral of \(\frac{9}{64}x^2y^2\) within the bounded region specified by the condition.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff0d90012-896b-4f52-8f0b-e38034636c35%2F5318cc20-c2c7-471d-8795-04f08ea8bc5f%2Ftwx4hdn_processed.png&w=3840&q=75)
Transcribed Image Text:### Joint Probability Density Function
**Problem 3: Joint Density Function**
Let \((X, Y)\) have the following joint density function:
\[
f(x, y) =
\begin{cases}
\frac{9}{64} x^2 y^2 & \text{if } 0 \leq x \leq 2, \, 0 \leq y \leq 2 \\
0 & \text{otherwise}
\end{cases}
\]
**Tasks:**
a. What is \(P(X = 0.6)\)?
b. Compute \(P(X + Y \leq 1)\).
### Explanation
The joint density function \(f(x, y)\) is defined for values of \(x\) and \(y\) between 0 and 2. For values outside this range, the function is zero. We need to determine the probabilities for specific conditions based on this function:
- **\(P(X = 0.6)\):**
The probability of a continuous random variable taking any specific value (like \(X = 0.6\)) in a continuous distribution is always zero.
- **\(P(X + Y \leq 1)\):**
To find this probability, integrate the joint density function over the region where \(X + Y \leq 1\). This would involve calculating the integral of \(\frac{9}{64}x^2y^2\) within the bounded region specified by the condition.
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