https://file.aiquickdraw.com/mathbot/file/1726810582396xngraltp.jpg

The question asks to solve a problem involving the joint probability density function \( g(x, y) \) for random variables \( X \) and \( Y \). From the image, we have:

### Joint Probability Density Function \( g(x, y) \):

\[
g(x, y) =
\begin{cases}
\frac{3}{2}, & x^2 \leq y \leq 1 \text{ and } 0 \leq x \leq 1 \\
0, & \text{elsewhere}
\end{cases}
\]

The first part of the question (part a) asks to find:

\[
P(0 \leq X \leq \frac{1}{2})
\]

### Solution Steps:

To find the probability \( P(0 \leq X \leq \frac{1}{2}) \), we need to integrate the joint probability density function over the appropriate region.

1. **Set up the double integral**: 

   \[
   P\left(0 \leq X \leq \frac{1}{2}\right) = \int_{x=0}^{\frac{1}{2}} \int_{y=x^2}^{1} g(x, y) \, dy \, dx
   \]

2. **Use the value of \( g(x, y) \)**: Since \( g(x, y) = \frac{3}{2} \) in the specified region, the integral becomes:

   \[
   P\left(0 \leq X \leq \frac{1}{2}\right) = \int_{x=0}^{\frac{1}{2}} \int_{y=x^2}^{1} \frac{3}{2} \, dy \, dx
   \]

3. **Evaluate the inner integral with respect to \( y \)**:

   \[
   \int_{y=x^2}^{1} \frac{3}{2} \, dy = \frac{3}{2} \left( 1 - x^2 \right)
   \]

4. **Substitute into the outer integral**:

   \[
   P\left(0 \leq X \leq \frac{1}{2}\right) = \int_{x=0}^{\frac{1}{2}} \frac{3}{2} (1 - x^2) \, dx
   \]

5. **Split the integral and solve**:

   \[
   P\left(0 \leq X \leq \frac{1}{2}\right) = \frac{3}{2} \left( \int_{x=0}^{\frac{1}{2}} 1 \, dx - \int_{x=0}^{\frac{1}{2}} x^2 \, dx \right)
   \]

   - For \( \int_{x=0}^{\frac{1}{2}} 1 \, dx = \frac{1}{2} \)
   - For \( \int_{x=0}^{\frac{1}{2}} x^2 \, dx = \frac{x^3}{3} \Big|_0^{\frac{1}{2}} = \frac{1}{24} \)

   So, we have:

   \[
   P\left(0 \leq X \leq \frac{1}{2}\right) = \frac{3}{2} \left( \frac{1}{2} - \frac{1}{24} \right)
   \]

6. **Simplify**:

   \[
   P\left(0 \leq X \leq \frac{1}{2}\right) = \frac{3}{2} \times \frac{11}{24} = \frac{33}{48} = \frac{11}{16}
   \]

Thus, the probability is \( \frac{11}{16} \).

---

Would you like further details on this problem or help with part b?

Here are 5 related questions to consider:
1. How do you find the marginal density function of \( X \) from the joint density \( g(x, y) \)?
2. What is the probability that \( X \) is greater than \( \frac{1}{2} \)?
3. How would you interpret the result geometrically in the \( xy \)-plane?
4. Can you derive the conditional density function of \( Y \) given \( X = x \)?
5. What changes if we modify the bounds for \( X \) or \( Y \) in the probability calculation?

**Tip**: For joint probability density functions, integrating over the region of interest is key to solving for probabilities. Always ensure the limits of integration correctly reflect the region defined by the problem.

Given the joint probability density function g(x, y) = (3/2) for x^2 ≤ y ≤ 1 and 0 ≤ x ≤ 1, find P(0 ≤ X ≤ 1/2).

Learn how to solve a joint probability density function problem with detailed steps, involving double integrals and probability theory. This problem is suitable for college-level students and covers key concepts such as joint density functions and integration over regions.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation