Let X and Y be continuous random variables with joint density function

fx,y(x, y)=3x/ 4  for 0 < x < 2 and 0<y<2-x

What is P(X > 1)?

To find the probability \( P(X > 1) \) for the continuous random variables \(X\) and \(Y\) with the given joint density function, we need to integrate the joint density over the region where \( X > 1 \).

### Step-by-Step Solution:

1. **Understand the Joint Density Function**:

   The joint density function is given by:
   \[
   f_{X,Y}(x, y) = \frac{3x}{4}, \quad \text{for } 0 < x < 2 \text{ and } 0 < y < 2 - x.
   \]

2. **Set up the Probability \( P(X > 1) \)**:

   We need to integrate the joint density function over the region where \( X > 1 \). Since \( X \) ranges from 1 to 2, and for each value of \( x \), \( y \) ranges from 0 to \( 2 - x \), we have:
   \[
   P(X > 1) = \int_{1}^{2} \int_{0}^{2-x} f_{X,Y}(x, y) \, dy \, dx.
   \]

3. **Substitute the Joint Density Function**:

   Substituting \( f_{X,Y}(x, y) = \frac{3x}{4} \), the probability becomes:
   \[
   P(X > 1) = \int_{1}^{2} \int_{0}^{2-x} \frac{3x}{4} \, dy \, dx.
   \]

4. **Integrate with Respect to \( y \) First**:

   \[
   P(X > 1) = \int_{1}^{2} \left( \frac{3x}{4} \int_{0}^{2-x} dy \right) dx.
   \]

   The inner integral with respect to \( y \) is:
   \[
   \int_{0}^{2-x} dy = (2 - x) - 0 = 2 - x.
   \]

   So, the probability becomes:
   \[
   P(X > 1) = \int_{1}^{2} \frac{3x}{4} (2 - x) \, dx.
   \]

5. **Simplify and Integrate with Respect to \( x \)**:

   Distribute \( \frac{3x}{4} \):
   \[
   P(X > 1) = \int_{1}^{2} \left( \frac{3x \cdot 2}{4} - \frac{3x^2}{4} \right) dx = \int_{1}^{2} \left( \frac{3x}{2} - \frac{3x^2}{4} \right) dx.
   \]

   Separate the integrals:
   \[
   P(X > 1) = \frac{3}{2} \int_{1}^{2} x \, dx - \frac{3}{4} \int_{1}^{2} x^2 \, dx.
   \]

   Compute each integral:
   - \(\int x \, dx = \frac{x^2}{2} \)
   - \(\int x^2 \, dx = \frac{x^3}{3} \)

   Plugging in the limits from 1 to 2:

   \[
   P(X > 1) = \frac{3}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} - \frac{3}{4} \left[ \frac{x^3}{3} \right]_{1}^{2}.
   \]

   Evaluate the limits:
   \[
   \left[ \frac{x^2}{2} \right]_{1}^{2} = \frac{2^2}{2} - \frac{1^2}{2} = \frac{4}{2} - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2}.
   \]

   \[
   \left[ \frac{x^3}{3} \right]_{1}^{2} = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}.
   \]

   Therefore:
   \[
   P(X > 1) = \frac{3}{2} \cdot \frac{3}{2} - \frac{3}{4} \cdot \frac{7}{3}.
   \]

   Simplify the terms:
   \[
   P(X > 1) = \frac{9}{4} - \frac{7}{4} = \frac{2}{4} = \frac{1}{2}.
   \]

### Final Answer:

\[
P(X > 1) = \frac{1}{2}.
\]

Would you like any further details or have any questions? 

### Relative Questions:
1. How do you find the marginal density function of \(X\) from a joint density function?
2. What is the meaning of a joint density function in probability theory?
3. How do you compute the probability \(P(X \leq 1)\) for this problem?
4. How do you verify if a joint density function is valid?
5. What is the difference between continuous and discrete random variables?

**Tip**: To find a marginal distribution from a joint density, integrate the joint density over all possible values of the other variable.

Learn how to calculate the probability P(X > 1) for continuous random variables X and Y with a given joint density function. Step-by-step solution involving integral calculus and probability theory explained.

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