Question

Let \( X \) and \( Y \) be continuous random variables with the joint probability density function
\[ f(x, y) = \begin{cases} x + y, & \text{if } 0 < x < 1, 0 < y < 1 \\ 0, & \text{otherwise} \end{cases} \] Then \( P(X + Y > \frac{1}{2}) \) equals

• A. \( \frac{23}{24} \)
• B. \( \frac{1}{12} \)
• C. \( \frac{11}{12} \)
• D. \( \frac{1}{24} \)

Hint:

When integrating over a region for continuous random variables, always carefully set the limits based on the given condition (here \( X + Y>\frac{1}{2} \)).

Answer 1

The Correct Option is A

Step 1: Understanding the joint density function.
The joint probability density function is given by \( f(x, y) = x + y \) for \( 0<x<1 \) and \( 0<y<1 \). The total probability over the unit square is 1, so we need to calculate the probability \( P(X + Y>\frac{1}{2}) \).
Step 2: Set up the integration.
To find \( P(X + Y>\frac{1}{2}) \), we integrate the joint density function over the region where \( X + Y>\frac{1}{2} \). This involves integrating \( f(x, y) \) over the appropriate bounds: \[ P(X + Y>\frac{1}{2}) = \int_0^1 \int_{\frac{1}{2}-x}^1 (x + y) \, dy \, dx \]
Step 3: Perform the integration.
After evaluating the integrals, we obtain the result \( P(X + Y>\frac{1}{2}) = \frac{23}{24} \).