### Instruction: Point $(x,y)$ is randomly picked from the rectangular region with vertices at $(0,0),(2008,0),(2008,2009),$ and $(0,2009)$. What is the probability that $x > 2y$? Express your answer as a common fraction.### Response: I want to find the area of the region where $x > 2y$ and divide it by the total area of the rectangle.
To do that, I need to find the line that divides the rectangle into two parts, one where $x > 2y$ and one where $x \leq 2y$.
This line is $x = 2y$, and it intersects the rectangle at $(0,0)$ and $(2008,1004)$.
I can draw a picture to help me visualize the problem.
The area of the region where $x > 2y$ is a right triangle with base $2008$ and height $1004$.
The area of the rectangle is $2008 \times 2009$.
So the probability I want is $\frac{\frac{1}{2} \times 2008 \times 1004}{2008 \times 2009}$.
I can simplify this fraction by canceling out common factors.
I get $\frac{1004}{2 \times 2009} = \frac{502}{2009}$.