2003 AMC 10A Problem 12

The point is picked evenly from the rectangle, so the probability is (area where $x < y$) divided by (area of the rectangle).

1. The rectangle is $4$ wide and $1$ tall, so its area is $4 \times 1 = 4$.
2. Test a few points to see where $x < y$ can happen. At $(0.2, 0.7)$, $0.2 < 0.7$, so it counts. At $(2, 0.5)$, $2 > 0.5$, so it does not. In fact $y$ is never more than $1$ in this rectangle, so any point with $x \ge 1$ fails. Only the left $1 \times 1$ square can have $x < y$.
3. Inside that square, draw the line $y = x$ from $(0,0)$ to $(1,1)$. The points with $x < y$ are above the line: the triangle with corners $(0,0)$, $(0,1)$ and $(1,1)$.
4. That triangle has two sides of length $1$ at a right angle, so its area is $\frac{1}{2} \times 1 \times 1 = \frac{1}{2}$.
5. The probability is
$$\frac{1/2}{4} = \frac{1}{8}.$$

The answer is $\boxed{\textbf{(A)}\ \frac{1}{8}}$.

When a point is chosen at random from a region, probability becomes a ratio of areas. The work is in drawing the boundary line and seeing where it leaves the region. Here most of the long rectangle lies where $x$ is already bigger than any possible $y$, so the good region is small.