A point $(x,y)$ is randomly picked from inside the rectangle with vertices  $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$.  What is the probability that  $x < y$?
The point $(x,y)$ satisfies $x < y$ if and only if it belongs to the shaded triangle bounded by the lines $x=y$, $y=1$, and $x=0$, the area of which is 1/2. The ratio of the area of the triangle to the area of the  rectangle is $\frac{1/2}{4} = \boxed{\frac{1}{8}}$.

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draw((-1,0)--(5,0),Arrow);
draw((0,-1)--(0,2),Arrow);
for (int i=1; i<5; ++i) {
draw((i,-0.3)--(i,0.3));
}
fill((0,0)--(0,1)--(1,1)--cycle,gray(0.7));
draw((-0.3,1)--(0.3,1));
draw((4,0)--(4,1)--(0,1),linewidth(0.7));
draw((-0.5,-0.5)--(1.8,1.8),dashed);
[/asy]
\frac{1}{8}
Level 4
Counting & Probability
Problem: A point $(x,y)$ is randomly picked from inside the rectangle with vertices  $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$.  What is the probability that  $x < y$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the probability that a randomly chosen point $(x,y)$ from the rectangle satisfies $x < y$, we first need to understand the area of the rectangle defined by the vertices $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. The rectangle has a width of 4 units and a height of 1 unit, giving it a total area of $4 \times 1 = 4$ square units.

Next, we need to identify the region within this rectangle where the condition $x < y$ holds true. The line $y = x$ will help us visualize this condition. However, since the rectangle only extends to $y = 1$, we need to find the intersection of the line $y = x$ with the top edge of the rectangle, which occurs at the point $(1,1)$.

The area of interest, where $x < y$, is bounded by the line $y = x$, the line $y = 1$, and the vertical line $x = 0$. This forms a triangular region with vertices at $(0,0)$, $(1,1)$, and $(0,1)$. 

To calculate the area of this triangle, we can use the formula for the area of a triangle: 
Area = $\frac{1}{2} \times \text{base} \times \text{height}$. 
Here, the base is the segment from $(0,0)$ to $(0,1)$, which has a length of 1, and the height is the segment from $(0,1)$ to $(1,1)$, which also has a length of 1. Thus, the area of the triangle is:
Area = $\frac{1}{2} \times 1 \times 1 = \frac{1}{2}$.

Now, to find the probability that a randomly chosen point $(x,y)$ satisfies $x < y$, we take the area of the triangular region where $x < y$ and divide it by the total area of the rectangle:
Probability = $\frac{\text{Area of triangle}}{\text{Area of rectangle}} = \frac{\frac{1}{2}}{4} = \frac{1}{8}$.

Thus, the final answer is $\frac{1}{8}$.<|END_OF_LATENT|> The final answer is $\frac{1}{8}$.