### Instruction: A point $(x,y)$ is randomly selected such that $0 \le x \le 3$ and $0 \le y \le 6$. What is the probability that $x+y \le 4$? Express your answer as a common fraction.### Response: I need to find the area of the region that satisfies $x+y \le 4$ and divide it by the area of the whole rectangle with dimensions $3 \times 6$.
To find the boundary of the region, I can set $x+y = 4$ and solve for $y$ in terms of $x$.
I get $y = 4 - x$.
The line $y = 4 - x$ intersects the $x$-axis at $(4,0)$ and the $y$-axis at $(0,4)$.
These are the vertices of the triangular region that satisfies $x+y \le 4$.
The area of the triangle is half the product of the base and the height.
The base is $4$ and the height is $4$, so the area is $8$.
The area of the rectangle is $3 \times 6 = 18$.
The probability that $x+y \le 4$ is the ratio of the areas: $\frac{8}{18} = \frac{4}{9}$.