+274 A point (x, y) is randomly and uniformly chosen inside the square with vertices (0, 0), (0, 2), (2, 2), and (2, 0). What is the probability that x+y<3?
The area of the square is 4. The points that satisfy x+y<3 form a triangle with vertices (0, 0), (1, 1), and (0, 2). The area of this triangle is 1/211 = 1/2. Therefore, the probability that x+y<3 is 1/2 / 4 = 1/8.
Alternatively, we can solve this problem by using the fact that the probability of a point falling in a certain region is equal to the ratio of the area of that region to the area of the entire square. The region that satisfies x+y<3 is a triangle with vertices (0, 0), (1, 1), and (0, 2). The area of this triangle is 1/211 = 1/2. The area of the entire square is 4. Therefore, the probability that x+y<3 is 1/2 / 4 = 1/8.
