Let X be a random variable uniformly distributed in the interval [0, 2] and Y be a random variable uniformly distributed in the interval [0, 2]. If X and Y are independent of each other, the probability P(X + Y <= 1) is __________ (rounded off to three decimal places).
GATE 2024 · Engineering Mathematics · Uniform Distribution · medium
Answer: 0.125
Identify the joint density: Since X and Y are independent, each Uniform(0,2) with density 1/2, the joint density is f(x,y) = 1/4 over the square [0,2] x [0,2].
Identify the favorable region: Since x, y >= 0 and x + y <= 1, the region is a right triangle with vertices at (0,0), (1,0), and (0,1). Note: the constraint x + y <= 1 with x,y >= 0 already forces x <= 1 and y <= 1, well within [0,2].
Compute the area of the triangle: The right triangle has base = 1 along the x-axis and height = 1 along the y-axis, so Area = (1/2)(1)(1) = 0.5.
Compute the probability: P(X + Y <= 1) = (area of triangle) / (area of square) = 0.5 / 4 = 0.125.