In x-y quadrant, there is a square with vertices (0,0) , (0,4) , (4,0) and (4,4) . What is the probability that the point on the square wil satisfy the condition x + 2y >= 3 .

Solution:

x + 2y > 3

x + 2y = 3

To get a line that cuts the square we need to substitute values for x and get y correspondingly.

Put x = 0, y = 3/2
x = 1, y = 1
x = 2, y = 1/2
x = 3, y = 0

We want region x + 2y > 3
That is 1/2 * 3/2 * 3 (Considering maximum value of x & y)

Now the probability.

(16 - 9/4)/16 = 53/64

Am I right here?