OA Explanation:

3x + 4y < 12
4y < -3x + 12
y < \(\frac{-3x}{4} + 3\)

Draw a line going down from left to right having slope \(\frac{-3}{4}\) and y-intercept at (0,3).
Since, we have < sign - we need to shade the region below the line

Now, let us find out the y-value on line when x = 2
Plug x = 2 in the above line;
\(y = \frac{-3}{4}(2) + 3\)
\(y = \frac{3}{2}\)

Probability that x > 2 (Green region in the figure below) = \(\frac{Ar. (△DAB)}{Ar. (△COB)} = \frac{(2)(1.5)}{(4)(3)} = \frac{1}{4}\)