Question

In the coordinate plane, rectangular region R has vertices at (0,0), (0,3), (4,3), and (4,0). If a point in region R is randomly selected, what is the probability that the point's y-coordinate will be greater than its x-coordinate?

Answer 1

Answer:

37.5%

Step-by-step explanation:

If you draw this rectangle you will see that it has a width of 4 (from (0,0) to (4,0)) and a height of 3 (from (0,0) to (0,3)). Now lets draw a line from (0,0) to (3,3) this is the group of points inside the rectangle where x=y. Notice that this line divide the initial rectangle in two figures, a triangle and a quadrilateral:

triangle:

with vertices in (0,0), (0,3) and (3,3)

quadrilateral:

with vertices in (0,0), (3,3), (4,0) and (4,3)

For each point of the triangle it is true that:

$y \geq x$

And for each point in the quadrilateral it is true that:

$x \geq y$

triangle area = [text] \frac{3*3}{2} = 4.5 [/text]

rectangle area = [text] 3 + \frac{3*3}{2} = 7.5 [/text]

Therefore:

$P(y>x) = \frac{4.5}{12} = 0.375 = 37.5%$