[email protected] wrote: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?

A. 7/12
B. 5/12
C. 3/8
D. 1/3
E. 1/4

Hmmm, this is a slight twist on a question I created back in 2009: https://www.beatthegmat.com/probability- ... 28353.html
In the original question, I asked students to find P(the point's x-coordinate is greater than its y-coordinate). This question asks us to find P(the point's y-coordinate is greater than its x-coordinate).


One approach is to draw the figure.
Here's my original solution..

Of course, this is the solution to find P(the point's x-coordinate is greater than its y-coordinate). So, P(x > y) = 5/8

This means that P(y > x) = 1 - 5/8 = [spoiler]3/8[/spoiler] = C

Cheers,
Brent

[email protected] wrote:But why did we choose (0,0) (3,3) as the points where Y is greater than X, why cant it be (0,0)(4,4)?

(0,0) and (3,3) are points on the line y = x (notice that the x-coordinate and the y-coordinate are equal for both points)

This line, y = x, divides the region into two parts:
- the part where the x-coordinate is greater than the y-coordinate
- the part where the y-coordinate is greater than the x-coordinate

Cheers,
Brent

macattack wrote:Brent please something is bothering me. Let's say the question stated P(Y>=x) how would the solution change

Interesting question, macattack. I think the answer would get us into a pretty esoteric area of math.
I have a strong feeling that the answer would not change since the line y = x does not have a measurable area.

Don't worry about that concept though.

Cheers,
Brent

macattack wrote:Just for fun.. what you are saying suggests that P(x>=y)=P(x>y) but knowing that the number of desired outcome of the former in definitely higher than the latter definitely weakens that argument.

You're right in that we are adding a LOT of desired outcomes. In fact, we are adding an infinite number of outcomes.
Since we're adding the points on the line segment, it appears that the probability should increase. By how much does it increase?
Well, let's find P(the point is on the line segment in region R)

Since there is an infinite number of points on the line segment, and since region R consists of an infinite number of points, it appears that P(the point is on the line segment in region R) = infinite/infinite.
At this point, I believe we're getting into something called "countably infinite" (although it may also be called something else).

Although I can't answer your question, I can assure you that you don't need to worry about things like the area of a line.

What you do need to know is that, when it comes to probabilities involving area, the probability of an event proportional to the target area.

Cheers,
Brent