probability problem

We know that the triangle has a base of 4 and a height of 5 which has an area of 10.

The area within the triangle where x-y>0 is bound by the lines y=x, y=0 and x=4. You can quickly see that this new, smaller triangle has a base of length 4 and a height of length 4... therefore the area is 8.

So, there is a 8/10 chance that the point will be x-y>0. Or 4/5. OA is E?

x can be any of the values 0,1,2,3,4
y can be any of the values 0,1,2,3,4,5

So the possible combinations where x-y > 0 are (1,0), (2,0), (2,1),(3,0),(3,1),(3,2), (4,0),(4,1),(4,2),(4,3) ==> 10 possible combinations out of a total of 30 combinations. So the probability is 10/30 or 1/3.

Is the answer A?

Correction to my previous post - Is the answer B?

This is incorrect for a few reasons... among them:
1) no reason x and y can only be integers... for example, (4,3.999999999999999) is perfectly fine point.
2) your logic on x being anything between 0 and 4 and y being anything between 0 and 5 is flawed. For example, the point (0,5) does not exist within the triangle... but you have counted it. What you have described is a rectangle.

The best approach is to consider the area of the two triangles described. See my description.

ramyaravindran wrote:x can be any of the values 0,1,2,3,4
y can be any of the values 0,1,2,3,4,5

So the possible combinations where x-y > 0 are (1,0), (2,0), (2,1),(3,0),(3,1),(3,2), (4,0),(4,1),(4,2),(4,3) ==> 10 possible combinations out of a total of 30 combinations. So the probability is 10/30 or 1/3.

Is the answer A?

egybs wrote:We know that the triangle has a base of 4 and a height of 5 which has an area of 10.

The area within the triangle where x-y>0 is bound by the lines y=x, y=0 and x=4. You can quickly see that this new, smaller triangle has a base of length 4 and a height of length 4... therefore the area is 8.

So, there is a 8/10 chance that the point will be x-y>0. Or 4/5. OA is E?

This is exactly how I solved it, my answer is 4/5 too.

The answer is 4/5