Probability

Looking at this problem, I see an interesting way to apply an integral (in some way) to this situation.

I started approaching this by thinking about how a triangle works. For its three sides to connect, no side can have a length equal to or greater than the sum of the other two sides.

This is the main rule we will use to determine what values are possible.

Now, how do we apply this? Let's think of the information we have. In this case, it's only one piece of information: one of the sides has length two.

Knowing this, we can start to use this in conjunction with the rule we have for a triangle.

x + y >= 2
|x - y| <= 2

If we add in our other limiting factors, we can make an enclosed area, from which we can calculate the area of!

0 < x < 3
0 < y < 3
With all of this, we end up with a shape whose vertices connect in the following order:
(0, 3), (0, 2), (2, 0), (3, 0), (3, 1), (1, 3), (0, 3)

While it's possible to locate the berices on paper by hand, we can also split this hexagon up into three shapes.

For x = (0, 1): (0, 3), (0, 2), (1, 1), (1, 3), (0, 3) [a right trapezoid, with area = 1.5]
For x = (1, 2): (1, 1), (2, 0), (2, 2), (1, 3), (1, 1) [a parallelogram, with area = 2]
For x = (2, 3): (3, 0), (2, 0), (2, 2), (3, 1), (3, 0) [another right trapezoid, with area = 1.5]

Now, we only have to sum up the area this shape takes up, and divide by the area enclosed by x = (0, 3), y = (0, 3) (a 3x3 square, with area = 9)

1.5 + 2 + 1.5 = 5
5/9 = 55.5555...%

You answer this type of question with a contour probability map.

I have done the map here:

https://www.geogebra.org/classic/epd8ewqe

x and y both have to be between 0 and 9 so the area for this is  3*3=9

Then for it to be a triangle, the sum of any 2 sides must be greater than the 3rd side

So I have graphed

x+y>2

x+2>y

y+2>x

The intersection of all those regions is the favourable outcomes.   Area  = 6

SO     the probability of a favourable outcome is  6/9  =  2/3