Two numbers, x and y are selected at random from the interval (0,3). What is the probability that a triangle with sides of length 1, x, and y exists?
As I have said before, I am terrible with Geometry.
we have a 1/4 chance of each number for x and y, so there is a 1/16 chance for any given pair of numbers.
a list of all possible pairs:
since 0 is an invalid number for a triangle, there is a 7/16 chance for an invalid triangle already. we continue to cross out invalids. also, if the sum of any two side lengths of a triangle is not greater than the length of the third side, it is an invalid triangle. 1,2 and 2,1 is invalid, so we have 9/16 invalid. 2,3 and 3,2 are invalid, 1,3 and 3,1 are invalid, now we have 13/16 invalid. 3,3 and 2,2 are isosceles so they are valid, and we already said 1,1 was valid. there is a 3/16 chance you would get a real triangle with side lengths 1, x, and y. this is an example with fair four sided dice. if it is done with unfair dice, the results could vary greatly.
im afraid not, by a certain geometry theorem, if any two side lengths, when added, are not greater than the length of the third side, it is not a triangle. this eliminates many of the options. i shall cross out the eliminated ones
0,0 0,1 0,2 0,3 1,0
1,2 1,3 2,0 2,1
2,3 3,0 3,1 3,2
as you can see, there are 3/16 possible options left. i have taken geometry and have an entire binder of postulates and theorems.
In the question it specifies that we are looking for the probobility that a triangle can exist with side lengths of 1, x and y.
We do know that x and y can be either 0, 1, 2 or 3. We can list that all out:
This gives us a 50/50 chance or 1/2 chance that a triangle with 1,x and y can exists.
Btw I am looking at the problem now and it says the answer is 1/2.
This is just my attempt to explain this but if you see issues with my solving of this problem please point them out.
If you are just here for the answer it is: 1/2