I got: .A random process starts from point
Point is obtained in a uniformly selected direction, 1 unit from
The same process is repeated at point to obtain point ,
and at point to obtain point
What is the probability that a triangle is obtained when point is generated?
Code:B o * * * * * * O o * * * * * * * * o A * o C
Segment can be any direction.
Segment must be within 60° on either side of .Code:B o \ \ \ \ 60° \ O o * * * * * o A 60° / / / / / o B
Otherwise, and we will have no triangle.
Segment must be within 60° ofCode:B o / * /60°* / * / * * O o * * * * * o A
As Captain was objecting, assuming "that segment BC intersect segment OA" the answer is NOT 1/18. I think it is because angle OAB can be bigger than 60 degrees. Assume it is 70. A triangle is still possible. I think AOB can have values upto 90 degrees. That will change the solution with new combinations.
Actually this can be simulated in Excel. And you should have an answer.
Here is my approach:
a: first angle randomly generated
b: second angle randomly generated
a in [0,60]
Then 0 < b < (180-a)/2 forms a triangle.
a in [60,90]
Then 0 < b < (180-2a) forms a triangle.
Compute probabilities in both case.
Then multiple by 2 (why?)
You should get the answer.