In order to be able to get back to this question easily, I am going to write a partial answer (since you have not had an answer in 12 hrs).
I believe the answer is 1/6...but I will re-work the problem and get back to you.
No, I was wrong but here is the correct answer with explanation.
Draw a figure and mark temporarily A, B and C on the circumference.
The angle ACB must be 60 degrees; this requires the smaller arc AB to be 120 degrees (since the 60 degree angle is measured by 1/2 of its intercepted arc).
Now you should redraw the circle and mark off on the circle the points on the circumference which form a regular hexagon, i.e. there are 60 degrees of arc between each pair of points. Mark one of these points A and 120 degrees away mark point B. Mark B' as the point at the opposite end of the diameter from point B and A' as the point at the opposite of the diameter from A.
Point C cannot be on the smaller arc between A and B because then angle ACB would be 120 degrees.
As long as point C is anywhere on the larger arc between point A and point B angle ACB will be 60 degrees.
Now start point C at point A and let it move toward B'. As long as C is between A and B' or between A' and B the center of the circle is outside the triangle ACB.
When point C is between point B' and point A' the center of the circle is inside the triangle ACB. The arc from point B' to A' is 120 degrees. Therefore, the probability that a point randomly selected as point C should be such that the center of the circle is inside triangle ACB is 120/360 = 1/3.
