Two distinct points $A$ and $B$ are chosen at random from 15 points equally spaced around a circle centered at $O$ such that each pair of points $A$ and $B$ has the same probability of being chosen. The probability that the perpendicular bisectors of $OA$ and $OB$ intersect strictly inside the circle can be expressed in the form $\frac{m}{n}$, where $m,n$ are relatively prime positive integers. Find $m+n$.

[i]Ray Li.[/i]
1. **Understanding the Problem:**
   We need to find the probability that the perpendicular bisectors of \(OA\) and \(OB\) intersect strictly inside the circle. This happens when the angle \(\angle AOB\) is less than \(120^\circ\).

2. **Total Number of Pairs:**
   There are 15 points on the circle. The total number of ways to choose 2 distinct points \(A\) and \(B\) from these 15 points is given by the combination formula:
   \[
   \binom{15}{2} = \frac{15 \times 14}{2} = 105
   \]

3. **Counting Favorable Pairs:**
   For the perpendicular bisectors of \(OA\) and \(OB\) to intersect strictly inside the circle, the angle \(\angle AOB\) must be less than \(120^\circ\). This means that the arc between \(A\) and \(B\) must be less than \(120^\circ\).

   Since the points are equally spaced, each point represents an angle of \(\frac{360^\circ}{15} = 24^\circ\). Therefore, we need to count the pairs where the angle \(\angle AOB\) is less than \(120^\circ\), which corresponds to fewer than 5 points between \(A\) and \(B\) (since \(24^\circ \times 5 = 120^\circ\)).

4. **Calculating Favorable Pairs:**
   For each point \(A\), there are 4 points on either side of \(A\) that form an angle less than \(120^\circ\) with \(A\). Thus, for each point \(A\), there are \(2 \times 4 = 8\) favorable points \(B\).

   Since there are 15 points, the total number of favorable pairs is:
   \[
   15 \times 8 = 120
   \]
   However, each pair \((A, B)\) is counted twice (once as \((A, B)\) and once as \((B, A)\)), so we need to divide by 2:
   \[
   \frac{120}{2} = 60
   \]

5. **Calculating the Probability:**
   The probability that the perpendicular bisectors of \(OA\) and \(OB\) intersect strictly inside the circle is the ratio of the number of favorable pairs to the total number of pairs:
   \[
   \frac{60}{105} = \frac{4}{7}
   \]

6. **Simplifying the Fraction:**
   The fraction \(\frac{4}{7}\) is already in its simplest form, where \(m = 4\) and \(n = 7\).

7. **Summing \(m\) and \(n\):**
   \[
   m + n = 4 + 7 = 11
   \]

The final answer is \( \boxed{ 11 } \).