Problem 1184

1. Plotting the Points:
- We start by plotting all 7 points on the circumference of the circle. Let's label these points as $A_1, A_2, A_3, A_4, A_5, A_6, A_7$.

2. Labeling the Points:
- Out of these 7 points, Bread selects 4 points to form a quadrilateral. Let's label these points as $Q_1, Q_2, Q_3, Q_4$.
- Kwu then selects the remaining 3 points to form a triangle. Let's label these points as $P_1, P_2, P_3$.

3. Condition for Non-Intersection:
- For Kwu's triangle to not intersect Bread's quadrilateral, the 3 points $P_1, P_2, P_3$ must be consecutive on the circle. This ensures that the triangle formed by $P_1, P_2, P_3$ lies entirely within one of the arcs formed by the quadrilateral $Q_1, Q_2, Q_3, Q_4$.

4. Counting Consecutive Points:
- There are 7 points on the circle. We need to count the number of ways to choose 3 consecutive points out of these 7 points.
- The number of ways to choose 3 consecutive points from 7 points is 7. This is because we can start at any of the 7 points and choose the next two points in a clockwise direction.

5. Total Combinations:
- The total number of ways to choose any 3 points out of 7 is given by the binomial coefficient $\binom{7}{3}$.
- $$\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$$

6. Calculating the Probability:
- The probability that Kwu's triangle does not intersect Bread's quadrilateral is the ratio of the number of favorable outcomes (choosing 3 consecutive points) to the total number of outcomes (choosing any 3 points out of 7).
- $$\text{Probability} = \frac{\text{Number of ways to choose 3 consecutive points}}{\text{Total number of ways to choose 3 points}} = \frac{7}{35} = \frac{1}{5}$$

The final answer is $\boxed{\frac{1}{5}}$