1. **State the problem:** We have 7 men and 6 women, total 13 runners. We want to form teams of 8 runners and find the proportion of teams with more women than men. 2. **Total number of teams:** The total ways to choose 8 runners from 13 is given by the combination formula: $$\binom{13}{8} = \frac{13!}{8!5!}$$ 3. **Teams with more women than men:** Since the team size is 8, more women than men means women > men. Possible women counts are 5, 6 (since max women is 6). 4. **Calculate teams with 5 women:** Choose 5 women from 6 and 3 men from 7: $$\binom{6}{5} \times \binom{7}{3}$$ 5. **Calculate teams with 6 women:** Choose 6 women from 6 and 2 men from 7: $$\binom{6}{6} \times \binom{7}{2}$$ 6. **Sum teams with more women:** $$\binom{6}{5} \times \binom{7}{3} + \binom{6}{6} \times \binom{7}{2}$$ 7. **Calculate values:** $$\binom{13}{8} = \frac{13 \times 12 \times 11 \times 10 \times 9}{5 \times 4 \times 3 \times 2 \times 1} = 1287$$ $$\binom{6}{5} = 6$$ $$\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$$ $$\binom{6}{6} = 1$$ $$\binom{7}{2} = \frac{7 \times 6}{2} = 21$$ 8. **Calculate numerator:** $$6 \times 35 + 1 \times 21 = 210 + 21 = 231$$ 9. **Calculate proportion:** $$\frac{231}{1287}$$ 10. **Simplify fraction:** $$\frac{231}{1287} = \frac{\cancel{231}^\times 1}{\cancel{231}^\times 5.57}$$ Since 231 divides 1287 exactly 5.57 times, let's find exact simplification: $$231 = 3 \times 7 \times 11$$ $$1287 = 3 \times 7 \times 61$$ Cancel common factors 3 and 7: $$\frac{3 \times 7 \times 11}{3 \times 7 \times 61} = \frac{11}{61}$$ **Final answer:** The proportion of teams with more women than men is $$\boxed{\frac{11}{61}}$$