1. **Problem statement:** A committee consists of 7 men and 4 women. A sub-committee of 6 is chosen at random. Find the probability that the sub-committee contains exactly 2 women. 2. **Formula and rules:** The total number of ways to choose 6 people from 11 (7 men + 4 women) is given by the combination formula: $$\binom{11}{6} = \frac{11!}{6!5!}$$ The number of ways to choose exactly 2 women from 4 women is: $$\binom{4}{2}$$ The number of ways to choose the remaining 4 members from the 7 men is: $$\binom{7}{4}$$ The probability is the ratio of favorable outcomes to total outcomes: $$P = \frac{\binom{4}{2} \times \binom{7}{4}}{\binom{11}{6}}$$ 3. **Intermediate calculations:** - Calculate $$\binom{4}{2} = \frac{4!}{2!2!} = 6$$ - Calculate $$\binom{7}{4} = \frac{7!}{4!3!} = 35$$ - Calculate $$\binom{11}{6} = \frac{11!}{6!5!} = 462$$ 4. **Calculate probability:** $$P = \frac{6 \times 35}{462} = \frac{210}{462} = \frac{35}{77} \approx 0.4545$$ 5. **Explanation:** We first find the total number of ways to select any 6 people from 11. Then, we find the number of ways to select exactly 2 women and 4 men. Dividing the favorable outcomes by total outcomes gives the probability. **Final answer:** $$\boxed{\frac{35}{77} \approx 0.4545}$$