From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men, is.

Total number of ways = ⁵C₁ × ¹⁰C₃ + ⁵C₂ × ¹⁰C₂ + ⁵C₃ × ¹⁰C₁ + ⁵C₄
= ⁵!/⁴!₁! × ¹⁰!/⁷!₃! + ⁵!/₂!₃! × ¹⁰!/₈!₂! + ⁵!/₃!₂! × ¹⁰!/⁹!₁! + ⁵!/₄!
= 5 × (10 × 9 × 8)/(3 × 2 × 1) + 5 × 4/(2 × 1) × (10 × 9)/(2 × 1) + 5 × 4/(2 × 1) × 10 + 5
= 5 × 120 + 10 × 45 + 10 × 10 + 5
= 600 + 450 + 100 + 5 = 1155
Total no. of ways = 1155
No. of ways to form a committee having more women than men = ⁵C₃ × ¹⁰C₁ + ⁵C₄
= ⁵!/₃!₂! × ¹⁰!/⁹!₁! + ⁵!/₄!
= 10 × 10 + 5 = 105
Probability = 105/1155 = 1/11