機會率問題，請各位大大解答，謝謝！?

a) 4C2 / 9C2 = 6/36 = 1/6
或 
4/9 × 3/8 = 1/6

b) (4C1 × 5C1)/ 9C2 = 20/36 = 5/9
或
4/9 × 5/8 + 5/9 × 4/8 = 5/9

c) (抽鎖A及非A匙種數 + 抽鎖B及非B匙種數 + 抽鎖C及非C匙種數)/(抽一鎖二匙種數)
= (1×6C2 + 1×5C2 + 1×7C2) / (3 × 9C2)
= (15 + 10 + 21) / (3 × 36)
= 23/54

或

P(抽到鎖A但抽不到匙A)+P(抽到鎖B但抽不到匙B)+P(抽到鎖C但抽不到匙C)
= 1/3×6/9×5/8 + 1/3×5/9×4/8 + 1/3×7/9×6/8
= 5/36 + 5/54 + 7/36
= (15+10+21)/108
= 23/54