1-19n=C ①k=CC;C P n-/33 ②k=Ccc2×P=213 ③k= ckclCICICIXP=6 32 n-/33 ④k=C6C10P=k/n=18/33 ⑤设A为事件“恰有1双配对”,则 A为事件“无1双配 对1- 19 4 C12 n = 1 2 1 5 1 8 1 k = C6 C C C 33 16 = = n P k ? 1 2 1 4 1 2 1 5 1 ③ k = C6 C C C C 33 32 = = n P k 1 4 1 5 1 k = C6 C C 33 8 = = n P k ④ 2 10 1 k = C6 C P = k / n =18/33 ⑤ 设A为事件“ 恰有1双配对 ”，则 为事件“ 无1双配 对 ”. ① ② A