中大數學系大學甄選入學試題[103](1)

P(兩球不同) = P(白紅) + P(紅白) + P(黑)
0.67 = 5/(5+2+3) × N/(7+N) + 2/(5+2+3) × 7/(7+N) + 3/(5+2+3)
0.67 = 1/2 × N/(7+N) + 1/5 × 7/(7+N) + 3/10
67(7+N) = 50N + 140 + 30(7+N)
469 + 67N = 80N + 350
119 = 13N
N = 119/13 = 9.1538...
正整數 N 應為 9。
驗算: 5/10 × 9/(7+9) + 2/10 × 7/(7+9) + 3/10 = 107/160 = 0.66875 ≈ 0.67
別解:
P(兩球不同) = 1 - P(兩球相同)
0.67 = 1 - P(白白) - P(紅紅)
0.67 = 1 - 5/(5+2+3) × 7/(7+N) - 2/(5+2+3) × N/(7+N)
0.67 = 1 - 1/2 × 7/(7+N) - 1/5 × N/(7+N)
67(7+N) = 100(7+N) - 350 - 20N
469 + 67N = 350 + 80N
119 = 13N
N = 119/13 = 9.1538...
正整數 N 應為 9。

2015-04-06 06:45:36 補充：
因版面吃了字,看漏了乙袋有5顆黑球,更正應為N=8:

P(兩球不同)=0.67
=1－P(兩白球)－P(兩紅球)－P(兩黑球)
=1－(5/10)*[7/(12+N)]－(2/10)*[N/(12+N)]－(3/10)*[5/(12+N)]
(5/10)*[7/(12+N)]+(2/10)*[N/(12+N)]+(3/10)*[5/(12+N)]=0.33
35+2N+15=0.33*(120+10N)
50+2N=39.6+3.3N
1.3N=10.4
N=8