數學概率問題??

Total 彈子= 3+4+5=12
Q1:
P(三枚均為藍色彈子)= (5/12)(4/11)(3/10)=1/22 (彈子不被放回,所以每次減少)
1st 2nd 3rd
Q2:
P(其中最少一枚紅色)= 1-P(一枚紅色都沒有)
= 1-(9/12)(8/11(7/10)
= 1-21/55
= 34/55
P(最少一枚紅色)係好難計,因為好長,所以要諗相反.
由於概率範圍係 0≦P(x)≦1, 所以要一減相反

Q3:
P(三枚彈子顏色皆不相同)= (3/12)(4/11)(5/10)(6)=6/22=3/11
1st 2nd 3rd
由於三枚彈子出現次序不同,
V1:RBY (3/12)(5/11)(4/10) V4:YBR (4/12)(5/11)(3/10)
V2:BYR (5/12)(4/11)(3/10) V5:RYB (3/12)(4/11)(5/10)
V3:YRB (4/12)(3/11)(5/10) V6:BRY (5/12)(3/11)(4/10)
出現6個組合,所以乘6

1. P(三枚都是藍色) = 5/12 * 4/11 * 3/10 = 1/22
2. P(其中最少一枚紅色) = 1 - P(所有都不是紅色)
= 1 - (9/12 * 8/11 * 7/10)
= 1 - 21/55
= 34/55
3. P(三枚頻色皆不相同) = 3 * (3/12) * (4/11) * (5/10) = 3/22