probability question 1

a) Assume each seat have a number.
P(seat order) = P(no.1, no.2....no.6)
P(FMFMFM) + P(MFMFMF)

P(FMFMFM)
=(1/6)(1/5)(1/4)(1/3)(1/2)(3)(2)(3)(2)
(因為3個女仔的次序可以掉轉, 即( F1F2F3; F1F3F2 , F2F1F3, F2F3F1, F3F1F2, F3F2F1),
所以乘 (3)(2), 男的亦是一様有 6個排列
=36/720 =1/20

P(MFMFMF)
=1/20

P(FMFMFM)+P(MFMFMF)
=2/20 =1/10 //

b) P(MMMFFF) +P(MMFFM)+ P(MFFFMM) + P(FFFMMM) +P(FFMMMF) +
P(FMMMFF)
=1/20 * 6 =6/20 =3/10 //

c) P(MMFFMF) + P(MMFMFF) +P(MFFMFM) + P(MFMFFM) + P(FMMFMF)
+ P(FMMFFM) + P(FFMMFM)
=1/20 * 7=7/20 //

2007-11-20 14:56:22 補充：
改正　c)P(1 man sit between 2 women) ┼ P( 2 men sit together) ┼ P(3 men sit together) =1P( 2 men sit together ) =1 - 1/10 - 3/10 = 6/10 =3/5 //

2007-11-20 14:57:49 補充：
Another method is calculating how many ways (more complicated) :since the pattern should be " FMMF" (exactly 2 men), and move toward the right around the table.

2007-11-20 14:58:13 補充：
(F M M F) M F(F M M F) F MF (F M M F ) MM (F M M F ) FM F (F M M F )F M (F M M F )F) M F (F M M F) F M (F M M M F) M F (F M M F) F M (F M M M F) F M (FM M F) M F (F= 1/20 * 12 = 6/10 =3/5 //

a) 3/6 * 3/5 * 2/4 * 2/3 * 1/2 * 1/1 = 1/20
M F M F M F

b) 3/6 * 2/5 * 1/4 * 3/3*2/3 * 1/3 = 1/90 * 6 = 1/15
M M M F F F
F M M M F F
F F M M M F
F F F M M M
M F F F M M
M M F F F M

c) 3/6 * 2/5 * 3/4 * 1/3 * 2/3 * 1/3 = 1/90 *2 = 1/45
M M (F) M F (F) (F) <-- fixed, otherwise three man sit together
M M (F) F M (F)