投掷三枚色子,求所得点数的最大值为最小值2倍的概率.

投掷三枚色子,求所得点数的最大值为最小值2倍的概率.
投掷三枚色子,求所得点数的最大值为最小值2倍的概率.

投掷三枚色子,求所得点数的最大值为最小值2倍的概率.
古典概型通法完整版
基本事件空间=D
D={(1,1,1)(1,1,2)(1,1,3)(1,1,4)(1,1,5)(1,1,6)(1,2,1)(1,2,2)(1,2,3)(1,2,4)(1,2,5)(1,2,6)(1,3,1)(1,3,2)(1,3,3)(1,3,4)(1,3,5)(1,3,6)(1,4,1)(1,4,2)(1,4,3)(1,4,4)(1,4,5)(1,4,6)(1,5,1)(1,5,2)(1,5,3)(1,5,4)(1,5,5)(1,5,6)(1,6,1)(1,6,2)(1,6,3)(1,6,4)(1,6,5)(1,6,6)(2,1,1)(2,1,2)(2,1,3)(2,1,4)(2,1,5)(2,1,6)(2,2,1)(2,2,2)(2,2,3)(2,2,4)(2,2,5)(2,2,6)(2,3,1)(2,3,2)(2,3,3)(2,3,4)(2,3,5)(2,3,6)(2,4,1)(2,4,2)(2,4,3)(2,4,4)(2,4,5)(2,4,6)(2,5,1)(2,5,2)(2,5,3)(2,5,4)(2,5,5)(2,5,6)(2,6,1)(2,6,2)(2,6,3)(2,6,4)(2,6,5)(2,6,6)（3,1,1)(3,1,2)(3,1,3)(3,1,4)(3,1,5)(3,1,6)(3,2,1)(3,2,2)(3,2,3)(3,2,4)(3,2,5)(3,2,6)(3,3,1)(3,3,2)(3,3,3)(3,3,4)(3,3,5)(3,3,6)(3,4,1)(3,4,2)(3,4,3)(3,4,4)(3,4,5)(3,4,6)(3,5,1)(3,5,2)(3,5,3)(3,5,4)(3,5,5)(3,5,6)(3,6,1)(3,6,2)(3,6,3)(3,6,4)(3,6,5)(3,6,6)(4,1,1)(4,1,2)(4,1,3)(4,1,4)(4,1,5)(4,1,6)(4,2,1)(4,2,2)(4,2,3)(4,2,4)(4,2,5)(4,2,6)(4,3,1)(4,3,2)(4,3,3)(4,3,4)(4,3,5)(4,3,6)(4,4,1)(4,4,2)(4,4,3)(4,4,4)(4,4,5)(4,4,6)(4,5,1)(4,5,2)(4,5,3)(4,5,4)(4,5,5)(4,5,6)(4,6,1)(4,6,2)(4,6,3)(4,6,4)(4,6,5)(4,6,6)（5,1,1)(5,1,2)(5,1,3)(5,1,4)(5,1,5)(5,1,6)(5,2,1)(5,2,2)(5,2,3)(5,2,4)(5,2,5)(5,2,6)(5,3,1)(5,3,2)(5,3,3)(5,3,4)(5,3,5)(5,3,6)(5,4,1)(5,4,2)(5,4,3)(5,4,4)(5,4,5)(5,4,6)(5,5,1)(5,5,2)(5,5,3)(5,5,4)(5,5,5)(5,5,6)(5,6,1)(5,6,2)(5,6,3)(5,6,4)(5,6,5)(5,6,6)（6,1,1)(6,1,2)(6,1,3)(6,1,4)(6,1,5)(6,1,6)(6,2,1)(6,2,2)(6,2,3)(6,2,4)(6,2,5)(6,2,6)(6,3,1)(6,3,2)(6,3,3)(6,3,4)(6,3,5)(6,3,6)(6,4,1)(6,4,2)(6,4,3)(6,4,4)(6,4,5)(6,4,6)(6,5,1)(6,5,2)(6,5,3)(6,5,4)(6,5,5)(6,5,6)(6,6,1)(6,6,2)(6,6,3)(6,6,4)(6,6,5)(6,6,6)}
N=216
事件“所得点数的最大值为最小值2倍”=A
A={(1,1,2) (1,2,1)(1,2,2) (2,1,1)(2,1,2) (2,2,1) (2,2,4) (2,3,4) (2,4,2)(2,4,3)(2,4,4) (3,2,4) (3,3,6) (3,4,2) (3,4,6) (3,5,6) (3,6,3)(3,6,4)(3,6,5)(3,6,6) (4,2,2)(4,2,3)(4,2,4) (4,3,2) (4,3,6) (4,4,2) (4,6,3) (5,3,6)(5,6,3) (6,3,3)(6,3,4)(6,3,5)(6,3,6) (6,4,3)((6,5,3) (6,6,3)}
M=36
所以P=36/216=1/6

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