随机数乘积概率问题试解(修改版)

从0到3随机取三个实数, 他们的乘积大于1并小于3的概率是多少?

试解：

P(1 < ZYX < 3) = Int(0,3)[P(1 < ZYX < 3| X = x)(1/3)dx]

因x最小值可以取到 1/9，所以

P(1 < ZYX < 3) = Int(1/9,3)[P(1 < ZYx < 3)(1/3)dx] =

Int(1/9,3)[Int(0,3)[P(1 < ZYx < 3| Y = y)(1/3)dy](1/3)dx] =

因yx最小值可以取到 1/3，所以对y积分的下限是1/(3x)

P(1 < ZYX < 3)  = Int(1/9,3)[Int(1/(3x),3)[P(1 < Zyx < 3)(1/3)dy](1/3)dx] =

(1/9)Int(1/9,3)[Int(1/(3x),3)[P(1 < Zyx < 3)dy]dx] =

(1/9)Int(1/9,3)[Int(1/(3x),3)[P(1/(xy) < Z < 3/(xy))dy]dx] =

因Z的取值在(0，3)，把对y的积分拆成两部分：
Int(1/(3x),1/x): (1/3) < xy < 1
Int(1/(1/x,3): 1 < xy < 3x

P(1 < ZYX < 3) = (1/9)Int(1/9,3)[Int(1/(3x),1/x)[P(1/(xy) < Z < 3/(xy))dy] +  
                                 Int(1/x,3)[P(1/(xy) < Z < 3/(xy))dy] dx] =

(1/9)Int(1/9,3)[Int(1/(3x),1/x)[P(1/(xy) < Z < 3)dy]dx] +
(1/9)Int(1/3,3)[Int(1/x,3)[P(1/(xy) < Z < 3/(xy))dy] dx] =

(1/27)Int(1/9,3)[Int(1/(3x),1/x)[(3 - 1/(xy))dy]dx] +  
(1/27)Int(1/3,3)[Int(1/x,3)[2/(xy))dy]dx] =

(1/27)Int(1/9,3)[ (3( 1/x - 1/3x) - (1/x)(ln(1/x) - ln(1/(3x))dx] +
(1/27)Int(1/3,3)[(2/x)(ln3 - ln(1/x) ) dx] =

(1/27)Int(1/9,3)[(2/x - (1/x)ln3)dx]
(1/27)Int(1/3,3)[(2/x)(ln3 + ln(x)) dx] =

(1/27)(Int(1/9,3)[(1/x)(2-ln3)dx] +
       Int(1/3,3)[(2/x)ln3dx] + Int(1/3,3)(2/x)ln(x)dx]) =

(1/27)((2-ln3)(ln3-ln(1/9)) + 2ln3(ln3-ln(1/3) + (ln(3))^2 - (ln(1/3))^2) =
(1/27)((2-ln3)3ln3 + 2ln3(ln3+ln3)) =

(1/27)(6ln3 - 3(ln3)^2 + 4(ln3)^2) =
(1/27)(6ln3 + (ln3)^2)