把编号为1到n的n个球放进编号为1到n的n个盒子,要求球的编号和盒子的编号不相同,并且每个盒子只能放一...

2024-12-24 19:53:01
推荐回答(2个)
回答1:

s(n)=(n-1) [ s(n-1)+s(n-2)]

总排列有C(n.n) 种。

1号球不入1号盒有(n-1)种 1号盒不入1号球有(n-1)种

只考虑1号球和1号盒有(n-1)^2种,在此排列中再考虑另外的n-2个球和(n-2)个盒,有(n-3)^2种

球号与盒号全不相同的概率:(n-1)^2*(n-3)^2*(n-5)^2*~~*1/C(n.n)

=(n-1)(n-3)(n-5)*~~1/n(n-2)(n-4)``1

恰有一个:球号与盒号相等的概率:

=(n-2)(n-4)(n-6)*~~*1/(n-1)(n-3)(n-5)``1 (n大于等于3)

扩展资料:

对事件发生可能性大小的量化引入“概率”。独立重复试验总次数n,事件A发生的频数μ,事件A发生的频率Fn(A)=μ/n,A的频率Fn(A)有没有稳定值?如果有,就称频率μ/n的稳定值p为事件A发生的概率,记作P(A)=p(概率的统计定义)。

P(A)是客观的,而Fn(A)是依赖经验的。统计中有时也用n很大的时候的Fn(A)值当概率的近似值。

设某一事件A(也是S中的某一区域),S包含A,它的量度大小为μ(A),若以P(A)表示事件A发生的概率,考虑到“均匀分布”性,事件A发生的概率取为:P(A)=μ(A)/μ(S),这样计算的概率称为几何概型。若Φ是不可能事件,即Φ为Ω中的空的区域,其量度大小为0,故其概率P(Φ)=0。

参考资料来源:百度百科-概率

回答2:

s(n)=(n-1) [ s(n-1)+s(n-2)]

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