solution

先考虑对于如何求出解出第道题的概率。直接求并不好求，但是解不出第道题的概率很好求，就是。那么解出第到题的概率就是。

因为题目数量非常小。所以可以枚举每道题是否解出来了，然后统计一下每种情况的概率，记录到答案里面即可。

code

/*
* @Author: wxyww
* @Date:   2020-05-18 14:04:38
* @Last Modified time: 2020-05-18 14:08:27
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
#include<cmath>
using namespace std;
typedef long long ll;
const int N = 20;
ll read() {
    ll x = 0,f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1; c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = x * 10 + c - '0'; c = getchar();
    }
    return x * f;
}

double ans[N],pre[N];

int main() {
    int n = 12;
    for(int i = 1;i <= n;++i) {
        double x;
        scanf("%lf",&x);
        ans[i] = 1 - x;
    }
    for(int k = 1;k <= 2;++k) {
        for(int i = 1;i <= n;++i) {
            double x;
            scanf("%lf",&x);
            ans[i] = ans[i] * (1 - x);
        }
    }

    for(int i = 0;i < (1 << n);++i) {
        int cnt = 0;double k = 1.0;
        for(int j = 0;j < n;++j) {
            if((i >> j) & 1) {
                cnt++;
                k = k * (1 - ans[j + 1]);
            }
            else k = k * ans[j + 1];
        }
        pre[cnt] += k;
    }
    for(int i = 0;i <= n;++i) printf("%.6lf\n",pre[i]);

    return 0;
}