2471 - 概率论与数理统计第3章 第二题
（1）盒子里装有4只黑球，3只红球、2只白球，在其中任取4只球，以X表示取到黑球的只数，以Y表示取到红球的只数。求X和Y的联合分布律。
样例输入0 ，输出1
测试输入1，将X和Y的联合分布律以及边缘分布律按照输出格式输出（注意输出的是分数，而不是实数，且分数是不可再约分的）。
题目输入
-1
题目输出
Y\X| 0 1 2 3| p{Y=j}
0| 0 0 3/35 2/35| 1/7
1| 0 6/35 12/35 2/35| 4/7
2| 1/35 6/35 3/35 0| 2/7
p{X=i}| 1/35 12/35 18/35 4/35| 1
输入/输出样例
题目输入
0
题目输出
1
C语言解答
#include <stdio.h> int main() { int a; float b; while(scanf("%d",&a) != EOF) {if (a==1) { printf("Y\\X| 0 1 2 3 4| p{Y=j}\n"); printf("0| 0 0 1/21 4/63 1/126| 5/42\n"); printf("1| 0 2/21 2/7 2/21 0| 10/21\n"); printf("2| 1/42 4/21 1/7 0 0| 5/14\n"); printf("3| 1/63 2/63 0 0 0| 1/21\n"); printf("p{X=i}| 5/126 20/63 10/21 10/63 1/126| 1"); } else {b=1; printf("%d\n",b);} } return 0; }
C++解答
#include <iostream> #include <string> using namespace std; int main() { string b[5][6] = { "0","0","1/21","4/63","1/126","5/42", "0","2/21","2/7","2/21","0","10/21", "1/42","4/21","1/7","0","0","5/14", "1/63","2/63","0","0","0","1/21", "5/126","20/63","10/21","10/63","1/126","1" }; int a; while(scanf("%d",&a) != EOF) { if(a==1) { cout<<"Y\\X| 0 1 2 3 4| p{Y=j}"<<endl; int i; for(i=0 ;i<4;i++) cout<<i<<"| "<<b[i][0]<<" "<<b[i][1]<<" "<<b[i][2]<<" "<<b[i][3]<<" "<<b[i][4]<<"| "<<b[i][5]<<endl; cout<<"p{X=i}"<<"| "<<b[i][0]<<" "<<b[i][1]<<" "<<b[i][2]<<" "<<b[i][3]<<" "<<b[i][4]<<"| "<<b[i][5]<<endl; } else if(a==0) { cout<<1<<endl; } } return 0; }