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2471 - 概率论与数理统计第3章 第二题

通过次数

0

提交次数

0

时间限制 : 1 秒 内存限制 : 128 MB

（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

输入格式

0

输出格式

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;
}

Java解答


import java.util.*;
public class Main {
	public static void main(String args[]) {
		Scanner cin = new Scanner(System.in);
		int a;
		int b;
		 
		while (cin.hasNext()) {
			a = cin.nextInt(); 
			if (a==1){
				
			int	c=1;
			int	d=2;
			int	e=4;
			int	f=7;
			int	g=21;
			int	h=42;
			int	i=63;
			int	j=126;
			int c1=5;
			int c2=10;
			int c3=14;
			int c4=20;
				System.out.printf("Y\\X| 0 1 2 3 4| p{Y=j}\n0| 0 0 "+c+"/"+g+" "+e+"/"+i+" "+c+"/"+j+"| "+c1+"/"+h+"\n1| 0"+" "+d+"/"+g+" "+d+"/"+f+" "+d+"/"+g+" 0| "+c2+"/"+g+"\n2| "+c+"/"+h+" "+e+"/"+g+" "+c+"/"+f +" 0 0| "+c1+"/"+c3+"\n3| "+c+"/"+i+" "+d+"/"+i+" 0 0 0| "+c+"/"+g+"\np{X=i}| "+c1+"/"+j+" "+c4+"/"+i+" "+c2+"/"+g+" "+c2+"/"+i+" "+c+"/"+j+"| 1\n");
  
   
}
else if (a==0){
  b=1; 
System.out.printf("%d", b);}
}
		}
	}
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