#偷懒系列#概统：组合数概率相关

缘起做概统作业的时候，需要计算一些二项分布相关的E(x)（数学期望）, V(x)（方差）等，由于涉及到组合数，计算量相当大，于是诞生了以下这个小工具。演示上图就是计算∑ i=0 20 C(20,i)p i (1−p) ( n−i ) \sum_{i=0}^{20} C(20, i)p^i(1-p)^(n-i^) 有关的。源代码可以到我的github：https://github.com

LoHiauFung

398人浏览 · 2016-09-29 00:40:16

LoHiauFung · 2016-09-29 00:40:16 发布

缘起

做概统作业的时候，需要计算一些二项分布相关的E(x)（数学期望）, V(x)（方差）等，由于涉及到组合数，计算量相当大，于是诞生了以下这个小工具。

演示

上图就是计算

\sum i = 0 20 C (20, i) p i (1 - p) (n - i)

$\sum_{i = 0}^{20} C(20, i)p^i(1-p)^(n-i^)$ 有关的。

源代码

可以到我的github：https://github.com/LoHiaufung/Lazy.git, 文件名为：CombinationalNumber.cpp
或者复制粘贴以下源代码:

#include <iostream>
#include <math.h>

using namespace std;

int main () {
    int n;
    cout << "Type in n: ";
    cin >> n;

    double* c = new double[n+1];
    c[0] = 1;
    for (int i = 1; i < n+1; i++) {
        c[i] = c[i-1] / i * (n - i + 1); 
    }

    double successProbability;
    cout << "Type in the success probability: ";
    cin >> successProbability;

    // 每个i对应的概率 
    double* sp = new double[n+1];
    for (int i = 0; i < n + 1; i++) {
        sp[i] = c[i] * pow(successProbability, i) * pow(1 - successProbability, n - i); 
    }

    //所求的加权、不加权部分和的起止[begin, end]
    int begin, end = 0;
    cout << "To get the sum,the begin index: ";
    cin >> begin;
    cout << "To get the sum,the end index: ";
    cin >> end;

    // 无权部分和
    double sum = 0;
    for (int i = begin; i < end + 1; i++) {
        sum += sp[i];
    }
    cout << "[begin, end] sum of probability is: " << sum << endl;

    // 加权部分和
    double sumWithWeight = 0;
    for (int i = begin; i < end + 1; i++) {
        sumWithWeight += sp[i] * i;
    }
    cout << "[begin, end] sum of probability with weight is: " << sumWithWeight << endl;

    // 数学期望
    double expectedValue = 0;
    for (int i = 0; i < n + 1; i++) {
        expectedValue += sp[i] * i;
    }
    cout << "[0, n] sum expected value is: " << expectedValue << endl;

    // 方差
    double variance = 0;
    for (int i = 0; i < n + 1; i++) {
        variance +=  (i - expectedValue) * (i - expectedValue) * sp[i];
    }
    cout << "[0, n] the variance is: " << variance << endl;
    cout << "[0, n] the standard deviation is: " << sqrt(variance) << endl;

    delete[] c;
    delete[] sp;
    return 0;
}