二项式随机变量

在本文中，我们将讨论二项式随机变量。

先决条件：随机变量

一种特定类型的离散随机变量，用于计算特定事件在固定次数的尝试或试验中发生的频率。

为了使变量成为二项式随机变量，必须满足以下所有条件：

1. 试验次数固定(样本量固定)。
2. 在每个试验中，感兴趣的事件要么发生，要么不发生。
3. 在每个试验中，发生(或不发生)的概率是相同的。
4. 审判是相互独立的。

数学符号

n = number of trials
p = probability of success in each trial
k = number of success in n trials

现在我们尝试找出n次试验中k次成功的概率。

在这里，每个试验的成功概率与其他试验无关，为p。
因此，我们首先选择将成功的k个试验，而在其余nk个试验中将失败的试验。这样做的方法数量是

由于所有n个事件都是独立的，因此n次试验中k次成功的概率等于每个试验的概率相乘。

在这里，它的k个成功和nk个失败，因此实现k个成功和nk个失败的每种方式的概率为

因此，最终概率为

(number of ways to achieve k success
 and n-k failures)
 *
(probability for each way to achieve k
 success and n-k failure)

然后，二项式随机变量概率由下式给出：

令X为二项式随机变量，试验次数为n，每次试验成功的概率为p。
预期的成功次数是

E[X] = np

成功次数的方差由下式给出：

Var[X] = np(1-p)

例1 ：考虑一个随机实验，其中将一枚有偏见的硬币(正面概率= 1/3)投掷10次。求出出现的正面个数为5的概率。

解决方案 ：

设X为n = 10且p = 1/3的二项式随机变量P(X = 5)=?

这是相同的实现

C++

// C++ program to compute Binomial Probability
#include 
#include 
using namespace std;
  
// function to calculate nCr i.e., number of 
// ways to choose r out of n objects
int nCr(int n, int r)
{
    // Since nCr is same as nC(n-r)
    // To decrease number of iterations
    if (r > n / 2)
        r = n - r;
  
    int answer = 1;
    for (int i = 1; i <= r; i++) {
        answer *= (n - r + i);
        answer /= i;
    }
  
    return answer;
}
  
// function to calculate binomial r.v. probability
float binomialProbability(int n, int k, float p)
{
    return nCr(n, k) * pow(p, k) *
                pow(1 - p, n - k);
}
  
// Driver code
int main()
{
  
    int n = 10;
    int k = 5;
    float p = 1.0 / 3;
  
    float probability = binomialProbability(n, k, p);
  
    cout << "Probability of " << k;
    cout << " heads when a coin is tossed " << n;
    cout << " times where probability of each head is " << p << endl;
    cout << " is = " << probability << endl;
}

Java

// Java program to compute Binomial Probability
  
import java.util.*;
  
class GFG
{
    // function to calculate nCr i.e., number of 
    // ways to choose r out of n objects
    static int nCr(int n, int r)
    {
        // Since nCr is same as nC(n-r)
        // To decrease number of iterations
        if (r > n / 2)
            r = n - r;
      
        int answer = 1;
        for (int i = 1; i <= r; i++) {
            answer *= (n - r + i);
            answer /= i;
        }
      
        return answer;
    }
      
    // function to calculate binomial r.v. probability
    static float binomialProbability(int n, int k, float p)
    {
        return nCr(n, k) * (float)Math.pow(p, k) * 
                        (float)Math.pow(1 - p, n - k);
    }
      
    // Driver code
    public static void main(String[] args)
    {
        int n = 10;
        int k = 5;
        float p = (float)1.0 / 3;
      
        float probability = binomialProbability(n, k, p);
          
        System.out.print("Probability of " +k);
        System.out.print(" heads when a coin is tossed " +n);
        System.out.println(" times where probability of each head is " +p);
        System.out.println( " is = " + probability );
    }
}
  
/* This code is contributed by Mr. Somesh Awasthi */

Python3

# Python3 program to compute Binomial 
# Probability
  
# function to calculate nCr i.e.,
# number of ways to choose r out
# of n objects
def nCr(n, r):
      
    # Since nCr is same as nC(n-r)
    # To decrease number of iterations
    if (r > n / 2):
        r = n - r;
  
    answer = 1;
    for i in range(1, r + 1):
        answer *= (n - r + i);
        answer /= i;
  
    return answer;
  
# function to calculate binomial r.v.
# probability
def binomialProbability(n, k, p):
  
    return (nCr(n, k) * pow(p, k) * 
                        pow(1 - p, n - k));
  
# Driver code
n = 10;
k = 5;
p = 1.0 / 3;
  
probability = binomialProbability(n, k, p);
  
print("Probability of", k, 
      "heads when a coin is tossed", end = " ");
print(n, "times where probability of each head is", 
                                      round(p, 6));
print("is = ", round(probability, 6));
  
# This code is contributed by mits

C#

// C# program to compute Binomial
// Probability.
using System;
  
class GFG {
      
    // function to calculate nCr
    // i.e., number of ways to 
    // choose r out of n objects
    static int nCr(int n, int r)
    {
          
        // Since nCr is same as
        // nC(n-r) To decrease 
        // number of iterations
        if (r > n / 2)
            r = n - r;
      
        int answer = 1;
        for (int i = 1; i <= r; i++)
        {
            answer *= (n - r + i);
            answer /= i;
        }
      
        return answer;
    }
      
    // function to calculate binomial
    // r.v. probability
    static float binomialProbability(
                  int n, int k, float p)
    {
        return nCr(n, k) * 
                   (float)Math.Pow(p, k)
                * (float)Math.Pow(1 - p,
                                 n - k);
    }
      
    // Driver code
    public static void Main()
    {
        int n = 10;
        int k = 5;
        float p = (float)1.0 / 3;
      
        float probability = 
              binomialProbability(n, k, p);
          
        Console.Write("Probability of "
                                      + k);
        Console.Write(" heads when a coin "
                       + "is tossed " + n);
        Console.Write(" times where "
           + "probability of each head is "
                                      + p);
        Console.Write( " is = "
                           + probability );
    }
}
  
// This code is contributed by nitin mittal.

PHP

 $n / 2)
        $r = $n - $r;
  
    $answer = 1;
    for ($i = 1; $i <= $r; $i++) {
        $answer *= ($n - $r + $i);
        $answer /= $i;
    }
  
    return $answer;
}
  
// function to calculate binomial r.v.
// probability
function binomialProbability($n, $k, $p)
{
    return nCr($n, $k) * pow($p, $k) *
                   pow(1 - $p, $n - $k);
}
  
// Driver code
    $n = 10;
    $k = 5;
    $p = 1.0 / 3;
  
    $probability = 
         binomialProbability($n, $k, $p);
  
    echo "Probability of " . $k;
    echo " heads when a coin is tossed "
                                    . $n;
    echo " times where probability of "
                 . "each head is " . $p ;
    echo " is = " . $probability ;
  
// This code is contributed by nitin mittal.
?>

输出：

Probability of 5 heads when a coin is tossed 10 times where probability of each head is 0.333333
 is = 0.136565

参考：
stat200