Probability of getting K heads in N coin tosses


Given two integers N and R. The task is to calculate the probability of getting exactly r heads in n successive tosses.

A fair coin has an equal probability of landing a head or a tail on each toss.

Examples:



Input : N = 1, R = 1 
Output : 0.500000 

Input : N = 4, R = 3
Output : 0.250000 

Approach
Probability of getting K heads in N coin tosses can be calculated using below formula:

     \[\frac{1}{2^n} * \frac{n!}{ r! * (n-r)!}\]

Below is the implementation of the above approach:

C++

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#include <bits/stdc++.h>
using namespace std;
  
// function to calculate factorial
int fact(int n)
{
    int res = 1;
    for (int i = 2; i <= n; i++)
        res = res * i;
    return res;
}
  
// apply the formula
double count_heads(int n, int r)
{
    double output;
    output = fact(n) / (fact(r) * fact(n - r));
    output = output / (pow(2, n));
    return output;
}
  
// Driver function
int main()
{
    int n = 4, r = 3;
      
    // call count_heads with n and r
    cout << count_heads(n, r);
    return 0;
}

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Java

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class GFG{
   
// function to calculate factorial
static int fact(int n)
{
    int res = 1;
    for (int i = 2; i <= n; i++)
        res = res * i;
    return res;
}
   
// apply the formula
static double count_heads(int n, int r)
{
    double output;
    output = fact(n) / (fact(r) * fact(n - r));
    output = output / (Math.pow(2, n));
    return output;
}
   
// Driver function
public static void main(String[] args)
{
    int n = 4, r = 3;
       
    // call count_heads with n and r
    System.out.print(count_heads(n, r));
}
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python3 program to find probability 
# of getting K heads in N coin tosses
  
# Function to calculate factorial 
def fact(n):
      
    res = 1
    for i in range(2, n + 1): 
        res = res * i
    return res
  
# Applying the formula 
def count_heads(n, r):
      
    output = fact(n) / (fact(r) * fact(n - r))
    output = output / (pow(2, n))
    return output
  
# Driver code
n = 4
r = 3
  
# Call count_heads with n and r
print (count_heads(n, r))
  
# This code is contributed by Pratik Basu 

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C#

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using System;
  
public class GFG{
    
// Function to calculate factorial
static int fact(int n)
{
    int res = 1;
    for (int i = 2; i <= n; i++)
        res = res * i;
    return res;
}
    
// Apply the formula
static double count_heads(int n, int r)
{
    double output;
    output = fact(n) / (fact(r) * fact(n - r));
    output = output / (Math.Pow(2, n));
    return output;
}
    
// Driver function
public static void Main(String[] args)
{
    int n = 4, r = 3;
        
    // Call count_heads with n and r
    Console.Write(count_heads(n, r));
}
}
// This code contributed by sapnasingh4991

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Output:

0.250000

Time Complexity: In this implementation, we have to calculate factorial based on the value n, so time complexity would be O(n)
Auxiliary Space: In this implementation, we are not using any extra space, so auxiliary space required is O(1)

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