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++

filter_none

edit
close

play_arrow

link
brightness_4
code

#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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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 

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

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

chevron_right


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)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.