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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