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 of binomial distribution of probability: 
where p = probability of getting head and q = probability of getting tail. p and q both are 1/2. So the equation becomes 

Below is the implementation of the above approach:  

C++

#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;
}

Java

import java.io.*;
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

Python3

# 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

C#

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

Javascript

<script>

// Function to calculate factorial
function fact(n)
{
    var res = 1;
    for(var i = 2; i <= n; i++)
        res = res * i;

    return res;
}

// Apply the formula
function count_heads(n, r)
{
    var output;
    output = fact(n) / (fact(r) * fact(n - r));
    output = output / (Math.pow(2, n));
    return output;
}

// Driver Code
var n = 4, r = 3;

// Call count_heads with n and r
document.write(count_heads(n, r));

// This code is contributed by noob2000

</script>

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)