Probability of getting K heads in N coin tosses

24 May, 2022
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++
Java
Python3
C#
Javascript
#include <bits/stdc++.h>using namespace std; // function to calculate factorialint fact(int n){    int res = 1;    for (int i = 2; i <= n; i++)        res = res * i;    return res;} // apply the formuladouble count_heads(int n, int r){    double output;    output = fact(n) / (fact(r) * fact(n - r));    output = output / (pow(2, n));    return output;} // Driver functionint main(){    int n = 4, r = 3;         // call count_heads with n and r    cout << count_heads(n, r);    return 0;}
class GFG{  // function to calculate factorialstatic int fact(int n){    int res = 1;    for (int i = 2; i <= n; i++)        res = res * i;    return res;}  // apply the formulastatic 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 functionpublic 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 program to find probability# of getting K heads in N coin tosses # Function to calculate factorialdef fact(n):         res = 1    for i in range(2, n + 1):        res = res * i    return res # Applying the formuladef count_heads(n, r):         output = fact(n) / (fact(r) * fact(n - r))    output = output / (pow(2, n))    return output # Driver coden = 4r = 3 # Call count_heads with n and rprint (count_heads(n, r)) # This code is contributed by Pratik Basu
using System; public class GFG{   // Function to calculate factorialstatic int fact(int n){    int res = 1;    for (int i = 2; i <= n; i++)        res = res * i;    return res;}   // Apply the formulastatic 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 functionpublic 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
<script> // Function to calculate factorialfunction fact(n){    var res = 1;    for(var i = 2; i <= n; i++)        res = res * i;             return res;} // Apply the formulafunction count_heads(n, r){    var output;    output = fact(n) / (fact(r) * fact(n - r));    output = output / (Math.pow(2, n));    return output;} // Driver Codevar n = 4, r = 3; // Call count_heads with n and rdocument.write(count_heads(n, r)); // This code is contributed by noob2000 </script>
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) 
29AjayKumar
sapnasingh4991
PratikBasu
noob2000
simmytarika5
Combinatorial
Mathematical
Mathematical
Combinatorial
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