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:
![\[\frac{1}{2^n} * \frac{n!}{ r! * (n-r)!}\]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-6e4e45e4ded2f05a8b5cab9cba5e3516_l3.png "Rendered by QuickLaTeX.com")
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
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 = 4r = 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 |
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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