Given four integers a, b, c and d. Player A & B try to score a penalty. Probability of A shooting the target is a / b while probability of B shooting the target is c / d. The player who scores the penalty first wins. The task is to find the probability of A winning the match.
Examples:
Input: a = 1, b = 3, c = 1, d = 3
Output: 0.6
Input: a = 1, b = 2, c = 10, d = 11
Output: 0.52381
Approach: If we consider variables K = a / b as the probability of A shooting the target and R = (1 – (a / b)) * (1 – (c / d)) as the probability that A as well as B both missing the target.
Therefore, the solution forms a Geometric progression K * R0 + K * R1 + K * R2 + ….. whose sum is (K / 1 – R). After putting the values of K and R we get the formula as K * (1 / (1 – (1 – r) * (1 – k))).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the probability of A winningdouble getProbability(int a, int b, int c, int d){ // p and q store the values // of fractions a / b and c / d double p = (double)a / (double)b; double q = (double)c / (double)d; // To store the winning probability of A double ans = p * (1 / (1 - (1 - q) * (1 - p))); return ans;}// Driver codeint main(){ int a = 1, b = 2, c = 10, d = 11; cout << getProbability(a, b, c, d); return 0;} |
Java
// Java implementation of the approachclass GFG {// Function to return the probability// of A winningstatic double getProbability(int a, int b, int c, int d) { // p and q store the values // of fractions a / b and c / d double p = (double) a / (double) b; double q = (double) c / (double) d; // To store the winning probability of A double ans = p * (1 / (1 - (1 - q) * (1 - p))); return ans;}// Driver codepublic static void main(String[] args){ int a = 1, b = 2, c = 10, d = 11; System.out.printf("%.5f", getProbability(a, b, c, d));}}// This code contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach # Function to return the probability# of A winning def getProbability(a, b, c, d) : # p and q store the values # of fractions a / b and c / d p = a / b; q = c / d; # To store the winning probability of A ans = p * (1 / (1 - (1 - q) * (1 - p))); return round(ans,5); # Driver code if __name__ == "__main__" : a = 1; b = 2; c = 10; d = 11; print(getProbability(a, b, c, d)); # This code is contributed by Ryuga |
C#
// C# implementation of the approach using System;class GFG{// Function to return the probability // of A winning public static double getProbability(int a, int b, int c, int d){ // p and q store the values // of fractions a / b and c / d double p = (double) a / (double) b; double q = (double) c / (double) d; // To store the winning probability of A double ans = p * (1 / (1 - (1 - q) * (1 - p))); return ans;}// Driver code public static void Main(string[] args){ int a = 1, b = 2, c = 10, d = 11; Console.Write("{0:F5}", getProbability(a, b, c, d));}}// This code is contributed by Shrikant13 |
PHP
<?php// PHP implementation of the approach// Function to return the probability // of A winningfunction getProbability($a, $b, $c, $d){ // p and q store the values // of fractions a / b and c / d $p = $a / $b; $q = $c / $d; // To store the winning probability of A $ans = $p * (1 / (1 - (1 - $q) * (1 - $p))); return round($ans,6);}// Driver code$a = 1;$b = 2;$c = 10;$d = 11;echo getProbability($a, $b, $c, $d);// This code is contributed by chandan_jnu?> |
Javascript
<script>// JavaScript implementation of the approach // Function to return the probability// of A winning function getProbability(a , b , c , d) { // p and q store the values // of fractions a / b and c / d var p = a / b; var q = c / d; // To store the winning probability of A var ans = p * (1 / (1 - (1 - q) * (1 - p))); return ans; } // Driver code var a = 1, b = 2, c = 10, d = 11; document.write( getProbability(a, b, c, d).toFixed(5));// This code contributed by aashish1995</script> |
0.52381
Time Complexity: O(1)
Auxiliary Space: O(1)
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