当个人击中目标的概率给定时，A 赢得比赛的概率

给定四个整数a 、 b 、 c和d 。球员A和B尝试点球。 A 击中目标的概率为a / b而 B 击中目标的概率为c / d 。先点球的玩家获胜。任务是找出 A 赢得比赛的概率。
例子：

Input: a = 1, b = 3, c = 1, d = 3 
Output: 0.6
Input: a = 1, b = 2, c = 10, d = 11 
Output: 0.52381 
 

方法：如果我们考虑变量K = a / b作为 A 射中目标的概率， R = (1 – (a / b)) * (1 – (c / d))作为 A 和 B 的概率都没有命中目标。
因此，该解形成几何级数K * R ⁰ + K * R ¹ + K * R ² + …..其总和为(K / 1 – R) 。将K和R的值放入后，我们得到公式K * (1 / (1 – (1 – r) * (1 – k))) 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the probability of A winning
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
int main()
{
    int a = 1, b = 2, c = 10, d = 11;
    cout << getProbability(a, b, c, d);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function to return the probability
// of A winning
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;
    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

Javascript

 
// 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
 

输出：

0.52381