# Maximizing Probability of one type from N containers

Given N containers consisting of N copies of number A and N copies of number B. We need to arrange the numbers in the N containers in such a way that the probability of choosing a container randomly and drawing one copy of number A is maximum.

Examples:

```
Input : N = 1
Output : 0.5,

Input : N = 2
Output : 0.667

```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We need to look for an optimal arrangement of the N copies of number A and N copies of number B in N containers.

```Pi = (Favorable Outcomes) / (Total Outcomes)

where Pi denotes the probability of an event i.
```

So to maximize this, either we minimize the denominator (i.e., the total outcomes) or maximise the numerator (i.e., the total favorable Outcomes) keeping the other constant.

Consider a possible arrangement where in the first (N-1) containers we put only copies of Number A, such that first (N-1) containers contain (N-1) copies of the Number A and no copy of number B. The left out copy of number A and the N copies of number B are then put in the last container.

Probability (copy of A from any of the first (N-1) containers) = Pn-1 containers = 1

Probability (copy of A from Nth container) = PN = 1(N+1)

Pmax = PN-1 containers * (N – 1) + PN

∴ Pmax = N / (N + 1)

## C++

 `// CPP program to find maximum probability of ` `// getting a copy 'A' from N ` `#include ` `using` `namespace` `std; ` ` `  `// Returns the Maximum probability for Drawing ` `// 1 copy of number A from N containers with N ` `// copies each of numbers A and B ` `double` `calculateProbability(``int` `N) ` `{ ` `    ``// Pmax = N/(N+1) ` `    ``double` `probability = (``double``)N / (N + 1); ` `    ``return` `probability; ` `} ` ` `  `int` `main() ` `{ ` `    ``int` `N; ` `    ``double` `probabilityMax; ` ` `  `    ``// 1. N = 1 ` `    ``N = 1; ` `    ``probabilityMax = calculateProbability(N); ` `    ``cout << ``"Maximum Probability for N = "`  `         ``<< N << ``" is, "` `<< setprecision(4)  ` `        ``<< fixed << probabilityMax << endl; ` ` `  `    ``// 2. N = 2 ` `    ``N = 2; ` `    ``probabilityMax = calculateProbability(N); ` `    ``cout << ``"Maximum Probability for N = "`  `         ``<< N << ``" is, "` `<< setprecision(4)  ` `         ``<< fixed << probabilityMax << endl; ` ` `  `    ``// 3. N = 10 ` `    ``N = 10; ` `    ``probabilityMax = calculateProbability(N); ` `    ``cout << ``"Maximum Probability for N = "`  `         ``<< N << ``" is, "` `<< setprecision(4)  ` `         ``<< fixed << probabilityMax << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find maximum probability of ` `// getting a copy 'A' from N ` `class` `GFG { ` `     `  `    ``// Returns the Maximum probability for Drawing ` `    ``// 1 copy of number A from N containers with N ` `    ``// copies each of numbers A and B ` `    ``static` `double` `calculateProbability(``int` `N) ` `    ``{ ` `         `  `        ``// Pmax = N/(N+1) ` `        ``double` `probability = (``double``)N / (N + ``1``); ` `         `  `        ``return` `probability; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `         `  `        ``int` `N; ` `        ``double` `probabilityMax; ` ` `  `        ``// 1. N = 1 ` `        ``N = ``1``; ` `        ``probabilityMax = calculateProbability(N); ` `         `  `        ``System.out.println(``"Maximum Probability for"` `                  ``+ ``" N = "` `+ N + ``" is, "` `+ Math.round( ` `                  ``probabilityMax * ``10000.0``) / ``10000.0``); ` ` `  `        ``// 2. N = 2 ` `        ``N = ``2``; ` `        ``probabilityMax = calculateProbability(N); ` `        ``System.out.println(``"Maximum Probability for N = "` `                          ``+ N + ``" is, "` `+ Math.round( ` `                     ``probabilityMax * ``10000.0``) / ``10000.0``); ` ` `  `        ``// 3. N = 10 ` `        ``N = ``10``; ` `        ``probabilityMax = calculateProbability(N); ` `        ``System.out.println(``"Maximum Probability for N = "` `                            ``+ N + ``" is, "` `+ Math.round( ` `                     ``probabilityMax * ``10000.0``) / ``10000.0``); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# Python3 program to find maximum ` `# probability of getting a copy 'A' from N ` ` `  `# Returns the Maximum probability for  ` `# Drawing 1 copy of number A from N  ` `# containers with N copies each of ` `# numbers A and B ` `def` `calculateProbability(N): ` ` `  `    ``# Pmax = N / (N+1) ` `    ``probability ``=` `N ``/` `(N ``+` `1``) ` `    ``return` `probability ` ` `  `# Driver code ` ` `  `# 1. N = 1 ` `N ``=` `1` `probabilityMax ``=` `calculateProbability(N) ` `print``(``"Maximum Probability for N = "``, ` `       ``N , ``"is, %.4f"` `%``probabilityMax) ` ` `  `# 2. N = 2 ` `N ``=` `2` `probabilityMax ``=` `calculateProbability(N); ` `print``(``"Maximum Probability for N ="``, ` `       ``N, ``"is, %.4f"` `%``probabilityMax) ` ` `  `# 3. N = 10 ` `N ``=` `10` `probabilityMax ``=` `calculateProbability(N); ` `print``(``"Maximum Probability for N ="``, ` `       ``N,``"is, %.4f"` `%``probabilityMax) ` ` `  `# This code is contributed by Anant Agarwal. `

## C#

 `// C# program to find maximum probability of ` `// getting a copy 'A' from N ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Returns the Maximum probability for Drawing ` `    ``// 1 copy of number A from N containers with N ` `    ``// copies each of numbers A and B ` `    ``static` `double` `calculateProbability(``int` `N) ` `    ``{ ` `         `  `        ``// Pmax = N/(N+1) ` `        ``double` `probability = (``double``)N / (N + 1); ` `        ``return` `probability; ` `    ``} ` `     `  `    ``//Driver code ` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``int` `N; ` `        ``double` `probabilityMax; ` `      `  `        ``// 1. N = 1 ` `        ``N = 1; ` `        ``probabilityMax = calculateProbability(N); ` `        ``Console.WriteLine(``"Maximum Probability for N = "` `             ``+ N + ``" is, "` `+ ` `        ``Math.Round(probabilityMax * 10000.0) / 10000.0); ` `      `  `        ``// 2. N = 2 ` `        ``N = 2; ` `        ``probabilityMax = calculateProbability(N); ` `        ``Console.WriteLine(``"Maximum Probability for N = "` `             ``+ N + ``" is, "` `+  ` `        ``Math.Round(probabilityMax * 10000.0) / 10000.0); ` `      `  `        ``// 3. N = 10 ` `        ``N = 10; ` `        ``probabilityMax = calculateProbability(N); ` `        ``Console.WriteLine(``"Maximum Probability for N = "` `             ``+ N + ``" is, "` `+ ` `        ``Math.Round(probabilityMax * 10000.0) / 10000.0); ` `      `  `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## PHP

 ` `

Output:

```Maximum Probability for N = 1 is, 0.5000
Maximum Probability for N = 2 is, 0.6667
Maximum Probability for N = 10 is, 0.9091
```

Time Complexity of the above program is O(1).

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