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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