Probability of a key K present in array

Given an array A[] and size of array is N and one another key K. The task is to find the probability that the Key K present in array.

Examples:

Input : N = 6
        A[] = { 4, 7, 2, 0, 8, 7, 5 }
        K = 3
Output :0
Since value of k = 3  is not present in array,
hence the probability of 0.

Input :N = 10
       A[] = { 2, 3, 5, 1, 9, 8, 0, 7, 6, 5 }
       K = 5
Output :0.2

The probability of can be found out using the below formula:



Probability = total number of K present /
                          size of array.

First, count the number of K’s and then the probability will be the number of K’s divided by N i.e. count / N.

Below is the implementation of the above approach:

C++

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// C++ code to find the probability of
// search key K present in array
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the probability
float kPresentProbability(int a[], int n, int k)
{
    float count = 0;
  
    for (int i = 0; i < n; i++) 
        if (a[i] == k)
            count++;
  
    // find probability
    return count / n;
}
  
// Driver Code
int main()
{
  
    int A[] = { 4, 7, 2, 0, 8, 7, 5 };
    int K = 3;
    int N = sizeof(A) / sizeof(A[0]);
    cout << kPresentProbability(A, N, K);
    return 0;
}

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Python3














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# Python3 code to find the  


# probability of search key 


# K present in 1D-array (list). 


  


# Function to find the probability 


def kPresentProbability(a, n, k) : 


  


    count = a.count(k) 


  


    # find probability upto 


    # 2 decimal places 


    return round(count / n , 2) 


  


# Driver Code 


if __name__ == "__main__" : 


      


    A = [ 4, 7, 2, 0, 8, 7, 5 ] 


    K = 2


    N = len(A) 


      


    print(kPresentProbability( A, N, K)) 


  


# This code is contributed 


# by AnkitRai1 



















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Java














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// Java code to find the probability  


// of search key K present in array 


class GFG 


{ 


  


// Function to find the probability 


static float kPresentProbability(int a[], 


                                 int n,  


                                 int k) 


{ 


    float count = 0; 


      


    for (int i = 0; i < n; i++)  


        if (a[i] == k) 


            count++; 


      


    // find probability 


    return count/ n; 


} 


  


// Driver Code 


public static void main(String[] args)  


{ 


    int A[] = { 4, 7, 2, 0, 8, 7, 5 }; 


    int K = 2; 


    int N = A.length; 


    double n = kPresentProbability(A, N, K); 


    double p = (double)Math.round(n * 100) / 100; 


    System.out.println(p); 


} 


} 


  


// This code is contributed 


// by ChitraNayal 



















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C#














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// C# code to find the probability  


// of search key K present in array 


using System; 


  


class GFG  


{ 


  


// Function to find the probability 


static float kPresentProbability(int[] a, 


                                 int n,  


                                 int k) 


{ 


    float count = 0; 


      


    for (int i = 0; i < n; i++)  


        if (a[i] == k) 


            count++; 


      


    // find probability 


    return count/ n; 


} 


  


// Driver Code 


public static void Main() 


{ 


    int[] A = { 4, 7, 2, 0, 8, 7, 5 }; 


    int K = 2; 


    int N = A.Length; 


    double n = kPresentProbability(A, N, K); 


    double p = (double)Math.Round(n * 100) / 100; 


    Console.Write(p); 


} 


} 


  


// This code is contributed 


// by ChitraNayal 



















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PHP














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<?php 


// PHP code to find the probability 


// of search key K present in array 


  


// Function to find the probability 


function kPresentProbability(&$a, $n, $k) 


{ 


    $count = 0; 


  


    for ($i = 0; $i < $n; $i++)  


        if ($a[$i] == $k) 


            $count++; 


  


    // find probability 


    return $count / $n; 


} 


  


// Driver Code 


$A = array( 4, 7, 2, 0, 8, 7, 5 ); 


$K = 2; 


$N = sizeof($A); 


echo round(kPresentProbability($A, $N, $K), 2); 


  


// This code is contributed 


// by ChitraNayal 


?> 



















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Output:


0.14





Time Complexity: O(N)



          
          
          
          

            

    

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Improved By :  AnkitRai01, chitranayal