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