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Probability of a key K present in array

  • Difficulty Level : Easy
  • Last Updated : 10 Mar, 2021

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++




// 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;
}

Python3




# 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

Java




// 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

C#




// 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

PHP




<?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
?>

Javascript




<script>
// JavaScript code to find the probability of
// search key K present in array
 
// Function to find the probability
function kPresentProbability(a,n,k)
{
    let count = 0;
 
    for (let i = 0; i < n; i++)
        if (a[i] == k)
            count+=1;
    // find probability
    return count / n;
}
 
// Driver Code
 
    let A = [ 4, 7, 2, 0, 8, 7, 5 ];
    let K = 3;
    let N = A.length;
    document.write(kPresentProbability(A, N,K));
     
// This code contributed by Rajput-Ji
 
</script>
Output



0

Time Complexity: O(N)
 

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