# Probability of a key K present in array

• Difficulty Level : Easy
• Last Updated : 30 Aug, 2022

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 Kâ€™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 ``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

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## Javascript

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Output

`0`

Complexity Analysis:

• Time Complexity: O(N)
• Auxiliary Space: O(1)

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