# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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); ` `    ``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

 ` `

Output:

```0.14
```

Time Complexity: O(N)

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Improved By : AnkitRai01, chitranayal