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# 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++

 `// 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

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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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