# 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 = 3Output :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 = 5Output :0.2

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend **live classes **with experts, please refer **DSA Live Classes for Working Professionals **and **Competitive Programming Live for Students**.

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 <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)**