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

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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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## 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 ` `?> ` |

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**Output:**

0.14

**Time Complexity: O(N)**

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