Probability for three randomly chosen numbers to be in AP

Given a number n and an array containing 1 to (2n+1) consecutive numbers. Three elements are chosen at random. Find the probability that the elements chosen are in A.P.

Examples:  

Input : n = 2
Output : 0.4
Explanation:
The array would be {1, 2, 3, 4, 5}
Out of all elements, triplets which are in AP: {1, 2, 3}, {2, 3, 4}, {3, 4, 5}, {1, 3, 5}
No of ways to choose elements from the array: 10 (⁵C₃) 
So, probability = 4/10 = 0.4

Input : n = 5
Output : 0.1515

The number of ways to select any 3 numbers from (2n+1) numbers are:^{(2n + 1)} C ₃ 
Now, for the numbers to be in AP: 
with common difference 1---{1, 2, 3}, {2, 3, 4}, {3, 4, 5}...{2n-1, 2n, 2n+1} 
with common difference 2---{1, 3, 5}, {2, 4, 6}, {3, 5, 7}...{2n-3, 2n-1, 2n+1} 
with common difference n--- {1, n+1, 2n+1} 
Therefore, Total number of AP group of 3 numbers in (2n+1) numbers are: 
(2n - 1)+(2n - 3)+(2n - 5) +...+ 3 + 1 = n * n (Sum of first n odd numbers is n * n ) 
So, probability for 3 randomly chosen numbers in (2n + 1) consecutive numbers to be in AP = (n * n) / ^{(2n + 1)} C ₃ = 3 n / (4 (n * n) - 1)
 

Below is the implementation of the above approach:

// CPP program to find probability that 
// 3 randomly chosen numbers form AP.
#include <bits/stdc++.h>
using namespace std;

// function to calculate probability
double procal(int n)
{
    return (3.0 * n) / (4.0 * (n * n) - 1);
}

// Driver code to run above function
int main()
{
    int a[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(a)/sizeof(a[0]);
    cout << procal(n);
    return 0;
}

// Java program to find probability that
// 3 randomly chosen numbers form AP.

class GFG {
    
    // function to calculate probability
    static double procal(int n)
    {
        return (3.0 * n) / (4.0 * (n * n) - 1);
    }

    // Driver code to run above function
    public static void main(String arg[])
    {
        int a[] = { 1, 2, 3, 4, 5 };
        int n = a.length;
        System.out.print(Math.round(procal(n) * 1000000.0) / 1000000.0);
    }
}

// This code is contributed by Anant Agarwal.

# Python3 program to find probability that 
# 3 randomly chosen numbers form AP.

# Function to calculate probability
def procal(n):

    return (3.0 * n) / (4.0 * (n * n) - 1)

# Driver code 
a = [1, 2, 3, 4, 5] 
n = len(a)
print(round(procal(n), 6))

# This code is contributed by Smitha Dinesh Semwal.

// C# program to find probability that
// 3 randomly chosen numbers form AP.
using System;

class GFG {
    
    // function to calculate probability
    static double procal(int n)
    {
        return (3.0 * n) / (4.0 * (n * n) - 1);
    }

    // Driver code
    public static void Main()
    {
        int []a = { 1, 2, 3, 4, 5 };
        int n = a.Length;
        Console.Write(Math.Round(procal(n) *
                    1000000.0) / 1000000.0);
    }
}

// This code is contributed by nitin mittal

<?php
// PHP program to find probability that 
// 3 randomly chosen numbers form AP.

// function to calculate probability
function procal($n)
{
    return (3.0 * $n) / 
           (4.0 * ($n * 
              $n) - 1);
}

    // Driver code 
    $a = array(1, 2, 3, 4, 5);
    $n = sizeof($a);
    echo procal($n);
    
// This code is contributed by aj_36
?>

<script>
// Javascript program to find probability that 
// 3 randomly chosen numbers form AP.

// function to calculate probability
function procal(n)
{
    return (3.0 * n) / 
           (4.0 * (n * 
              n) - 1);
}

    // Driver code 
    let a = [1, 2, 3, 4, 5];
    let n = a.length;
    document.write(procal(n));
    
// This code is contributed by _saurabh_jaiswal
</script>

0.151515

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