Probability for three randomly chosen numbers to be in AP

06 Apr, 2021
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
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) 
C++
Java
Python3
C#
PHP
Javascript
// CPP program to find probability that// 3 randomly chosen numbers form AP.#include <bits/stdc++.h>using namespace std; // function to calculate probabilitydouble procal(int n){    return (3.0 * n) / (4.0 * (n * n) - 1);} // Driver code to run above functionint 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 probabilitydef procal(n):     return (3.0 * n) / (4.0 * (n * n) - 1) # Driver codea = [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 probabilityfunction 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 probabilityfunction 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>
Output:  
0.151515
Time Complexity : O(1) 
nitin mittal
jit_t
_saurabh_jaiswal
arithmetic progression
Probability
Combinatorial
Mathematical
Mathematical
Combinatorial
Probability
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