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Probability for three randomly chosen numbers to be in AP

  • Difficulty Level : Medium
  • Last Updated : 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

 

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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++




// 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




// 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




# 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#




// 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
// 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
?>

Javascript




<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>

Output: 
 

0.151515

Time Complexity : O(1)
 




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