Question

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.

Answer 1

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 $=1—\{1,2,3\},\{2,3,4\},\{3,4,5\}\dots \{2n-1,2n,2n+1\}$

with common difference $=2—\{1,3,5\},\{2,4,6\},\{3,5,7\}\dots \{2n-3,2n-1,2n+1\}$

with common difference $=n—\{1,n+1,2n+1\}$

Therefore, Total number of AP group of $3$ numbers in $(2n+1)$ numbers are:

$=(2n–1)+(2n–3)+(2n–5)+\dots +3+1=n\times n$ (Sum of first $n$ odd numbers is $n\times n$ )

So, probability for $3$ randomly chosen numbers in $(2n+1)$ consecutive numbers to be in AP $=\frac{n\times n}{{}^{(2n+1)}{C}_3}=\frac{3n}{4n^2-1}$