CGPET 2022 Solved Paper

Out of (2n+1) tickets, we have (n+1)tickets having odd number written on them and n tickets have even number. For three numbers to be in AP either two number from odd or two numbers from even numbers must be selected.
∴ Required probability =

‌n+1C2+‌nC2

‌2n+1C3

=

(n+1)n 2×1 + n(n−1) 2×1

(2n+1)(2n)(2n−1) 3×2×1

=

3[n(n+1)+n(n−1)]

2n(2n+1)(2n−1)

=

3n(n+1+n−1)

2n(2n+1)(2n−1)

=

3n

4n2−1