Probability of AP

Suppose there are $(2n+1)$ tickets.

then,The Total number of elements in sample space is ${(2n+1) \choose 3}$

Now,we have to select any two numbers.the third will be the average of those two.

Now the restriction is that the mean of those two numbers should be an integer. Therefore we select two odd numbers or two even numbers.

This can be done in ${n \choose 2}+{n+1 \choose 2}$ ways.

therfore,answer is $\frac {{n \choose 2}+{n+1 \choose 2}}{{(2n+1) \choose 3}}$

The tickets are numbered as 1, 2, 3, .... (2n-1), n, (2n+1).

Number of ways to draw 3 tickets at random is (2n+1) C3.

Numbers that are in AP---

1) when common difference d=1, there are (2n-1) groups of 3 numbers each, which are in AP. (1,2,3), (2,3,4)...... {(2n-1), n, (2n+1)}

2) When difference d =2, we have (2n-3) groups in AP such as (1,3,5), (2,4,6)... {(2n-3),(2n-1), (2n+1)}

3)When d=3, we have (2n-5) groups..... and so on...

4) when d = n-1, we have 3 groups.

5) when d = n we have 1 group. {1, (n+1), (2n+1)}

Thus total number of groups which are in AP are in the form

1+3+ 5+..... (2n-5) + (2n-3) + (2n-1) which is itself an AP with first term 1 and common difference d= 2.

This sums up to n^2.

Required Probability = n^2 / (2n+1)C3

= 3n / (4n^2)-1

Sorry I am busy at the moment . But I'll just post the general from for $2n+1$ tickets .

P = $\dfrac{3n}{4n^{2} - 1 }$

Now just put $n=5000$ .

When we have 2n+1 tickets the number of triplets of tickets which satisfy the asked property is $n^2$ it's easly demonstrable. Then we just have to divide n^2 for $\binom{2n+1}{3}$