probability of sum of two integers less than an integer Two integers [not necessarily distinct] are chosen from the set {1,2,3,...,n}. What is the probability that their sum is <=k?
My approach is as follows. Let a and b be two integers. First we calculate the probability of the sum of a+b being equal to x [1<=x<=n]. WLOG let a be chosen first. For b= x-a to be positive, we must have 1<=a < x. This gives (x-1) possible values for a out of total n possible values. Probability of valid selection of a= (x-1)/n. For each valid selection of a, we have one and only one possible value of b. Only 1 value of b is then valid out of total n possible values. Thus probability of valid selection of b= 1/n. Thus probability of (a+b= x) = (x-1)/n(n-1).
Now probability of (a+b<=k)
= Probability of (a+b= 2) + probability of (a+b= 3) + ... + probability of (a+b= k) 
= {1+2+3+4+5+...+(k-1)}n(n-1)
= k(k-1)/n(n-1).
Can anybody please check if my approach is correct here?

Notice if $k\le 1$ the probability is $0$, and if $k\ge 2n$ the probability is $1$, so let's assume $2\le k\le 2n-1$.  For some $i$ satisfying $2\le i\le 2n-1$, how many ways can we choose $2$ numbers to add up to $i$?  If $i\le n+1$, there are $i-1$ ways.  If $i\ge n+2$, there are $2n-i+1$ ways.
Now, suppose $k\le n+1$, so by summing we find:
$$\sum_{i=2}^{k}i-1=\frac{k(k-1)}{2}$$
If $k\ge n+2$, if we sum from $i=2$ to $i=n+1$ we get $\frac{(n+1)n}{2}$, and then from $n+1$ to $k$ we get:
$$\sum_{i=n+2}^k2n-i+1=\frac{1}{2}(3n-k)(k-n-1)$$
Adding the amount for $i\le n+1$ we get:
$$2kn-\frac{k^2}{2}+\frac{k}{2}-n^2-n$$
Since there are $n^2$ choices altogether, we arrive at the following probabilities:
$$\begin{cases}\frac{k(k-1)}{2n^2}&1\le k\le n+1\\\frac{4kn-k^2+k-2n^2-2n}{2n^2}&n+2\le k\le 2n\end{cases}$$
