what is the probability that drawn balls have the same number? In the urn, there is one ball with number $1$, $2$ with number $2$, and so on until $n$ balls with the number $n$.  
From the urn, we draw two balls. 
Calculate the probability that the two drawn balls have the same number.
 A: Let $X$ denote the first chosen number and let $Y$ denote the second chosen number.
Then for $n\geq2$:
$$P(X=Y)=\sum_{k=2}^nP(X=k)P(Y=k\mid X=k)=\sum_{k=2}^n\frac{k}{\frac12n(n+1)}\frac{k-1}{\frac12n(n+1)-1}=$$$$\frac8{n(n+1)[n(n+1)-2]}\sum_{k=2}^n\binom{k}2=\frac8{(n-1)n(n+1)(n+2)}\binom{n+1}3=\frac{4}{3n+6}$$
A: Total number of balls is $$\frac{(n)\cdot (n+1)}{2}$$
So, sample space is $$^{(\frac{(n)\cdot (n+1)}{2})}C_2$$
You want to  select 2 balls of similar number. This starts from ball numbered 2.
Number of ways to select 2 balls from each group of same numbered balls is
$$^2C_2+^3C_2+.....+^nC_2$$ 

So, final solution is:
$$\frac{(^2C_2+^3C_2+.....+^nC_2)}{^{(\frac{(n)\cdot (n+1)}{2})}C_2}$$ 

Further simplification: $$ (^2C_2+^3C_2+.....+^nC_2)$$ 
can be written as $$\frac{1}{2} \cdot (2\cdot1+3\cdot2+4\cdot3+....+n\cdot(n-1))$$
Let $$2S=(2\cdot1+3\cdot2+4\cdot3+....+n\cdot(n-1))$$
$$2S=\Sigma_1^n  n\cdot(n-1)$$
$$2S=\frac{n\cdot(n^2-1)}{3}$$
So, $$S=\frac{n\cdot(n^2-1)}{6}$$
$$ ^\frac {n\cdot(n+1)}{2}C_2=\frac{1}{2}\cdot(\frac{n\cdot (n+1)}{2})\cdot(\frac{n^2+n-1}{2})$$

Giving you $$\frac{4}{3({n+2})} $$ 

A: Overall there are $N=1+2+\cdots+n=\frac{n(n+1)}{2}$ balls.
You need to calculate:
$$\begin{align}P(2\cap 2)+P(3\cap 3)+\cdots +P(n\cap n)=\\
\frac{2}{N}\cdot \frac1{N-1}+\frac{3}{N}\cdot \frac2{N-1}+\cdots+\frac{n}{N}\cdot \frac{n-1}{N-1}=\\
\frac{1}{N(N-1)}\sum_{k=2}^n k(k-1)=\\
\frac{1}{N(N-1)}\sum_{k=1}^n k(k-1)=\\
\frac1{N(N-1)} \left[\sum_{k=1}^n k^2-\sum_{k=1}^nk\right]=\\
\frac1{N(N-1)} \left[N\cdot \frac{2n+1}{3}-N\right]=\\
\frac{2(n-1)}{3(N-1)}=\\
\frac{4(n-1)}{3(n(n+1)-2)}=\\
\frac{4}{3(n+2)}.\\
\end{align}$$
