<s>[INST] 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.
 [/INST] 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 $$\frac{4}{3({n+2})} $$ 

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