What is the probability that these two objects are of the same color? We have $11$ bins with $10$ objects each. Every object is either black or white, and the $i$th bin ($1 \le i \le 11$) has precisely $(i -1)$ black objects in it. Someone selects, uniformly at random, one of those bins and then selects, also uniformly at random, two objects from it. What is the probability that these two objects are of the same color?
 A: \begin{align}P(WW)&=\sum_{i=1}^{11}P(WW\mid bin=i)P(bin=i)=\sum_{i=1}^{9}\frac{\dbinom{10-(i-1)}{2}}{\dbinom{10}{2}}\frac{1}{11}\\[0.2cm]&=\frac{1}{11\cdot45}\sum_{i=1}^9\dbinom{11-i}{2}=\frac{1}{11\cdot 45}\sum_{i=0}^{8}\dbinom{2+i}{2}\\[0.4cm]P(BB)&=\sum_{i=1}^{11}P(BB\mid bin=i)P(bin=i)=\sum_{i=3}^{11}\frac{\dbinom{i-1}{2}}{\dbinom{10}{2}}\frac{1}{11}\\[0.2cm]&=\frac{1}{11\cdot45}\sum_{i=1}^{9}\dbinom{11-i}{2}=P(WW)\end{align} and the probability $p$ that the two objects are of the same color is equal to $$p=P(WW)+P(BB)=2P(WW)$$

The sum of the binomials can be calculated as \begin{align}\sum_{i=0}^8\dbinom{2+i}{2}&=\sum_{i=0}^8\frac{(2+i)!}{2!i!}=\sum_{i=0}^8\frac{(2+i)(1+i)}{2}=\sum_{i=0}^81+\frac32\sum_{i=0}^8i+\frac12\sum_{i=0}^8i^2\\&=9+\frac32\cdot\frac{8(9)}{2}+\frac12\cdot\frac{8(9)(17)}{6}=165\end{align} and therefore $$p=2\cdot \frac{165}{11(45)}=\frac{2(15)}{45}=\frac23$$ Note: the result (if the calculations are correct) is actually 1/3 WW, 1/3 BB and 1/3 BW, (order does not matter in BW) and indicates perhaps that there might be a faster way to solve this by observing some symmetry. In other words, if one could argue that $p(WW)=p(BB)=P(BW)$, then this would solve the problem immediately, without all these calculations.
A: The number of ways of getting $2$ black balls $= \binom{0}{2} + \binom12 +\binom22 +..... +\binom{10}2 = \binom{11}{3}$
by properties of binomial coefficients (sum of binomial coefficients over upper index),
Thus $Pr$(both balls are of the same color) $= \dfrac{2\binom{11}{3}}{11\binom{10}2} = \dfrac23$ 
PS:
$$\text{Although the link gives it as} \sum_{j=0}^n \binom{j}{m} = \binom{n+1}{m+1}$$
$$\text{you can note that it reduces to} \sum_{j=m}^n \binom{j}{m} = \binom{n+1}{m+1}$$
which is the famous hockey stick identity 
