Three squares of a chess board being randomly chosen at random, what is the chance that two are of one color and one of other? I applied this concept;  
$\Rightarrow$ there are total of 64 squares out of which 32 are white and others are black.  
Now I considered two cases that, (1) the two squares of same colour are white and the other is black. (2) the two squares of same colour are black and the other is white.
thus P(E) = (2*${32 \choose 2}$*${32 \choose 1}$)/(${64 \choose 3}$).
Am I right?
 A: The number of ways to choose two black and one white square is ${32 \choose 2}32$ with the first factor from choosing the two black squares and the second from choosing the white square.  Two white and one black is the same by symmetry.  There are ${64 \choose 3}$ ways to choose three squares, so the chance you want is 
$$\frac {2\cdot {32 \choose 2}\cdot 32}{{64 \choose 3}}=\frac {16}{21}$$
which is nicely less than $1$.
A: There are four cases:


*

*three white square, which happens with probability $\binom{32}{3}/\binom{64}{3} = \frac{5}{42}$,

*two white squares and one black square, which happens with probability $\binom{32}{2} \binom{32}{1} / \binom{64}{3} = \frac{8}{21}$, 

*one white square and two black squares, which happens with probability $\binom{32}{1} \binom{32}{2} / \binom{64}{3} = \frac{8}{21}$, and 

*three black squares, which happens with probability $\binom{32}{3} / \binom{64}{3} = \frac{5}{42}$.


These four probabilities sum to $1$.
A: Another approach is to consider the complement that all are one color, yielding:
$$1-2\binom{32}{3}/\binom{64}{3}=\frac{16}{21}.$$
