Expectation from a colour-matching withdrawal game 
A person draws 3 balls from a bag containing 3 white, 4 red and 5 black balls. He is offered 10 bucks, 5 bucks, 2 bucks if he draws 3 ball of same color, 2 balls of same color and 1 ball of each color respectively. Find how much he expects to earn.

I tried to solve this question by calculating probability of getting 3 balls of same color that came $3/44$ but for solving balls of 2 same colors there are so many combination and calculation becomes lengthy so is there any way.

As you presumably found, there are ${12\choose3}=220$ ways to choose $3$ balls altogether, of which there are ${3\choose3}+{4\choose3}+{5\choose3}=1+4+10=15$ ways to choose them all of the same color.  It's also straightforward to count the number of ways of choose one ball of each color:  ${3\choose1}{4\choose1}{5\choose1}=60$.  This leaves
$$220-15-60=145$$
as the number of ways to choose two balls of the same color (and the third of another).  Thus the expected value for your game is
$$10\cdot{15\over220}+5\cdot{145\over220}+2\cdot{60\over220}={199\over44}$$
The key thing to realize here is that the three possibilities are mutually exclusive and account for everything that can happen, so if you can easily compute the probability for two of them, the probability for the third can be computed by complementarity.  
