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.

You had a good start. The expected value is calculated from the sum of the probability and the value associated with that probability. 
So, the expected value here would be the following:
$10$ $ 3 \choose 3 $ $12 \choose 3$^-1  + $10$$4 \choose 3$ $2 \choose 3$^-1 + $10$ $5 \choose 3$ $12 \choose 3$^-1  + $5$$ 3 \choose 2 $ $ 9 \choose 1$ $12 \choose 3$^-1 + $5$$4 \choose 2$ $8 \choose 1 $ $12 \choose 3$^-1 + $5$ $  5 \choose 2$ $7 \choose 1$ $12 \choose 3$^-1 + $ 3 \choose 1 $ $4 \choose 1$ $5 \choose 1$) $*12 \choose 3$^-1. 
Which, all laid out is:
$10 times the probability of drawing the three balls respectively, 
$5 times the probability of drawing two of the same and one different ball respectively,
and $1 times the probability of drawing all different balls.
Happy studying!
