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.
 A: The calculation for "2 balls of the same colour" isn't that hard. Suppose the same-coloured balls are white; there are $\binom32$ ways to pick the white balls and 9 choices for the last, non-white ball. Similarly for the red and black balls, leading to
$$\frac{9\binom32+8\binom42+7\binom52}{\binom{12}3}=\frac{29}{44}$$
By complementation, the probability that three different-colour balls are picked is $\frac3{11}$ and the expectation falls quickly:
$$E=10\cdot\frac3{44}+5\cdot\frac{29}{44}+2\cdot\frac3{11}=\frac{199}{44}=4.5227\dots$$
A: 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$$4 \choose 2$ $8 \choose 1 $ $12 \choose 3$^-1 + $5$ $  5 \choose 2$ $7 \choose 1$ $12 \choose 3$^-1 + $ 3 \choose 1 $ $4 \choose 1$ $5 \choose 1$) $*12 \choose 3$^-1. 
Which, all laid out is:
$10 times the probability of drawing the three balls respectively, 
$5 times the probability of drawing two of the same and one different ball respectively,
and $1 times the probability of drawing all different balls.
Happy studying!
A: As you presumably found, there are ${12\choose3}=220$ ways to choose $3$ balls altogether, of which there are ${3\choose3}+{4\choose3}+{5\choose3}=1+4+10=15$ ways to choose them all of the same color.  It's also straightforward to count the number of ways of choose one ball of each color:  ${3\choose1}{4\choose1}{5\choose1}=60$.  This leaves
$$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.  
