Get the probability out of a combination of events I'm learning probability. I'm having troubles with combinations. I think I'm not taking the right events of combinations. Please read the problem and tell me if what I did is correct. If not I appreciate your help.
Problem: One bag has 10 pairs of balls. If 8 are chosen, what's the probability of: a)No pair is chosen. b)1 pair is chosen,
For b), my solution: 
Ways of choosing balls: $\dbinom{20}{16}$ 
On pair is chosen: $\dbinom{16}{2}$
The rest of balls: $\dbinom{14}{14}$ 
The solution: $\dbinom{16}{2}$$\dbinom{14}{14}$/$\dbinom{20}{16}$ = 0.024


For a), I'm have no idea how to deal with it. Same combination of choosing balls 
Selecting half different pairs $\dbinom{8}{4}$, the other half $\dbinom{8}{4}$ 
The solution: $\dbinom{8}{4}$$\dbinom{8}{4}$/$\dbinom{20}{16}$ = 1.01, which is wrong


Thanks for your help
 A: It looks like eight balls are chosen from twenty, and those twenty are arranged in pairs (ten colors of balls, say).
If there are no pairs chosen, then all colors must be different.  To count the possibilities, first choose the eight colors that you pick out: ${10 \choose 8} = 45$ possibilities.  Then, pick which ball of each pair you pick out: $2^8 = 256$ possibilities.
This gives the probability of picking no pairs as
$$P_0 = \frac{45 \cdot 256}{{20 \choose 8}} \approx 0.0915. $$
For the second part, if the probability asked for is at least one pair, then you're practically done:
$$P_{\geq 1} = 1 - P_0.$$
If instead you're looking for exactly one pair, then it's a bit more work.
Pick the color for the pair $(10)$, pick the remaining six colors $({9 \choose 6} = 84)$, and pick which ball from each pair for those six $(2^6 = 64)$.
This gives
$$P_1 = \frac{10 \cdot 84 \cdot 64}{{20 \choose 8}} \approx 0.4268.$$
As a check, you can verify that
$$\sum_{n=0}^4 {10 \choose n} {10 - n \choose 8 - 2n} 2^{8-2n} = {20 \choose 8}.$$
A: I'm a little bit confused by your wording. But I'll do my best to answer anyway and you can tell me if any of my assumptions are wrong.
So we're given a bag with 20 distinct balls total. There are 2 balls each of 10 different colors. We want to draw 8 individual balls from the bag and find the probability that exactly two are the same color.
The first mistake I think you made is that I think they mean 8 balls are chosen, but you thought they meant 8 pairs were chosen. (16 balls) If that were the case then the pigeonhole principle would tell us that the probability of exactly 2 being the same color was 0.
The actual number of ways of choosing balls is probably $20 \choose 8$
I'm not sure where you got $16 \choose 2$, since there are 10 different colors, there are 10 ways to choose a single pair. Then there are 9 colors left, and we have to choose 1 ball from 6 of them, and 0 from the others, giving us $2^6{9 \choose 6}$ ways to pick the unpaired balls. so b) should be $10*2^6{9 \choose 6}{20 \choose 8}^{-1}$
Part a is actually a little bit simpler, we have to choose 8 balls of 20 but we can't choose more than one of the same color. So we choose 8 pairs of the 10 to select a ball from, and then select 1 of the 2 balls in each, giving us $2^8{10 \choose 8}$ possibilities.
