Probability of getting exactly 2 heads in 3 coins tossed with order not important? I have been thinking of this problem for the post 3-4 hours, I have come up with this problem it is not a home work exercise

Let's say I have 3 coins and I toss them, Here order is not important 

so possible sample space should be

0 H, 1 H, 2 HH, 3 HHH (H being heads)
  TTT, HTT, HHT, HHH

since P(T) and P(H) =1/2;
Here we have fair coins only, Since each and every outcome is equally likely, answer should be 

1/4 (is this correct)

and if that is correct, all of the probabilities don't add up to one, will I have to do the manipulation to make it add up to one, or I am doing anything wrong. 
EDIT
In my opinion, with order being not important, there should be only 4 possible outcomes. All of the answers have ignored that condition.

The outcomes you are looking for are either THH, HTH or HHT. Taking a look at for example THH: the possibility to toss T or H is $0.5$. Thus the possibility to throw T and then H and then H is $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$. But since we have three ways to "achieve" the desired result, the possibility of throwing exactly tow heads in three tosses is $\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$.
If you continue you like this, you'll find that the possibilites of all possible outcomes (THH, TTH, TTT, HTT ...) add up to 1 indeed.
