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.

I believe there is a 50% chance of getting a toss of more then 1 head in 3 tosses. 
If you look at a chart you will see that the possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. As you can see if you add up all the possible outcomes with more then 1 head you will get 4/8.
Therefore I believe it is an even chance of getting a toss of three dice with more than 1 head.
