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 sample space has size $2^3 = 8$ and consists of triples
$$
\begin{array}{*{3}{c}}
H&H&H \\
H&H&T \\
H&T&H \\
H&T&T \\
T&H&H \\
T&H&T \\
T&T&H \\
T&T&T
\end{array}
$$
The events
$$
\begin{align}
\{ 0 \text{ heads} \} &= \{TTT\}, \\
\{ 1 \text{ head} \} &= \{HTT, THT, TTH\},
\end{align}
$$
and I'll let you figure out the other two.
The probabilities are, for example,
$$
P(\{ 1 \text{ head} \}) = \frac{3}{8}.
$$
This is called a binomial distribution, and the sizes of the events "got $k$ heads out of $n$ coin flips" are called binomial coefficients.
