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.
 A: Consider all the possible ways to get two heads, $\rm HHT, HTH \; and \; THH$. There are $2 \cdot 2 \cdot 2 = 8$ possible combinations in total. Therefore, the  answer is $3/8$.
Your answer is wrong because the number of ways of changing around $\rm HHT$ (3) is not the same as the number of ways of changing around $\rm HHH$ (1). Can you see why this would invalidate your argument?
General solution: Binomial distribution. The probability of getting $k$ successes (here $2$) in $n$ trials (here $3$) is given by:
$$ \Pr(x=k) = \binom{n}{k} p^k (1-p)^{n-k}$$
Where $p$ is the probability of success (here, $p=1/2$), and $\binom{n}{k} = n!/(k!(n-k)!)$. This gives us:
$$ \binom{3}{2} \left(\frac12\right)^2 \left(1 - \frac12\right)^{3-2} $$
$$ 3 \cdot \frac14 \cdot \frac12 $$
$$ \frac38 $$
A: 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.
A: 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.
A: 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.
A: You are looking for two heads and nothing more. List the possibilities:
HHH HTH HHT HTT THH THT TTH TTH

There are 3 cases out of the 8 with two heads: 
HTH HHT THH

The answer is $\frac38$.
A: Saying "order is not important" does not mean that each scenario (0, 1, 2, 3 heads) is equally likely.  The probability of getting one head, for example, means that you got one head first, one head second, or one head third.  There are 8 possible outcomes, so the probability of one head must be 3/8.
A: 
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

So far, okay.

Since each and every outcome is equally likely

This statement is not true. If you flip three coins, then HHT is going to occur more often than HHH. The four outcomes are not equally likely.
The formula to calculate the probabilities is given in the answer by George V. Williams:
Pr(x = k) = (n C k) * [ p^k ] * [ (1-p)^(n-k) ]
Where:
(n C k) = n!/(k! * (n-k)!)
In the case of three fair coins, n = 3 and p = 0.5:
TTT (k=0 and HHH (k=3) both have probability 1/8 each.
HTT (k=1) and HHT (k=2) each have probability 3/8 each.
Summary:
If order is not important, then there are four outcomes, but with different probabilities.
If order was important, then there would be eight outcomes, with equal probability.
