Cumulative Probability Distribution Example Question 
Flip a fair coin three times.  let T be the random variable that denotes the number of tails that occur given that at least one head occurred.  Calculate the Probability and Cumulative Distribution Functions.

I was having a little trouble with this question so I decided to enumerate all the possibilities to get started and labeled the value of T in each.  
I thought that the outcome "ttt" is removed from the sample space because the question stem says "given that at least one head occurred."  Is this an incorrect assumption?
So the total number of possibilites is $2^3 -1 = 7$.
$
\begin{array}{c|c}
\text{outcome} & \text{T} \\
\hline
hhh & 0 & \binom{3}{0}=1\\
\hline
thh & 1 \\
hth & 1 & \binom{3}{1}=3 \\
hht & 1 \\
\hline
tth & 2 \\
tht & 2 & \binom{3}{2}=3 \\
htt & 2 \\
\hline
ttt & N/A & N/A\\
\end{array}
$
Since this is fair coin toss, each coin toss has a probility of 0.5 and each outcome has a probability of $0.5^3$.  
So, I thought the Probability function would be $\Pr(T=t)= \binom{3}{t} \cdot 0.5^3$, but this is incorrect.  The solutions list:
$\Pr[T=t]=\Pr[H=3-t \mid H \ge 1] = \cfrac{ \binom{3}{3-t} \cdot \frac{1}{8}}{\frac{7}{8}}\tag{1}$
I did another example which I thought was very similar:  

Flip a loaded coin four times.  Assume that Pr(head) = 0.7.  Let the random variable H denote the number of heads that result.  Find the Probability Distribution Function.

I correctly got $\Pr(H=h)=0.7^h \cdot 0.3^{4-h} \cdot \binom{4}{h}$ for $0 \le h \le 4$.  Here there's $\binom{4}{h}$ ways of getting a set for a particular value of heads and you multiply by the probabilities of heads and tails.  
I realize that the first question is a conditional probability question while the 2nd is not.  However, if I limit the sample space as I did from the beginning, I thought my approach was legitimate.  What is going on?  Thanks in advance.
 A: In general, if $A$ and $B$ are events, then the conditional probability of $A$, given that $B$ has happened, (in symbols, $\Pr(A|B)$), is defined by the formula
$$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}.$$
In our case, we have $B$ is the event "at least $1$ head." Note that $\Pr(B)=\frac{7}{8}$.
To find the probability of exactly $1$ tail, given that there is at least $1$ head, let $A$ be the event "exactly $1$ tail." Then $A\cap B$ is also the event "exactly $1$ tail," and, as you calculated, has probability $\frac{3}{8}$. It follows that the conditional probability of exactly $1$ tail, given there is at least one, is, by the displayed formula above, equal to
$$\frac{\frac{3}{8}}{\frac{7}{8}},$$
which simplifies to $\frac{3}{7}$.
Or we can think of the events HHH, HHT, HTH, HTT, THH, THT, and TTH as equally likely, given there is at least one head. So each has (conditional) probability $\frac{1}{7}$. Now for the conditional probabilities we can count. I do not advocate this approach, since the defining formula is far more versatile.  
In the case of your random variable $T$, it turns out to have values $0$, $1$, or $2$, with probabilities $\frac{1}{7}$, $\frac{3}{7}$, and $\frac{3}{7}$ respectively. Note that $T$ is an ordinary random variable, but not the same random variable as the (unconditional) number of heads, since we are disregarding any experiment in which we got $0$ heads. In particular, the probabilities add to $1$. (The probabilities you computed did not.) 
Now we calculate the cumulative distribution function $F_T(t)$, which, as you know, is $\Pr(T\le t)$.
If $t\lt 0$, then $F_T(t)=0$.
If $0\le t\lt 1$, then $F_T(t)=\frac{1}{7}$.
If $1\le t\lt 2$, then $F_T(t)=\frac{4}{7}$.
Finally, if $t\ge 2$, then $F_T(t)=1$.  
