Help with Poisson process and conditional probability $N = \{N(t) | t \ge 0\}$ is a Poisson process with intensity $\lambda > 0$. Let T be the time of occurrence of the first event. Show that $T | N(t) = 1 \ \sim U(0,t)$.
Hint: It may be a good idea to start with
$P(T \le x | N(t) = 1), \ 0\le x\le t$ 
In the solution they say "It is clear by defintion of T that"
$P(T\le x | N(t) = 1) = 0 $ if $x< 0$ and $P(T\le x | N(t) = 1) = 1 $ if $x > t$ Why is this clear?
So take $0\le x \le t$ and observe that
$P(T \le x | N(t) = 1) = \frac{P(N(x) = 1, N(t) = 1)}{P(N(t) = 1}$
since $\{T \le x\} = \{N(x) \ge 1 \}$, but as $N(t)=1 $, we have $N(x) = 1$ for $x\le t$. How do we conclude this last part and what does it actually mean? 
The rest of the calculation to conclude the sought distribution are quite straight forward. 
 A: On your first question, $T$ is defined to be the time of the fist event, which must occur not before the initial time, hence $T \ge 0$. Also you are assuming that $N(t)=1$ so by time $t$, the first event happened already, hence $T \le t$. Thus,
$$
\mathbb{P}[T \le x | N(t) = 1]
 = \begin{cases}
   0, & x < 0 \\
   1, & x > t
   \end{cases}
$$
A: Let $N(t) \sim P(\lambda t)$. You are interested in the distribution of $\{T|N(t)=1\}$ on $0\le x\le t$. Hence, we can try to compute its cdf:
\begin{align}
F_T(x) &= P\{T\le x|N(t)=1\} = \frac{P(\{T\le x\}\cap \{N(t)=1\})}{P(\{N(t)=1\})},
\end{align} 
in the nominator we want to compute the probability that the first event happen before time $x$ and that the total number of events in the interval $[0,t]$ was $1$, i.e., the first and the only arrival was in $[0,x]$, so
\begin{align}
P(\{T\le x\}\cap \{N(t)=1\}) &= P(\{N(x)=1\} \cap \{N(t-x)=0\})\\
 &= \lambda xe^{-\lambda x} \lambda e^{-\lambda (t-x)}\\
&=\lambda x e^{-\lambda t}.
\end{align} 
Plugging the result in the cdf we get,   
\begin{align}
F_T(x) &=  \frac{\lambda x e^{-\lambda t}}{\lambda t e^{-\lambda t}}=\frac{x}{t}, & 0\le x\le t.
\end{align}
Hence, $\{T|N(t)=1\}\sim \mathrm{U}[0,t]$.
