Exponential distribution and probability of a single occurance. The occurrence of events are exponentially distributed. I want to know what is the probability that there will be exactly one event between $0$ and $T$
Now, there are two events $x_1$ and $x_2$ and i want to make sure that only $x_1$ happens between $0$ and $T$
so i can write $\Pr(x_1≤t\mid x_1≤T, x_1+x_2>T)$ 
when $0≤t≤T$ and $t$ is the moment when $x_1$ occurs.
Now $\Pr(x_1≤t,x_1+x_2>T)=\underbrace{\Pr(x_1≥t,x_2>T-x_1)}_\text{is this correct?}$
The problem here is that im not sure if the under braced probability is correct. Any thoughts ?
 A: The event of interest is $A=[X\leqslant T\lt X+Y]$ where $X$ and $Y$ are i.i.d. and exponentially distributed. For every $x$ in $(0,T)$, $P(A\mid X=x)=P(Y\gt T-x\mid X=x)=P(Y\gt T-x)$. For every $x\gt T$, $P(A\mid X=x)=0$. Hence, $$P(A)=\int_0^\infty P(A\mid X=x)f_X(x)\mathrm dx=\int_0^TP(Y\gt T-x)f_X(x)\mathrm dx.$$ Can you finish?
As a way of checking your result, note that the number $N_T$ of points of the process falling in $(0,T)$ is Poisson with parameter $\lambda T$ and that you are asking for $P(A)=P(N_T=1)$.
A: "The occurrence of events are exponentially distributed." is a very vague way to say anything.  I'm going to go out on a limb and surmise (simply because the following is a commonly considered problem) that you mean the waiting time until the next even is exponentially distributed.
In that case, the number of events in a specified time interval has a Poisson distribution, so that that number $N$ satisfies
$$
\Pr(N=n) = \frac{e^{-\lambda T}(\lambda T)^n}{n!}
$$
where $\lambda$ is the average number of "events" per unit of time.  So plug in $n=1$ and you've got it.
"the probability that there will be exactly one event between $0$ and $T$" is not at all the same thing as $Pr(x_1≤t\mid x_1≤T,x_1+x_2>T)$.
