Specific Question about the Exponential Distribution I think the best way to ask this question is to make up a scenario. 

While Jeremy is sitting in a park he notices bees land on a particular type of flower according to a Poisson process with a rate of $0.4$. As time passes Jeremy nods off. $20$ minutes later Jeremy 'comes to',  and notices a flower with a bee on it. What is the probability that the bee landed on the flower between $t = 18$ and $t = 20$?

Take the point where Jeremy falls asleep to be $t = 0$. 
We have $T\sim \text{Exp}(0.4)$ where $T$ is the event that a bee lands on a flower in time $t$. We also have $B(t) \sim \text{Pois}(0.4)$ where $B(t)$ is the number of bees to pollinate a flower in time $t$?
Since we know that a bee has landed on the flower, do we have to consider a conditional probability? i.e. $Pr( [18 < T < 20]| B(20) = 1)$ or does it suffice to just consider  $Pr(18 < T < 20)$?

EDIT:
Consider, 
$Pr(B(18) = 0 | B(20) = 1) = \dfrac{Pr(B(2) = 1 )Pr( B(18) = 0)}{Pr(B(20) = 1)}$
 A: Yes, you need to consider the conditional probability. To see this, think about the alternative question "what is the probability that the bee was already there at time $18$?"
If $P(18<T<20)$ was the right answer to the original question, presumably the answer to this new question would be $P(T\leq 18)$. Also, we should have the two answers summing to $1$, since one of them must have happened. But $P(18<T<20)+P(T\leq 18)=P(T<20)\neq 1$.
Given that a bee has arrived, for that bee the time of arrival isn't an exponential distribution any more, because it must be between $0$ and $20$. Also, you know that no other bees have arrived in between. So the right probability is $P(B_{18}=0\mid B_{20}=1)$, where $B_t$ is the number of bees arrived by time $t$. 
This is $\frac{P(B_{18}=0,B_{20}=1)}{P(B_{20}=1)}$. We can write $B_{20}=B_{18}+X$, where $X\sim\text{Poi}(2\lambda)$ and $B_{18}\sim\text{Poi}(18\lambda)$ are independent; then $B_{20}=1$ and $B_{18}=0$ iff $B_{18}=0$ and $X=1$.
So we have $P(B_{18}=0,B_{20}=1)=P(B_{18}=0,X=1)=e^{-18\lambda}\cdot 2\lambda e^{-2\lambda}$, and $P(B_{20}=1)=20\lambda e^{-20\lambda}$, so the conditional probability is $\frac{2}{20}$ (the $\lambda$'s cancel). 
