Conditional probability with continuous random variables I am trying to solve this question:

Let $f_T (t\mid \lambda) = \lambda e^{−\lambda t}$, $t ≥ 0$ 
So $T$ is exponentially distributed when conditioned
  on $\lambda$. Assume that we have a prior for $\lambda$ that is continuous uniform between $a$
  and $b$.
(a) Given the event $E = \{T > \tau\}$, what can we say about $\lambda$? In other words, find $f(\lambda\mid E)$. Recall that Bayes rule allows for mixing probabilities and densities.
In particular:
  $$f(\lambda\mid E) = \frac {P(E\mid\lambda)f(\lambda)}{P(E)}.$$

I am a bit confused as to how I would go about solving this problem. I understand that $T$ is a random variable which has an exponential distribution given by $F_T$. I'm a little confused as to what "prior for $\lambda$ is continuous uniform between $a$ and $b$ means" my interpretation is that $\tau$ is the prior and is itself a random variable which is uniformly distributed. 
Does this then imply that event $E$ is also uniform?
I also can't figure out what the distribution of $f(\lambda)$ is (is it just constant?). 
The main problem I have is understanding what $F_T$ actually is. This question is given in the context of a random event happening within a time interval. $\lambda$ I believe is the base probability of an event happening (in which case its distribution would be constant) and $t$ is the amount of time before an event (in which case it would make sense that it is uniform, since time is measured in uniformly distributed equal intervals). 
Then, $F_T$ I interpret as, given that a random event has happened, the amount of time that has passed till that event, is exponentially distributed. Therefore, for example, large values of $t|\lambda$ would be unlikely to occur, which makes sense since the probability that the event will happen will quickly increase with time. 
I'm having the most difficult understanding event E, what does $T>\tau$ mean? And what does $f(\lambda \mid E)$ represent?
 A: First off, an event is a subset of the sample space; it doesn't make sense to talk about a probability distribution for $E$.
As for what "prior" means, $T$ is an exponential random variable with unknown parameter $\lambda$. The problem is saying that $\lambda$ itself is a random variable, with $U(a,b)$ distribution. So the probability density function of is $\lambda$ is $f(x) = \mathsf 1_{(a,b)}(x)$ (with $\mathsf 1$ denoting the indicator function). Now, given a value of $\lambda$, we have
\begin{align}
\mathbb P(E\mid\lambda) &= \mathbb P(T > \tau\mid \lambda)\\
&= \int_\tau^\infty \lambda e^{-\lambda t}\ \mathsf dt\\
&= e^{-\lambda\tau}.
\end{align}
Note that $\tau$ is just an arbitrary positive real number. Further,
\begin{align}
\mathbb P(E) &= \int_{\mathbb R} \mathbb P(T>\tau\mid \lambda)f(\lambda)\ \mathsf d\lambda\\
&= \int_{\mathbb R} \int_{\tau}^\infty f(t\mid\lambda)f(\lambda)\ \mathsf dt\ \mathsf d\lambda\\
&= \int_{\mathbb R} \int_{\tau}^\infty \lambda e^{-\lambda t}\mathsf 1_{(a,b)}(\lambda)\ \mathsf dt \ \mathsf d\lambda\\
&= \int_a^b e^{-\lambda\tau}\ \mathsf d\lambda\\
&= \frac1\tau (e^{-\lambda a}-e^{-\lambda b}).
\end{align}
Hence
$$
f(\lambda\mid E) = \left(\frac{\tau e^{-\lambda\tau}}{e^{-\lambda a}-e^{-\lambda b}}\right)\mathsf 1_{(a,b)}(\lambda).
$$
