Distribution and knowledge about a state of the environment Let's assume I have a frequency distribution for train travel times, estimated based on one month of empirical data:


*

*[1:00h, 1:20[  : 20%

*[1:20h, 1:40h[ : 50%

*[1:40h, 2:00]  : 30%


Edit:
The data is put into bins, because it is very unlikely that we have two exact data points and a difference from e.g. one second isn't relevant. For someone looking at the distribution it is such easier to see relevant details. The 20mins bins are just an example.
And I know at time t that the train already travels for 1:25h.
Does this information help me to update my knowledge about how long the train will travel?
Maybe conditional probabilities, but I don't know how to use them in this case.
 A: This question could be posed in a different way in the form of survival analysis. To make the example clearer, we can phrase "travel time" as "survival time until termination". 
With respect to your problem, 

And I know at time t that the train already travels for 1:25h. Does this information help me to update my knowledge about how long the train will travel? Maybe conditional probabilities, but I don't know how to use them in this case.

In terms of "does this information matter", it will all boil down to the assumptions you have on your data. If we use nonparametric approaches which do not make any assumptions about distribution of failure times, and just attach Kaplan Meier estimator, then using the  information above (assuming 10 cases for simplicity)
+----------+-------------+------------------+---------------------------+-------------+
| Time (m) | total cases | number of events | expected number of events | Hazard rate |
+----------+-------------+------------------+---------------------------+-------------+
| 60-80    | 10          | 2                | 0.2                       | 0.2         |
+----------+-------------+------------------+---------------------------+-------------+
| 80-100   | 8           | 5                | 0.62                      | 0.82        |
+----------+-------------+------------------+---------------------------+-------------+
| 100+     | 3           | 3                | 1.00                      | 1.82        |
+----------+-------------+------------------+---------------------------+-------------+

We can then calculate the survival rates, at that time, we have
$$
S( 1:25h ) = ((10-2)/10)*((8-5)/8) = 0.3
$$
Alternatively, if we use the estimated cumulative hazard rate in the table above, we can express survival rate as:
$$
S(t) = Pr(T> t) = e^{-H(t)}
$$
Which if we simplify it, would be $e^{-0.82} = 0.4404317$. 

Ultimately your model depends the various assumptions you might have, but outlined here is one possible way to approach the problem from a survival analysis perspective. 
A: $T:$Random Variable denoting travel time with limits $60\le t\le120$
we need $E[T|T>85]$
we have,
$T|60\le T\le80\sim U(60,80)~and~P(60\le T\le80)=0.2$
$T|80\le T\le100\sim U(80,100)~and~P(80\le T\le100)=0.5$
$T|100\le T\le120\sim U(100,120)~and~P(100\le T\le120)=0.3$
Now, for the train that travelled $85~min$, that is train survived up to 85.
$E[T|T>85]=\frac{100+85}{2}\times\frac{.5}{.5+.3}+\frac{120+110}{2}\times\frac{.3}{.5+.3}=99.0625$
So, expected time that the train will travel is $\sim14~min$ more.
Probability:- we can find it from available conditional distributions, say, you need $P(100\le T\le120|T>85)$ here we care about distribution for $T>85$ only.... Hence the Pr will be $\frac{.3}{.3+.5}$
For example,  $P(90\le T\le110|T>85)=\frac{90-85}{100-85}\times\frac{5}{8}+\frac{110-100}{120-100}\times\frac{3}{8}$
