Q:

how to get 2/15 as the probability?

Question: A car and a bus arrive at a railroad crossing at times independently and uniformly 
distributed between 7:15 and 7:30.  A train arrives at the crossing at 7:20 and halts traffic 
at the crossing for five minutes. Calculate the probability that the waiting time of the car or the bus at the crossing 
exceeds three minutes.
Solution shows this: To be delayed over three minutes, either the car or the bus must arrive between 7:20 and 7:22. The probability for each is 2/15. The probability they both arrive in that interval is (2/15)(2/15).
Thus, the probability of at least one being delayed is 2/15 + 2/15 – (2/15)(2/15) = 56/225 = 0.25. 
Can someone explain how they got 2/15?

A:

Here's how I would approach problems like these:

Reduce the problem into something you can describe via probability. Below, we use a diagram to turn some text about traffic into a question about how often a vehicle arrives in a specific time window.

Notice that question is about finding the probability of observing any value in a range. These sort of questions can be solved via cumulative distribution functions.
These steps gave us the probability of a single event (here, a delayed car) occurring. However, the actual problem asks about a combination of independent events, so use the rules of and-ing and or-ing probabilities to find that instead. 

See if that's enough to get you started. Otherwise, a more complete description follows.

Drawing a digram often helps. Here's one for this problem, explained below. 

The entire problem takes place over fifteen minutes (7:15-7:30). At 7:20, the train arrives, blocking the road for 5 minutes (black interval) from 7:20 to 7:25. 
A vehicle is only delayed if it arrives at the crossing while the train is blocking the road, so arrivals before 7:20 or after 7:25 don't result in delays; only arrivals between 7:20 and 7:25 do. However, we're only interested in delays of more than three minutes, so we can ignore the last three minutes of the train obstruction too (grey interval), since arriving then can't cause more than a three minute delay. What's left? Just the green interval, which is 5-3, or 2 minutes long.
So the question reduces down to determining how often the vehicle arrives during the green interval. The question stipulates that each vehicle's arrival time is uniformly distributed between 7:15 and 7:30, which makes this particularly easy--you just compare the length of the green interval (?? min) to the length of the entire interval (?? min). 
For more complicated distributions, you would need to do something a little more involved. The cumulative distribution function tells you what proportion of values are less than or equal to a given value. You can evaluate it twice to find the proportion of values that are between two values. For example, the proportion of times that are between 7:25 and 7:30 is the difference between the proportion of times that are before 7:30 and the proportion of times that are before 7:25.  For a uniform distribution, the cumulative distribution function is 
$$P(X\le x) = \frac{x-a}{x-b}$$
where $a$ and $b$ are the two endpoints and $x$ is some value between them. You should plug in the numbers and convince yourself this works (hint: It might be easier to do the math if you adjust the times from hr:min format and instead make them relative to some event). 
Both ways give us the probability that one vehicle was delayed  for more than three minutes. However, the question asks us for the probability that either the car or the truck was delayed by more than three minutes. Unfortunately, the question isn't totally clear here: do we want to calculate the probability that the car or the bus or both were delayed, or are we more interested in the probability that exactly one vehicle was delayed? Let's tackle the first one
In general, this is tricky, but the fact that the car and bus are independent make this much easier. When two events $A$, occurring with probability $P(A)$, and $B$ (occurring with probability $P(B)$), are independent and mutually exclusive, the probability of observing $A$ or $B$ is simply $P(A) + P(B)$. These events are independent, but not mutually exclusive--the bus being delayed doesn't somehow block the car from being late, or vice versa. When this occurs, the probability of seeing $A$ or $B$ needs is instead $P(A) + P(B) - P(A \& B)$. This avoids double-counting the events that occur in both. To find $P(A \& B)$ (for indepenent A and B, you just multiply the two probabilities: $P(A \& B) = P(A) \times P(B)$. 
And ... you are done.

