Find the probability that no more than three attempts will be required to open the lock 
Of the five keys, one is suitable for the lock. The key that did not
  fit when trying to open the lock is put aside. 
We need to find the probability that no more than three attempts will
  be required to open the lock.

I tried to use the following idea:
Let's write down the events:


*

*A = lock opened by the first key 

*B = lock opened by the second key 

*C = lock opened by the third key


Then we have probabilities:
We have 1 good and 4 bad keys (5 total):
P(A) = 1/5 = 0.2
Then we have the probability 4/5 that we take the wrong key.
Now we have 1 good and 3 bad keys (4 total):
P(B) = 4/5 * 1/4 = 1/5 = 0.2
Then we have the probability 3/4 that we take the wrong key.
Now we have 1 good and 2 bad keys (3 total):
P(C) = 4/5 * 3/4 * 1/3 = 1/5 = 0.2
This way we can sum it up and get a probability is 0.6.
I fear that my solution is poor and unfounded, and I ask for help to valid and improve it. 
You can also show your solution with your thought

There are 5 keys, 1 opens the lock and we choose 3 randomly without replacement. 
The probability that the "good" key is among the 3 chosen ones is
$$ \frac{\binom11\binom{4}{2}}{\binom{5}{3}}= 0.6 $$
The same as your answer. Your solution is fine.
