Probability of opening the valut in third try There is a secret vault which demands a particular 3-digit to open. If Alex starts entering the code randomly, what is the probability that he will open the vault on third try.
My approach:
Probabilty of of opening the vault in third attempt would be same as the product of probability of failed first attempt, probability of failed second attempt and probability of sucessful third attempt.
Probability of failed first attempt=$\frac{999}{1000}$ since 999 are incorrect and only 1 is correct.
Probability of failed second attempt=$\frac{998}{999}$
since he's already tried 1 of the wrong codes he will choose only from the remaining 998.
Probability of third succesful attempt=$\frac{1}{998}$
Therefore, the required probabilty=$\frac{999}{1000}.\frac{998}{999}.\frac{1}{998}$=$\frac{1}{1000}$
Is the above mentioned approach correct? If not please let me know how I can do it correctly

The result is $\frac{999}{1000}\frac{998}{999}\frac{1}{998}=\frac{1}{1000}$
