Question 1157548
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Let's consider these 5-digit codes with no repetitions.


In the left-most position, any of 10 digits can be placed (the problem does not prohibit 0 in this position).  It gives 10 options.


In the next position from the left, only one of remaining 9 digits can be placed (9 options).


In the 3-rd position from the left, only one of remaining 8 digits can be placed (8 options).


 . . . And so on . . . 


In all, there are 10*9*8*7*6 = 30240 such codes.


The specific code 93152 is unique. It is only one among 30240 all possible codes.


Therefore, the probability to randomly select this code is  {{{1/30240}}}.
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Solved, explained and completed.